专栏首页米虫的家LeetCode数据库题目集合

LeetCode数据库题目集合

https://leetcode-cn.com/problemset/database/ 题目都是leetcode 上了可以点击题目会有相应的链接 由于个人比较喜欢用开窗函数,所以都优先用了开窗 ,当然这些并不一定都是最优解,答案仅供参考 每道题后面都应相应的难度等级,如果没时间做的话 可以在leetcode 按出题频率刷题 我是安顺序刷的题,后续还会继续更新 祝大家面试取得好的成绩

175. 组合两个表

难度简单

SQL架构

表1: Person

Code

+-------------+---------+
| 列名         | 类型     |
+-------------+---------+
| PersonId    | int     |
| FirstName   | varchar |
| LastName    | varchar |
+-------------+---------+
PersonId 是上表主键

表2: Address

Code

+-------------+---------+
| 列名         | 类型    |
+-------------+---------+
| AddressId   | int     |
| PersonId    | int     |
| City        | varchar |
| State       | varchar |
+-------------+---------+
AddressId 是上表主键

编写一个 SQL 查询,满足条件:无论 person 是否有地址信息,都需要基于上述两表提供 person 的以下信息:

Code

FirstName, LastName, City, State

sql

select FirstName,LastName,City,State
from Person  p
left join  Address a 
on a.PersonId = p.PersonId

176. 第二高的薪水

难度简单

SQL架构

编写一个 SQL 查询,获取 Employee 表中第二高的薪水(Salary) 。

Code

+----+--------+
| Id | Salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+

例如上述 Employee 表,SQL查询应该返回 200 作为第二高的薪水。如果不存在第二高的薪水,那么查询应返回 null

Code

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200                 |
+---------------------+

sql

SELECT
    IFNULL(
      (SELECT DISTINCT Salary
       FROM Employee
       ORDER BY Salary DESC
        LIMIT 1 OFFSET 1),
    NULL) AS SecondHighestSalary

177. 第N高的薪水

难度中等

编写一个 SQL 查询,获取 Employee 表中第 n 高的薪水(Salary)。

Code

+----+--------+
| Id | Salary |
+----+--------+
| 1  | 100    |
| 2  | 200    |
| 3  | 300    |
+----+--------+

例如上述 Employee 表,n = 2 时,应返回第二高的薪水 200。如果不存在第 n 高的薪水,那么查询应返回 null

Code

+------------------------+
| getNthHighestSalary(2) |
+------------------------+
| 200                    |
+------------------------+

sql

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
  RETURN (
      SELECT IFNULL(
		(select salary  
		from(
		select salary,
		rank() over(order by salary desc) rk
		from Employee
        group by salary
		)t1
		where rk=N),NULL) SecondHighestSalary
  );
END

178. 分数排名

难度中等

SQL架构

编写一个 SQL 查询来实现分数排名。

如果两个分数相同,则两个分数排名(Rank)相同。请注意,平分后的下一个名次应该是下一个连续的整数值。换句话说,名次之间不应该有“间隔”。

Code

+----+-------+
| Id | Score |
+----+-------+
| 1  | 3.50  |
| 2  | 3.65  |
| 3  | 4.00  |
| 4  | 3.85  |
| 5  | 4.00  |
| 6  | 3.65  |
+----+-------+

例如,根据上述给定的 Scores 表,你的查询应该返回(按分数从高到低排列):

Code

+-------+------+
| Score | Rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 2    |
| 3.65  | 3    |
| 3.65  | 3    |
| 3.50  | 4    |
+-------+------+

重要提示:对于 MySQL 解决方案,如果要转义用作列名的保留字,可以在关键字之前和之后使用撇号。例如 Rank

sql

select Score,
dense_rank() over(order by Score desc) `rank`
from Scores

180. 连续出现的数字

难度中等

SQL架构

编写一个 SQL 查询,查找所有至少连续出现三次的数字。

Code

+----+-----+
| Id | Num |
+----+-----+
| 1  |  1  |
| 2  |  1  |
| 3  |  1  |
| 4  |  2  |
| 5  |  1  |
| 6  |  2  |
| 7  |  2  |
+----+-----+

例如,给定上面的 Logs 表, 1 是唯一连续出现至少三次的数字。

sql

+-----------------+
| ConsecutiveNums |
+-----------------+
| 1               |
+-----------------+

sql

select distinct Num ConsecutiveNums
from 
(
select
Num,
lead(Num,1,null) over(order by id) n2,
lead(Num,2,null) over(order by id) n3
from Logs
)t1
where Num = n2 and Num = n3

181. 超过经理收入的员工

难度简单

SQL架构

Employee 表包含所有员工,他们的经理也属于员工。每个员工都有一个 Id,此外还有一列对应员工的经理的 Id。

Code

+----+-------+--------+-----------+
| Id | Name  | Salary | ManagerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | NULL      |
| 4  | Max   | 90000  | NULL      |
+----+-------+--------+-----------+

给定 Employee 表,编写一个 SQL 查询,该查询可以获取收入超过他们经理的员工的姓名。在上面的表格中,Joe 是唯一一个收入超过他的经理的员工。

sql

+----------+
| Employee |
+----------+
| Joe      |
+----------+

sql

select a.Name  Employee 
from Employee a 
join Employee b
on a.ManagerId = b.id
where a.Salary>b.Salary

182. 查找重复的电子邮箱

难度简单

SQL架构

编写一个 SQL 查询,查找 Person 表中所有重复的电子邮箱。

示例:

Code

+----+---------+
| Id | Email   |
+----+---------+
| 1  | a@b.com |
| 2  | c@d.com |
| 3  | a@b.com |
+----+---------+

根据以上输入,你的查询应返回以下结果:

Code

+---------+
| Email   |
+---------+
| a@b.com |
+---------+

说明:所有电子邮箱都是小写字母。

sql

select Email
from Person
group by Email
having count(*)>1

183. 从不订购的客户

难度简单

SQL架构

某网站包含两个表,Customers 表和 Orders 表。编写一个 SQL 查询,找出所有从不订购任何东西的客户。

Customers 表:

Code

+----+-------+
| Id | Name  |
+----+-------+
| 1  | Joe   |
| 2  | Henry |
| 3  | Sam   |
| 4  | Max   |
+----+-------+

Orders 表:

Code

+----+------------+
| Id | CustomerId |
+----+------------+
| 1  | 3          |
| 2  | 1          |
+----+------------+

例如给定上述表格,你的查询应返回:

sql

+-----------+
| Customers |
+-----------+
| Henry     |
| Max       |
+-----------+

sql

select  c.Name Customers
from Customers  c left join Orders  o
on c.id = o.CustomerId
where o.id is null

184. 部门工资最高的员工

难度中等

SQL架构

Employee 表包含所有员工信息,每个员工有其对应的 Id, salary 和 department Id。

Code

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
+----+-------+--------+--------------+

Department 表包含公司所有部门的信息。

Code

+----+----------+
| Id | Name     |
+----+----------+
| 1  | IT       |
| 2  | Sales    |
+----+----------+

编写一个 SQL 查询,找出每个部门工资最高的员工。例如,根据上述给定的表格,Max 在 IT 部门有最高工资,Henry 在 Sales 部门有最高工资。

sql

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| Sales      | Henry    | 80000  |
+------------+----------+--------+

sql

select Department,Employee,Salary
from (
select d.Name  Department,e.Name Employee, e.Salary,
rank() over(partition by d.id order by Salary desc) rk
from Employee e join Department d
on e.DepartmentId=d.id
)tmp
where rk = 1

185. 部门工资前三高的所有员工

难度困难

SQL架构

Employee 表包含所有员工信息,每个员工有其对应的工号 Id,姓名 Name,工资 Salary 和部门编号 DepartmentId

Code

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 85000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 1            |
| 7  | Will  | 70000  | 1            |
+----+-------+--------+--------------+

Department 表包含公司所有部门的信息。

Code

+----+----------+
| Id | Name     |
+----+----------+
| 1  | IT       |
| 2  | Sales    |
+----+----------+

编写一个 SQL 查询,找出每个部门获得前三高工资的所有员工。例如,根据上述给定的表,查询结果应返回:

Code

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Randy    | 85000  |
| IT         | Joe      | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

解释:

IT 部门中,Max 获得了最高的工资,Randy 和 Joe 都拿到了第二高的工资,Will 的工资排第三。销售部门(Sales)只有两名员工,Henry 的工资最高,Sam 的工资排第二。

sql

select Department,Employee,Salary
from (
select d.Name  Department,e.Name Employee, e.Salary,
dense_rank() over(partition by d.id order by Salary desc) rk
from Employee e join Department d
on e.DepartmentId=d.id
)tmp
where rk <=3

196. 删除重复的电子邮箱

难度简单

编写一个 SQL 查询,来删除 Person 表中所有重复的电子邮箱,重复的邮箱里只保留 Id 最小 的那个。

Code

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
| 3  | john@example.com |
+----+------------------+
Id 是这个表的主键。

例如,在运行你的查询语句之后,上面的 Person 表应返回以下几行:

Code

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
+----+------------------+

提示:

  • 执行 SQL 之后,输出是整个 Person 表。
  • 使用 delete 语句。

sql

DELETE p1 FROM Person p1,
    Person p2
WHERE
    p1.Email = p2.Email AND p1.Id > p2.Id

注意是删除 ,不是查询

197. 上升的温度

难度简单

SQL架构

给定一个 Weather 表,编写一个 SQL 查询,来查找与之前(昨天的)日期相比温度更高的所有日期的 Id。

Code

+---------+------------------+------------------+
| Id(INT) | RecordDate(DATE) | Temperature(INT) |
+---------+------------------+------------------+
|       1 |       2015-01-01 |               10 |
|       2 |       2015-01-02 |               25 |
|       3 |       2015-01-03 |               20 |
|       4 |       2015-01-04 |               30 |
+---------+------------------+------------------+

例如,根据上述给定的 Weather 表格,返回如下 Id:

sql

+----+
| Id |
+----+
|  2 |
|  4 |
+----+

sql

select 
Id
from 
(
select Id,RecordDate,Temperature,
lag(RecordDate,1,9999-99-99) over (order by RecordDate) yd,
lag(Temperature,1,999) over(order by RecordDate ) yt
from Weather 
)tmp
where Temperature >yt
and datediff(RecordDate,yd)=1

262. 行程和用户

难度困难

SQL架构

Trips 表中存所有出租车的行程信息。每段行程有唯一键 Id,Client_Id 和 Driver_Id 是 Users 表中 Users_Id 的外键。Status 是枚举类型,枚举成员为 (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’)。

Code

+----+-----------+-----------+---------+--------------------+----------+
| Id | Client_Id | Driver_Id | City_Id |        Status      |Request_at|
+----+-----------+-----------+---------+--------------------+----------+
| 1  |     1     |    10     |    1    |     completed      |2013-10-01|
| 2  |     2     |    11     |    1    | cancelled_by_driver|2013-10-01|
| 3  |     3     |    12     |    6    |     completed      |2013-10-01|
| 4  |     4     |    13     |    6    | cancelled_by_client|2013-10-01|
| 5  |     1     |    10     |    1    |     completed      |2013-10-02|
| 6  |     2     |    11     |    6    |     completed      |2013-10-02|
| 7  |     3     |    12     |    6    |     completed      |2013-10-02|
| 8  |     2     |    12     |    12   |     completed      |2013-10-03|
| 9  |     3     |    10     |    12   |     completed      |2013-10-03| 
| 10 |     4     |    13     |    12   | cancelled_by_driver|2013-10-03|
+----+-----------+-----------+---------+--------------------+----------+

Users 表存所有用户。每个用户有唯一键 Users_Id。Banned 表示这个用户是否被禁止,Role 则是一个表示(‘client’, ‘driver’, ‘partner’)的枚举类型。

Code

+----------+--------+--------+
| Users_Id | Banned |  Role  |
+----------+--------+--------+
|    1     |   No   | client |
|    2     |   Yes  | client |
|    3     |   No   | client |
|    4     |   No   | client |
|    10    |   No   | driver |
|    11    |   No   | driver |
|    12    |   No   | driver |
|    13    |   No   | driver |
+----------+--------+--------+

写一段 SQL 语句查出 2013年10月1日2013年10月3日 期间非禁止用户的取消率。基于上表,你的 SQL 语句应返回如下结果,取消率(Cancellation Rate)保留两位小数。

取消率的计算方式如下:(被司机或乘客取消的非禁止用户生成的订单数量) / (非禁止用户生成的订单总数)

sql

+------------+-------------------+
|     Day    | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 |       0.33        |
| 2013-10-02 |       0.00        |
| 2013-10-03 |       0.50        |
+------------+-------------------+

sql

SELECT T.request_at AS `Day`, 
	ROUND(
			SUM(
				IF(T.STATUS = 'completed',0,1)
			)
			/ 
			COUNT(T.STATUS),
			2
	) AS `Cancellation Rate`
FROM trips AS T
WHERE 
T.Client_Id NOT IN (
	SELECT users_id
	FROM users
	WHERE banned = 'Yes'
)
AND
T.Driver_Id NOT IN (
	SELECT users_id
	FROM users
	WHERE banned = 'Yes'
)
AND T.request_at BETWEEN '2013-10-01' AND '2013-10-03'
GROUP BY T.request_at

511. 游戏玩法分析 I

难度简单

SQL架构

活动表 Activity

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| player_id    | int     |
| device_id    | int     |
| event_date   | date    |
| games_played | int     |
+--------------+---------+
表的主键是 (player_id, event_date)。
这张表展示了一些游戏玩家在游戏平台上的行为活动。
每行数据记录了一名玩家在退出平台之前,当天使用同一台设备登录平台后打开的游戏的数目(可能是 0 个)。

写一条 SQL 查询语句获取每位玩家 第一次登陆平台的日期

查询结果的格式如下所示:

Code

Activity 表:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1         | 2         | 2016-03-01 | 5            |
| 1         | 2         | 2016-05-02 | 6            |
| 2         | 3         | 2017-06-25 | 1            |
| 3         | 1         | 2016-03-02 | 0            |
| 3         | 4         | 2018-07-03 | 5            |
+-----------+-----------+------------+--------------+

Result 表:
+-----------+-------------+
| player_id | first_login |
+-----------+-------------+
| 1         | 2016-03-01  |
| 2         | 2017-06-25  |
| 3         | 2016-03-02  |
+-----------+-------------+

1.

sql

select player_id ,event_date first_login
from (
select player_id ,event_date,
rank() over(partition by player_id order by event_date) rk
from Activity
) tmp
where rk = 1

2.最优 (选最小日期)

sql

select player_id ,min(event_date) first_login
from Activity
group by player_id

512. 游戏玩法分析 II

难度简单

SQL架构

Table: Activity

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| player_id    | int     |
| device_id    | int     |
| event_date   | date    |
| games_played | int     |
+--------------+---------+
(player_id, event_date) 是这个表的两个主键
这个表显示的是某些游戏玩家的游戏活动情况
每一行是在某天使用某个设备登出之前登录并玩多个游戏(可能为0)的玩家的记录

请编写一个 SQL 查询,描述每一个玩家首次登陆的设备名称

查询结果格式在以下示例中:

Code

Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1         | 2         | 2016-03-01 | 5            |
| 1         | 2         | 2016-05-02 | 6            |
| 2         | 3         | 2017-06-25 | 1            |
| 3         | 1         | 2016-03-02 | 0            |
| 3         | 4         | 2018-07-03 | 5            |
+-----------+-----------+------------+--------------+

Result table:
+-----------+-----------+
| player_id | device_id |
+-----------+-----------+
| 1         | 2         |
| 2         | 3         |
| 3         | 1         |
+-----------+-----------+

sql

select player_id ,device_id
from (
select player_id ,event_date,device_id,
rank() over(partition by player_id order by event_date) rk
from Activity
) tmp
where rk = 1

534. 游戏玩法分析 III

难度中等20收藏分享切换为英文关注反馈

SQL架构

Table: Activity

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| player_id    | int     |
| device_id    | int     |
| event_date   | date    |
| games_played | int     |
+--------------+---------+
(player_id,event_date)是此表的主键。
这张表显示了某些游戏的玩家的活动情况。
每一行是一个玩家的记录,他在某一天使用某个设备注销之前登录并玩了很多游戏(可能是 0 )。

编写一个 SQL 查询,同时报告每组玩家和日期,以及玩家到目前为止玩了多少游戏。也就是说,在此日期之前玩家所玩的游戏总数。详细情况请查看示例。

查询结果格式如下所示:

Code

Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1         | 2         | 2016-03-01 | 5            |
| 1         | 2         | 2016-05-02 | 6            |
| 1         | 3         | 2017-06-25 | 1            |
| 3         | 1         | 2016-03-02 | 0            |
| 3         | 4         | 2018-07-03 | 5            |
+-----------+-----------+------------+--------------+

Result table:
+-----------+------------+---------------------+
| player_id | event_date | games_played_so_far |
+-----------+------------+---------------------+
| 1         | 2016-03-01 | 5                   |
| 1         | 2016-05-02 | 11                  |
| 1         | 2017-06-25 | 12                  |
| 3         | 2016-03-02 | 0                   |
| 3         | 2018-07-03 | 5                   |
+-----------+------------+---------------------+
对于 ID 为 1 的玩家,2016-05-02 共玩了 5+6=11 个游戏,2017-06-25 共玩了 5+6+1=12 个游戏。
对于 ID 为 3 的玩家,2018-07-03 共玩了 0+5=5 个游戏。
请注意,对于每个玩家,我们只关心玩家的登录日期。

开窗

sql

select player_id,event_date ,
sum(games_played) over(partition by player_id order by  event_date )games_played_so_far
from Activity

自连接

sql

select 
    a1.player_id,
    a1.event_date,
    sum(a2.games_played) games_played_so_far
from Activity a1,Activity a2
where a1.player_id=a2.player_id and 
      a1.event_date>=a2.event_date
group by 1,2;

550. 游戏玩法分析 IV

难度中等17收藏分享切换为英文关注反馈

SQL架构

Table: Activity

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| player_id    | int     |
| device_id    | int     |
| event_date   | date    |
| games_played | int     |
+--------------+---------+
(player_id,event_date)是此表的主键。
这张表显示了某些游戏的玩家的活动情况。
每一行是一个玩家的记录,他在某一天使用某个设备注销之前登录并玩了很多游戏(可能是 0)。

编写一个 SQL 查询,报告在首次登录的第二天再次登录的玩家的分数,四舍五入到小数点后两位。换句话说,您需要计算从首次登录日期开始至少连续两天登录的玩家的数量,然后除以玩家总数。

查询结果格式如下所示:

Code

Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1         | 2         | 2016-03-01 | 5            |
| 1         | 2         | 2016-03-02 | 6            |
| 2         | 3         | 2017-06-25 | 1            |
| 3         | 1         | 2016-03-02 | 0            |
| 3         | 4         | 2018-07-03 | 5            |
+-----------+-----------+------------+--------------+

Result table:
+-----------+
| fraction  |
+-----------+
| 0.33      |
+-----------+
只有 ID 为 1 的玩家在第一天登录后才重新登录,所以答案是 1/3 = 0.33

sql

select round(avg(a.event_date is not null), 2) fraction
from 
    (select player_id, min(event_date) as login
    from activity
    group by player_id) p 
left join activity a 
on p.player_id=a.player_id and datediff(a.event_date, p.login)=1

这个avg用的妙 is not null判断后,有eventdate值的返回1,null的返回0,avg相当于求和后(即符合条件的id个数)除以总id数即所求比例

569. 员工薪水中位数

难度困难

SQL架构

Employee 表包含所有员工。Employee 表有三列:员工Id,公司名和薪水。

Code

+-----+------------+--------+
|Id   | Company    | Salary |
+-----+------------+--------+
|1    | A          | 2341   |
|2    | A          | 341    |
|3    | A          | 15     |
|4    | A          | 15314  |
|5    | A          | 451    |
|6    | A          | 513    |
|7    | B          | 15     |
|8    | B          | 13     |
|9    | B          | 1154   |
|10   | B          | 1345   |
|11   | B          | 1221   |
|12   | B          | 234    |
|13   | C          | 2345   |
|14   | C          | 2645   |
|15   | C          | 2645   |
|16   | C          | 2652   |
|17   | C          | 65     |
+-----+------------+--------+

请编写SQL查询来查找每个公司的薪水中位数。挑战点:你是否可以在不使用任何内置的SQL函数的情况下解决此问题。

Code

+-----+------------+--------+
|Id   | Company    | Salary |
+-----+------------+--------+
|5    | A          | 451    |
|6    | A          | 513    |
|12   | B          | 234    |
|9    | B          | 1154   |
|14   | C          | 2645   |
+-----+------------+--------+

sql

select Id,Company,Salary
from (
select Id,Company,Salary,
ROW_NUMBER() over(partition by Company order by Salary) rk,
count(*) over(partition by Company) cnt
from Employee
)t1
where rk IN (FLOOR((cnt + 1)/2), FLOOR((cnt + 2)/2))

中位数: +1向下取整 +2 向下取整数

570. 至少有5名直接下属的经理

难度中等

SQL架构

Employee 表包含所有员工和他们的经理。每个员工都有一个 Id,并且还有一列是经理的 Id。

Code

+------+----------+-----------+----------+
|Id    |Name 	  |Department |ManagerId |
+------+----------+-----------+----------+
|101   |John 	  |A 	      |null      |
|102   |Dan 	  |A 	      |101       |
|103   |James 	  |A 	      |101       |
|104   |Amy 	  |A 	      |101       |
|105   |Anne 	  |A 	      |101       |
|106   |Ron 	  |B 	      |101       |
+------+----------+-----------+----------+

给定 Employee 表,请编写一个SQL查询来查找至少有5名直接下属的经理。对于上表,您的SQL查询应该返回:

Code

+-------+
| Name  |
+-------+
| John  |
+-------+

注意: 没有人是自己的下属。

sql

select Name
from Employee
where Id in (
select ManagerId
from Employee
group by ManagerId
having count(*)>=5
)

571. 给定数字的频率查询中位数

难度困难

SQL架构

Numbers 表保存数字的值及其频率。

Code

+----------+-------------+
|  Number  |  Frequency  |
+----------+-------------|
|  0       |  7          |
|  1       |  1          |
|  2       |  3          |
|  3       |  1          |
+----------+-------------+

在此表中,数字为 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3,所以中位数是 (0 + 0) / 2 = 0

Code

+--------+
| median |
+--------|
| 0.0000 |
+--------+

请编写一个查询来查找所有数字的中位数并将结果命名为 median

sql

select 
    avg(cast(number as float)) median
from 
    (
        select 
            Number,
            Frequency,
            sum(Frequency) over(order by Number) - Frequency prev_sum,
            sum(Frequency) over(order by Number) curr_sum
        from Numbers
    ) t1,
    (
        select 
            sum(Frequency) total_sum
        from Numbers
    ) t2
where 
    t1.prev_sum <= (cast(t2.total_sum as float) / 2) and 
    t1.curr_sum >= (cast(t2.total_sum as float) / 2)

如果 n1.Number 为中位数,n1.Number(包含本身)前累计的数字应大于等于总数/2 同时n1.Number(不包含本身)前累计数字应小于等于总数/2

例如:0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 3 共12个数

中位数0(包含本身)前累计的数字 7 >=6 0(不包含本身)前累计数字 0 <=6 例如:0,0,0,3,3,3 共6个数

中位数0(包含本身)前累计的数字 3 >=3 0(不包含本身)前累计数字 0 <=3

中位数3(包含本身)前累计的数字 6 >=3 3(不包含本身)前累计数字 3 <=3

sql

SELECT 
AVG(Number)median 
FROM
(SELECT n1.Number FROM Numbers n1 JOIN Numbers n2 ON n1.Number>=n2.Number 
 GROUP BY 
 n1.Number 
 HAVING 
 SUM(n2.Frequency)>=(SELECT SUM(Frequency) FROM Numbers)/2 
 AND 
 SUM(n2.Frequency)-AVG(n1.Frequency)<=(SELECT SUM(Frequency) FROM Numbers)/2
)s

574. 当选者

难度中等

SQL架构

表: Candidate

Code

+-----+---------+
| id  | Name    |
+-----+---------+
| 1   | A       |
| 2   | B       |
| 3   | C       |
| 4   | D       |
| 5   | E       |
+-----+---------+

表: Vote

Code

+-----+--------------+
| id  | CandidateId  |
+-----+--------------+
| 1   |     2        |
| 2   |     4        |
| 3   |     3        |
| 4   |     2        |
| 5   |     5        |
+-----+--------------+
id 是自动递增的主键,
CandidateId 是 Candidate 表中的 id.

请编写 sql 语句来找到当选者的名字,上面的例子将返回当选者 B.

Code

+------+
| Name |
+------+
| B    |
+------+

用了order by 全局排序 不够好

sql

select 
    Name
from Candidate c
    left join Vote v
    on c.id = v.CandidateId
    group by Name
    order by count(*) desc
    limit 1

先过滤再 效率高很多

sql

select Name
from Candidate
where id =
(
select CandidateId
    from    
        (
        select CandidateId, 
        count(*) over(partition by CandidateId ) cnt
        from Vote
        order by cnt desc 
        limit 1
        )t1
)

577. 员工奖金

难度简单

SQL架构

选出所有 bonus < 1000 的员工的 name 及其 bonus。

Employee 表单

Code

+-------+--------+-----------+--------+
| empId |  name  | supervisor| salary |
+-------+--------+-----------+--------+
|   1   | John   |  3        | 1000   |
|   2   | Dan    |  3        | 2000   |
|   3   | Brad   |  null     | 4000   |
|   4   | Thomas |  3        | 4000   |
+-------+--------+-----------+--------+
empId 是这张表单的主关键字

Bonus 表单

Code

+-------+-------+
| empId | bonus |
+-------+-------+
| 2     | 500   |
| 4     | 2000  |
+-------+-------+
empId 是这张表单的主关键字

输出示例:

Code

+-------+-------+
| name  | bonus |
+-------+-------+
| John  | null  |
| Dan   | 500   |
| Brad  | null  |
+-------+-------+

sql

select name,bonus
from Employee e
left join Bonus b on  e.empId=b.empId
where bonus<1000 or bonus is null

578. 查询回答率最高的问题

难度中等3收藏分享切换为英文关注反馈

SQL架构

survey_log 表中获得回答率最高的问题,survey_log 表包含这些列:id, action, question_id, answer_id, q_num, timestamp

id 表示用户 id;action 有以下几种值:”show”,”answer”,”skip”;当 action 值为 “answer” 时 answer_id 非空,而 action 值为 “show” 或者 “skip” 时 answer_id 为空;q_num 表示当前会话中问题的编号。

请编写 SQL 查询来找到具有最高回答率的问题。

示例:

Code

输入:
+------+-----------+--------------+------------+-----------+------------+
| id   | action    | question_id  | answer_id  | q_num     | timestamp  |
+------+-----------+--------------+------------+-----------+------------+
| 5    | show      | 285          | null       | 1         | 123        |
| 5    | answer    | 285          | 124124     | 1         | 124        |
| 5    | show      | 369          | null       | 2         | 125        |
| 5    | skip      | 369          | null       | 2         | 126        |
+------+-----------+--------------+------------+-----------+------------+
输出:
+-------------+
| survey_log  |
+-------------+
|    285      |
+-------------+
解释:
问题 285 的回答率为 1/1,而问题 369 回答率为 0/1,因此输出 285 。

sql

select question_id  survey_log
from (
  select
      question_id,
      sum(if(action = 'answer', 1, 0)) as AnswerCnt,
      sum(if(action = 'show', 1, 0)) as ShowCnt
  from
      survey_log
  group by question_id
) as tbl
order by (AnswerCnt / ShowCnt) desc
limit 1

直接不嵌套

sql

select question_id  survey_log
from survey_log
group by question_id
order by sum(if(action = 'answer', 1, 0)) / sum(if(action = 'show', 1, 0)) desc
limit 1

579. 查询员工的累计薪水

难度困难

SQL架构

Employee 表保存了一年内的薪水信息。

请你编写 SQL 语句,对于每个员工,查询他除最近一个月(即最大月)之外,剩下每个月的近三个月的累计薪水(不足三个月也要计算)。

结果请按 Id 升序,然后按 Month 降序显示。

示例: 输入:

Id

Month

Salary

1

1

20

2

1

20

1

2

30

2

2

30

3

2

40

1

3

40

3

3

60

1

4

60

3

4

70

输出:

Id

Month

Salary

1

3

90

1

2

50

1

1

20

2

1

20

3

3

100

3

2

40

解释:

员工 ‘1’ 除去最近一个月(月份 ‘4’),有三个月的薪水记录:月份 ‘3’ 薪水为 40,月份 ‘2’ 薪水为 30,月份 ‘1’ 薪水为 20。

所以近 3 个月的薪水累计分别为 (40 + 30 + 20) = 90,(30 + 20) = 50 和 20。

Id

Month

Salary

1

3

90

1

2

50

1

1

20

员工 ‘2’ 除去最近的一个月(月份 ‘2’)的话,只有月份 ‘1’ 这一个月的薪水记录。

Id

Month

Salary

2

1

20

员工 ‘3’ 除去最近一个月(月份 ‘4’)后有两个月,分别为:月份 ‘4’ 薪水为 60 和 月份 ‘2’ 薪水为 40。所以各月的累计情况如下:

Id

Month

Salary

3

3

100

3

2

40

sql

select Id,Month,
sum(Salary) over(partition by Id order by Month ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) Salary
from 
(
    select Id,Month,Salary,
    lead(Month,1,0) over(partition by Id order by Month) lm
    from Employee 
)t1
where lm != 0
order by  Id,Month desc

580. 统计各专业学生人数

难度中等

SQL架构

一所大学有 2 个数据表,分别是 studentdepartment ,这两个表保存着每个专业的学生数据和院系数据。

写一个查询语句,查询 department 表中每个专业的学生人数 (即使没有学生的专业也需列出)。

将你的查询结果按照学生人数降序排列。 如果有两个或两个以上专业有相同的学生数目,将这些部门按照部门名字的字典序从小到大排列。

student\ 表格如下:

Column Name

Type

student_id

Integer

student_name

String

gender

Character

dept_id

Integer

其中, student_id 是学生的学号, student_name 是学生的姓名, gender 是学生的性别, dept_id 是学生所属专业的专业编号。

department\ 表格如下:

Column Name

Type

dept_id

Integer

dept_name

String

dept_id 是专业编号, dept_name 是专业名字。

这里是一个示例输入: student\ 表格:

student_id

student_name

gender

dept_id

1

Jack

M

1

2

Jane

F

1

3

Mark

M

2

department\ 表格:

dept_id

dept_name

1

Engineering

2

Science

3

Law

示例输出为:

dept_name

student_number

Engineering

2

Science

1

Law

0

sql

select  dept_name ,count(student_id) student_number
from department d left join student s
on d.dept_id=s.dept_id 
group by dept_name
order by student_number desc

584. 寻找用户推荐人

难度简单9收藏分享切换为英文关注反馈

SQL架构

给定表 customer ,里面保存了所有客户信息和他们的推荐人。

Code

+------+------+-----------+
| id   | name | referee_id|
+------+------+-----------+
|    1 | Will |      NULL |
|    2 | Jane |      NULL |
|    3 | Alex |         2 |
|    4 | Bill |      NULL |
|    5 | Zack |         1 |
|    6 | Mark |         2 |
+------+------+-----------+

写一个查询语句,返回一个编号列表,列表中编号的推荐人的编号都 不是 2。

对于上面的示例数据,结果为:

Code

+------+
| name |
+------+
| Will |
| Jane |
| Bill |
| Zack |
+------+

sql

SELECT name FROM customer WHERE referee_id != 2 OR referee_id IS NULL;

MySQL 使用三值逻辑 —— TRUE, FALSE 和 UNKNOWN。任何与 NULL 值进行的比较都会与第三种值 UNKNOWN 做比较。这个“任何值”包括 NULL 本身!这就是为什么 MySQL 提供 IS NULL 和 IS NOT NULL 两种操作来对 NULL 特殊判断。 因此,在 WHERE 语句中我们需要做一个额外的条件判断 `referee_id IS NULL’。

585. 2016年的投资

难度中等14收藏分享切换为英文关注反馈

SQL架构

写一个查询语句,将 2016 年 (TIV_2016) 所有成功投资的金额加起来,保留 2 位小数。

对于一个投保人,他在 2016 年成功投资的条件是:

  1. 他在 2015 年的投保额 (TIV_2015) 至少跟一个其他投保人在 2015 年的投保额相同。
  2. 他所在的城市必须与其他投保人都不同(也就是说维度和经度不能跟其他任何一个投保人完全相同)。

输入格式:insurance\ 格式如下:

Column Name

Type

PID

INTEGER(11)

TIV_2015

NUMERIC(15,2)

TIV_2016

NUMERIC(15,2)

LAT

NUMERIC(5,2)

LON

NUMERIC(5,2)

PID 字段是投保人的投保编号, TIV_2015 是该投保人在2015年的总投保金额, TIV_2016 是该投保人在2016年的投保金额, LAT 是投保人所在城市的维度, LON 是投保人所在城市的经度。

样例输入

PID

TIV_2015

TIV_2016

LAT

LON

1

10

5

10

10

2

20

20

20

20

3

10

30

20

20

4

10

40

40

40

样例输出

TIV_2016

45.00

解释

Code

就如最后一个投保人,第一个投保人同时满足两个条件:
1. 他在 2015 年的投保金额 TIV_2015 为 '10' ,与第三个和第四个投保人在 2015 年的投保金额相同。
2. 他所在城市的经纬度是独一无二的。

第二个投保人两个条件都不满足。他在 2015 年的投资 TIV_2015 与其他任何投保人都不相同。
且他所在城市的经纬度与第三个投保人相同。基于同样的原因,第三个投保人投资失败。

所以返回的结果是第一个投保人和最后一个投保人的 TIV_2016 之和,结果是 45 。

sql

select sum(TIV_2016) TIV_2016
from (
    select PID,TIV_2016,cnt,
    count(*) over(partition by loc ) lcnt
    from (
        select PID,TIV_2016,
        count(TIV_2015) over(partition by TIV_2015 ) cnt,
        concat_ws(",",LAT,LON) loc
        from insurance 
    )t1
)t2
where lcnt=1 and cnt!=1

注意去重顺序 不要先对TIV_2015去重 不然 local去重时会丢失数据

优化 窗口

sql

SELECT 
    ROUND(SUM(TIV_2016), 2) as TIV_2016
FROM(
    SELECT
        *,
        count(*) over(partition by TIV_2015) as cnt_1,
        count(*) over(partition by LAT, LON) as cnt_2
    FROM
        insurance
) a 
WHERE a.cnt_1 > 1 AND a.cnt_2 < 2

586. 订单最多的客户

难度简单

SQL架构

在表 orders 中找到订单数最多客户对应的 customer_number

数据保证订单数最多的顾客恰好只有一位。

orders\ 定义如下:

Column

Type

order_number (PK)

int

customer_number

int

order_date

date

required_date

date

shipped_date

date

status

char(15)

comment

char(200)

样例输入

order_number

customer_number

order_date

required_date

shipped_date

status

comment

1

1

2017-04-09

2017-04-13

2017-04-12

Closed

2

2

2017-04-15

2017-04-20

2017-04-18

Closed

3

3

2017-04-16

2017-04-25

2017-04-20

Closed

4

3

2017-04-18

2017-04-28

2017-04-25

Closed

样例输出

customer_number

3

解释

Code

customer_number 为 '3' 的顾客有两个订单,比顾客 '1' 或者 '2' 都要多,因为他们只有一个订单
所以结果是该顾客的 customer_number ,也就是 3 。

进阶: 如果有多位顾客订单数并列最多,你能找到他们所有的 customer_number 吗?

sql

select customer_number
from orders
group by customer_number 
order by count(*)  desc
limit 1

如果 数据量很大 order by 不太好

595. 大的国家

难度简单

SQL架构

这里有张 World

Code

+-----------------+------------+------------+--------------+---------------+
| name            | continent  | area       | population   | gdp           |
+-----------------+------------+------------+--------------+---------------+
| Afghanistan     | Asia       | 652230     | 25500100     | 20343000      |
| Albania         | Europe     | 28748      | 2831741      | 12960000      |
| Algeria         | Africa     | 2381741    | 37100000     | 188681000     |
| Andorra         | Europe     | 468        | 78115        | 3712000       |
| Angola          | Africa     | 1246700    | 20609294     | 100990000     |
+-----------------+------------+------------+--------------+---------------+

如果一个国家的面积超过300万平方公里,或者人口超过2500万,那么这个国家就是大国家。

编写一个SQL查询,输出表中所有大国家的名称、人口和面积。

例如,根据上表,我们应该输出:

Code

+--------------+-------------+--------------+
| name         | population  | area         |
+--------------+-------------+--------------+
| Afghanistan  | 25500100    | 652230       |
| Algeria      | 37100000    | 2381741      |
+--------------+-------------+--------------+

sql

select  name ,population,area  
from World
where area  >3000000 or population >25000000

596. 超过5名学生的课

难度简单

SQL架构

有一个courses 表 ,有: student (学生)class (课程)

请列出所有超过或等于5名学生的课。

例如,表:

Code

+---------+------------+
| student | class      |
+---------+------------+
| A       | Math       |
| B       | English    |
| C       | Math       |
| D       | Biology    |
| E       | Math       |
| F       | Computer   |
| G       | Math       |
| H       | Math       |
| I       | Math       |
+---------+------------+

应该输出:

Code

+---------+
| class   |
+---------+
| Math    |
+---------+

Note: 学生在每个课中不应被重复计算。

sql

select  class 
from courses
group by class 
having count(distinct student)>=5

一个学生可能多次选课。。记得distinct

597. 好友申请 I :总体通过率

难度简单21收藏分享切换为英文关注反馈

SQL架构

在 Facebook 或者 Twitter 这样的社交应用中,人们经常会发好友申请也会收到其他人的好友申请。现在给如下两个表:

表: friend_request

sender_id

send_to_id

request_date

1

2

2016_06-01

1

3

2016_06-01

1

4

2016_06-01

2

3

2016_06-02

3

4

2016-06-09

表: request_accepted

requester_id

accepter_id

accept_date

1

2

2016_06-03

1

3

2016-06-08

2

3

2016-06-08

3

4

2016-06-09

3

4

2016-06-10

写一个查询语句,求出好友申请的通过率,用 2 位小数表示。通过率由接受好友申请的数目除以申请总数。

对于上面的样例数据,你的查询语句应该返回如下结果。

accept_rate

0.80

注意:

通过的好友申请不一定都在表 friend_request 中。在这种情况下,你只需要统计总的被通过的申请数(不管它们在不在原来的申请中),并将它除以申请总数,得到通过率 一个好友申请发送者有可能会给接受者发几条好友申请,也有可能一个好友申请会被通过好几次。这种情况下,重复的好友申请只统计一次。 如果一个好友申请都没有,通过率为 0.00 。

解释: 总共有 5 个申请,其中 4 个是不重复且被通过的好友申请,所以成功率是 0.80 。

进阶:

你能写一个查询语句得到每个月的通过率吗? 你能求出每一天的累计通过率吗?

sql

select
round(
    ifnull(
    (select count(*) from (select distinct requester_id, accepter_id from request_accepted) as A)
    /
    (select count(*) from (select distinct sender_id, send_to_id from friend_request) as B),
    0)
, 2) as accept_rate;

601. 体育馆的人流量

难度困难113收藏分享切换为英文关注反馈

SQL架构

X 市建了一个新的体育馆,每日人流量信息被记录在这三列信息中:序号 (id)、日期 (visit_date)、 人流量 (people)。

请编写一个查询语句,找出人流量的高峰期。高峰期时,至少连续三行记录中的人流量不少于100。

例如,表 stadium

Code

+------+------------+-----------+
| id   | visit_date | people    |
+------+------------+-----------+
| 1    | 2017-01-01 | 10        |
| 2    | 2017-01-02 | 109       |
| 3    | 2017-01-03 | 150       |
| 4    | 2017-01-04 | 99        |
| 5    | 2017-01-05 | 145       |
| 6    | 2017-01-06 | 1455      |
| 7    | 2017-01-07 | 199       |
| 8    | 2017-01-08 | 188       |
+------+------------+-----------+

对于上面的示例数据,输出为:

Code

+------+------------+-----------+
| id   | visit_date | people    |
+------+------------+-----------+
| 5    | 2017-01-05 | 145       |
| 6    | 2017-01-06 | 1455      |
| 7    | 2017-01-07 | 199       |
| 8    | 2017-01-08 | 188       |
+------+------------+-----------+

提示: 每天只有一行记录,日期随着 id 的增加而增加。

3表相连(244 ms)

sql

select distinct t1.*
from stadium t1, stadium t2, stadium t3
where t1.people >= 100 and t2.people >= 100 and t3.people >= 100
and
(
		(t1.id - t2.id = 1 and t1.id - t3.id = 2 and t2.id - t3.id =1)  -- t1, t2, t3
    or
    (t2.id - t1.id = 1 and t2.id - t3.id = 2 and t1.id - t3.id =1) -- t2, t1, t3
    or
    (t3.id - t2.id = 1 and t2.id - t1.id =1 and t3.id - t1.id = 2) -- t3, t2, t1
)
order by t1.id

窗口函数(272 ms)

sql

select id,visit_date,people from
(
    select id
    ,lead(people,1) over(order by id) ld
    ,lead(people,2) over(order by id) ld2
    ,visit_date
    ,lag(people,1) over(order by id) lg
    ,lag(people,2) over(order by id) lg2
    ,people
    from stadium
    ) a
where (a.ld>=100 and a.lg>=100 and a.people>=100)
or (a.ld>=100 and a.ld2>=100 and a.people>=100)
or (a.lg>=100 and a.lg2>=100 and a.people>=100)

602. 好友申请 II :谁有最多的好友

难度中等

SQL架构

在 Facebook 或者 Twitter 这样的社交应用中,人们经常会发好友申请也会收到其他人的好友申请。

request_accepted 存储了所有好友申请通过的数据记录,其中, requester_idaccepter_id 都是用户的编号。

requester_id

accepter_id

accept_date

1

2

2016_06-03

1

3

2016-06-08

2

3

2016-06-08

3

4

2016-06-09

写一个查询语句,求出谁拥有最多的好友和他拥有的好友数目。对于上面的样例数据,结果为:

id

num

3

3

注意:

  • 保证拥有最多好友数目的只有 1 个人。
  • 好友申请只会被接受一次,所以不会有 requester_idaccepter_id 值都相同的重复记录。

解释:

编号为 ‘3’ 的人是编号为 ‘1’,’2’ 和 ‘4’ 的好友,所以他总共有 3 个好友,比其他人都多。

进阶:

在真实世界里,可能会有多个人拥有好友数相同且最多,你能找到所有这些人吗?

sql

select rid as `id`,count(aid) as `num`
from
(
    select R1.requester_id as rid,R1.accepter_id as aid
    from request_accepted as R1
    UNION all
    select R2.accepter_id as rid,R2.requester_id as aid
    from request_accepted as R2
) as A
group by rid
order by num desc
limit 0,1

603. 连续空余座位

难度简单

SQL架构

几个朋友来到电影院的售票处,准备预约连续空余座位。

你能利用表 cinema ,帮他们写一个查询语句,获取所有空余座位,并将它们按照 seat_id 排序后返回吗?

seat_id

free

1

1

2

0

3

1

4

1

5

1

对于如上样例,你的查询语句应该返回如下结果。

seat_id

3

4

5

注意:

  • seat_id 字段是一个自增的整数,free 字段是布尔类型(’1’ 表示空余, ‘0’ 表示已被占据)。
  • 连续空余座位的定义是大于等于 2 个连续空余的座位。

sql

select seat_id
from (
select seat_id,
lag(seat_id,1,-99) over(order by seat_id) ls,
lead(seat_id,1,-99) over(order by seat_id) rs
from cinema
where free=1
)t1
where  seat_id-ls = 1 or rs-seat_id =1

607. 销售员

难度简单

SQL架构

描述

给定 3 个表: salespersoncompanyorders。 输出所有表 salesperson 中,没有向公司 ‘RED’ 销售任何东西的销售员。

示例: 输入

表: salesperson

Code

+----------+------+--------+-----------------+-----------+
| sales_id | name | salary | commission_rate | hire_date |
+----------+------+--------+-----------------+-----------+
|   1      | John | 100000 |     6           | 4/1/2006  |
|   2      | Amy  | 120000 |     5           | 5/1/2010  |
|   3      | Mark | 65000  |     12          | 12/25/2008|
|   4      | Pam  | 25000  |     25          | 1/1/2005  |
|   5      | Alex | 50000  |     10          | 2/3/2007  |
+----------+------+--------+-----------------+-----------+

salesperson 存储了所有销售员的信息。每个销售员都有一个销售员编号 sales_id 和他的名字 name

表: company

Code

+---------+--------+------------+
| com_id  |  name  |    city    |
+---------+--------+------------+
|   1     |  RED   |   Boston   |
|   2     | ORANGE |   New York |
|   3     | YELLOW |   Boston   |
|   4     | GREEN  |   Austin   |
+---------+--------+------------+

company 存储了所有公司的信息。每个公司都有一个公司编号 com_id 和它的名字 name

表: orders

Code

+----------+------------+---------+----------+--------+
| order_id | order_date | com_id  | sales_id | amount |
+----------+------------+---------+----------+--------+
| 1        |   1/1/2014 |    3    |    4     | 100000 |
| 2        |   2/1/2014 |    4    |    5     | 5000   |
| 3        |   3/1/2014 |    1    |    1     | 50000  |
| 4        |   4/1/2014 |    1    |    4     | 25000  |
+----------+----------+---------+----------+--------+

orders 存储了所有的销售数据,包括销售员编号 sales_id 和公司编号 com_id

输出

Code

+------+
| name | 
+------+
| Amy  | 
| Mark | 
| Alex |
+------+

解释

根据表 orders 中的订单 ‘3’ 和 ‘4’ ,容易看出只有 ‘John’ 和 ‘Pam’ 两个销售员曾经向公司 ‘RED’ 销售过。

所以我们需要输出表 salesperson 中所有其他人的名字。\

sql

select name
from salesperson
where sales_id not in
(
    select sales_id
    from orders
    where com_id =
            (
            select com_id 
            from company
            where name ='RED'
            )
)

608. 树节点

难度中等

SQL架构

给定一个表 treeid 是树节点的编号, p_id 是它父节点的 id 。

Code

+----+------+
| id | p_id |
+----+------+
| 1  | null |
| 2  | 1    |
| 3  | 1    |
| 4  | 2    |
| 5  | 2    |
+----+------+

树中每个节点属于以下三种类型之一:

  • 叶子:如果这个节点没有任何孩子节点。
  • 根:如果这个节点是整棵树的根,即没有父节点。
  • 内部节点:如果这个节点既不是叶子节点也不是根节点。

写一个查询语句,输出所有节点的编号和节点的类型,并将结果按照节点编号排序。上面样例的结果为:

Code

+----+------+
| id | Type |
+----+------+
| 1  | Root |
| 2  | Inner|
| 3  | Leaf |
| 4  | Leaf |
| 5  | Leaf |
+----+------+

解释

  • 节点 ‘1’ 是根节点,因为它的父节点是 NULL ,同时它有孩子节点 ‘2’ 和 ‘3’ 。
  • 节点 ‘2’ 是内部节点,因为它有父节点 ‘1’ ,也有孩子节点 ‘4’ 和 ‘5’ 。
  • 节点 ‘3’, ‘4’ 和 ‘5’ 都是叶子节点,因为它们都有父节点同时没有孩子节点。
  • 样例中树的形态如下: ​ 1 ​ / ​ 2 3 ​ / ​ 4 5

注意

如果树中只有一个节点,你只需要输出它的根属性。

sql

select id,
    (case when p_id is null then "Root"
    when id not in (select ifnull(p_id,0) from tree) then "Leaf"
    else "Inner" end)  Type
from tree

610. 判断三角形

难度简单

SQL架构

一个小学生 Tim 的作业是判断三条线段是否能形成一个三角形。

然而,这个作业非常繁重,因为有几百组线段需要判断。

假设表 triangle 保存了所有三条线段的三元组 x, y, z ,你能帮 Tim 写一个查询语句,来判断每个三元组是否可以组成一个三角形吗?

x

y

z

13

15

30

10

20

15

对于如上样例数据,你的查询语句应该返回如下结果:

x

y

z

triangle

13

15

30

No

10

20

15

Yes

sql

select x,y,z, 
if(x+y>z && x+z>y && y+z>x,'Yes','No') triangle
from triangle

612. 平面上的最近距离

难度中等

SQL架构

point_2d 保存了所有点(多于 2 个点)的坐标 (x,y) ,这些点在平面上两两不重合。

写一个查询语句找到两点之间的最近距离,保留 2 位小数。

x

y

-1

-1

0

0

-1

-2

最近距离在点 (-1,-1) 和(-1,2) 之间,距离为 1.00 。所以输出应该为:

shortest

1.00

注意:任意点之间的最远距离小于 10000 。

sql

SELECT
    ROUND(SQRT(MIN((POW(p1.x - p2.x, 2) + POW(p1.y - p2.y, 2)))), 2) AS shortest
FROM
    point_2d p1
        JOIN
    point_2d p2 ON p1.x != p2.x OR p1.y != p2.y

优化 :减少重复计算

sql

SELECT
    ROUND(SQRT(MIN((POW(p1.x - p2.x, 2) + POW(p1.y - p2.y, 2)))),2) AS shortest
FROM
    point_2d p1
        JOIN
    point_2d p2 ON (p1.x <= p2.x AND p1.y < p2.y)
        OR (p1.x <= p2.x AND p1.y > p2.y)
        OR (p1.x < p2.x AND p1.y = p2.y)

613. 直线上的最近距离

难度简单

SQL架构

point 保存了一些点在 x 轴上的坐标,这些坐标都是整数。

写一个查询语句,找到这些点中最近两个点之间的距离。

x

-1

0

2

最近距离显然是 ‘1’ ,是点 ‘-1’ 和 ‘0’ 之间的距离。所以输出应该如下:

shortest

1

注意:每个点都与其他点坐标不同,表 table 不会有重复坐标出现。

进阶:如果这些点在 x 轴上从左到右都有一个编号,输出结果时需要输出最近点对的编号呢?

开窗方法 178m

sql

select min(l-x) shortest
from( 
select x,lead(x,1,null) over(order by x) l
from point
)t1

join方法 268m

sql

SELECT
    MIN(ABS(p1.x - p2.x)) AS shortest
FROM
    point p1
        JOIN
    point p2 ON p1.x != p2.x
;

614. 二级关注者

难度中等

SQL架构

在 facebook 中,表 follow 会有 2 个字段: followee, follower ,分别表示被关注者和关注者。

请写一个 sql 查询语句,对每一个关注者,查询关注他的关注者的数目。

比方说:

Code

+-------------+------------+
| followee    | follower   |
+-------------+------------+
|     A       |     B      |
|     B       |     C      |
|     B       |     D      |
|     D       |     E      |
+-------------+------------+

应该输出:

Code

+-------------+------------+
| follower    | num        |
+-------------+------------+
|     B       |  2         |
|     D       |  1         |
+-------------+------------+

解释:

B 和 D 都在在 follower 字段中出现,作为被关注者,B 被 C 和 D 关注,D 被 E 关注。A 不在 follower 字段内,所以A不在输出列表中。

注意:

  • 被关注者永远不会被他 / 她自己关注。
  • 将结果按照字典序返回。

sql

select followee follower,count(distinct follower) num
from follow
where followee  in (
    select follower
    from follow
    group by follower
)
group by followee
order by follower

这里出现了重复关注 ,需要去重

615. 平均工资:部门与公司比较

难度困难

SQL架构

给如下两个表,写一个查询语句,求出在每一个工资发放日,每个部门的平均工资与公司的平均工资的比较结果 (高 / 低 / 相同)。

表: salary

id

employee_id

amount

pay_date

1

1

9000

2017-03-31

2

2

6000

2017-03-31

3

3

10000

2017-03-31

4

1

7000

2017-02-28

5

2

6000

2017-02-28

6

3

8000

2017-02-28

employee_id 字段是表 employeeemployee_id 字段的外键。

employee_id

department_id

1

1

2

2

3

2

对于如上样例数据,结果为:

pay_month

department_id

comparison

2017-03

1

higher

2017-03

2

lower

2017-02

1

same

2017-02

2

same

解释

在三月,公司的平均工资是 (9000+6000+10000)/3 = 8333.33…

由于部门 ‘1’ 里只有一个 employee_id 为 ‘1’ 的员工,所以部门 ‘1’ 的平均工资就是此人的工资 9000 。因为 9000 > 8333.33 ,所以比较结果是 ‘higher’。

第二个部门的平均工资为 employee_id 为 ‘2’ 和 ‘3’ 两个人的平均工资,为 (6000+10000)/2=8000 。因为 8000 < 8333.33 ,所以比较结果是 ‘lower’ 。

在二月用同样的公式求平均工资并比较,比较结果为 ‘same’ ,因为部门 ‘1’ 和部门 ‘2’ 的平均工资与公司的平均工资相同,都是 7000 。

sql

select 
    pay_month,
    department_id,
     (case when avgs>ts then 'higher'
            when avgs<ts then 'lower'
            else 'same' end) as comparison
from 
(
    select 
        date_format(pay_date,'%Y-%m')pay_month,
        department_id,
        avg(amount) over(partition by date_format(pay_date,'%Y-%m') )ts,
        avg(amount) over(partition by date_format(pay_date,'%Y-%m'),department_id) avgs
    from salary s
    left join employee e
    on s.employee_id = e.employee_id
   
)t1
 group by pay_month, department_id

也可以用if

IF(avgs>ts,'higher',IF(avgs=ts,'same','lower')) AS comparison

618. 学生地理信息报告

难度困难

SQL架构

一所美国大学有来自亚洲、欧洲和美洲的学生,他们的地理信息存放在如下 student 表中。

name

continent

Jack

America

Pascal

Europe

Xi

Asia

Jane

America

写一个查询语句实现对大洲(continent)列的 透视表 操作,使得每个学生按照姓名的字母顺序依次排列在对应的大洲下面。输出的标题应依次为美洲(America)、亚洲(Asia)和欧洲(Europe)。数据保证来自美洲的学生不少于来自亚洲或者欧洲的学生。

对于样例输入,它的对应输出是:

America

Asia

Europe

Jack

Xi

Pascal

Jane

进阶:如果不能确定哪个大洲的学生数最多,你可以写出一个查询去生成上述学生报告吗?

开窗

sql

select
max(if(continent='America',name,null)) America,
max(if(continent='Asia',name,null)) Asia,
max(if(continent='Europe',name,null)) Europe
from 
    (select *, row_number() over(partition by continent order by name) rk
    from student) t
group by rk

变量

sql

SELECT 
    America, Asia, Europe
FROM
    (SELECT @as:=0, @am:=0, @eu:=0) t,
    (SELECT 
        @as:=@as + 1 AS asid, name AS Asia
    FROM
        student
    WHERE
        continent = 'Asia'
    ORDER BY Asia) AS t1
        RIGHT JOIN
    (SELECT 
        @am:=@am + 1 AS amid, name AS America
    FROM
        student
    WHERE
        continent = 'America'
    ORDER BY America) AS t2 ON asid = amid
        LEFT JOIN
    (SELECT 
        @eu:=@eu + 1 AS euid, name AS Europe
    FROM
        student
    WHERE
        continent = 'Europe'
    ORDER BY Europe) AS t3 ON amid = euid

官方给出的。。同下方开窗

sql

select America,Asia,Europe 
from(
    select row_number() over(order by name) as rn,name as America from student
    where continent='America'
) a
left join(
    select row_number() over(order by name) as rn,name as Asia from student
    where continent='Asia'
) b on a.rn=b.rn
left join(
    select row_number() over(order by name) as rn,name as Europe from student
    where continent='Europe'
) c on a.rn=c.rn

619. 只出现一次的最大数字

难度简单

SQL架构

my_numbersnum 字段包含很多数字,其中包括很多重复的数字。

你能写一个 SQL 查询语句,找到只出现过一次的数字中,最大的一个数字吗?

Code

+---+
|num|
+---+
| 8 |
| 8 |
| 3 |
| 3 |
| 1 |
| 4 |
| 5 |
| 6 |

对于上面给出的样例数据,你的查询语句应该返回如下结果:

Code

+---+
|num|
+---+
| 6 |

注意:

如果没有只出现一次的数字,输出 null

sql

SELECT
    MAX(num) AS num
FROM
    (SELECT
        num
    FROM
        my_numbers
    GROUP BY num
    HAVING COUNT(num) = 1) t1

620. 有趣的电影

难度简单86收藏分享切换为英文关注反馈

SQL架构

某城市开了一家新的电影院,吸引了很多人过来看电影。该电影院特别注意用户体验,专门有个 LED显示板做电影推荐,上面公布着影评和相关电影描述。

作为该电影院的信息部主管,您需要编写一个 SQL查询,找出所有影片描述为 boring (不无聊) 的并且 id 为奇数 的影片,结果请按等级 rating 排列。

例如,下表 cinema:

Code

+---------+-----------+--------------+-----------+
|   id    | movie     |  description |  rating   |
+---------+-----------+--------------+-----------+
|   1     | War       |   great 3D   |   8.9     |
|   2     | Science   |   fiction    |   8.5     |
|   3     | irish     |   boring     |   6.2     |
|   4     | Ice song  |   Fantacy    |   8.6     |
|   5     | House card|   Interesting|   9.1     |
+---------+-----------+--------------+-----------+

对于上面的例子,则正确的输出是为:

Code

+---------+-----------+--------------+-----------+
|   id    | movie     |  description |  rating   |
+---------+-----------+--------------+-----------+
|   5     | House card|   Interesting|   9.1     |
|   1     | War       |   great 3D   |   8.9     |
+---------+-----------+--------------+-----------+

sql

select  id,movie,description,rating 
from cinema
where id%2=1 and description !='boring'
order by rating desc,id,movie,description

626. 换座位

难度中等

SQL架构

小美是一所中学的信息科技老师,她有一张 seat 座位表,平时用来储存学生名字和与他们相对应的座位 id。

其中纵列的 id 是连续递增的

小美想改变相邻俩学生的座位。

你能不能帮她写一个 SQL query 来输出小美想要的结果呢?

示例:

Code

+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Abbot   |
|    2    | Doris   |
|    3    | Emerson |
|    4    | Green   |
|    5    | Jeames  |
+---------+---------+

假如数据输入的是上表,则输出结果如下:

Code

+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Doris   |
|    2    | Abbot   |
|    3    | Green   |
|    4    | Emerson |
|    5    | Jeames  |
+---------+---------+

注意:

如果学生人数是奇数,则不需要改变最后一个同学的座位。

开窗

sql

select id,
(case when id%2=0 then f
      when id%2=1 && b is not null then b
      else student end) student
from(
    select id,student,
    lag(student,1,null) over(order by id) f,
    lead(student,1,null) over(order by id) b
    from seat
)t1

非嵌套

sql

select 
    if(id%2=0,
        id-1,
        if(id=(select count(distinct id) from seat),
            id,
            id+1)) 
    as id,student 
from seat 
order by id;

用异或

sql

select b.id,a.student from 
seat as a,seat as b,(select count(*) as cnt from seat) as c 
where b.id=1^(a.id-1)+1
-- where a.id=1^(b.id-1)+1; 也可以这样写,更容易理解
 || (c.cnt%2 && b.id=c.cnt && a.id=c.cnt);

627. 交换工资

难度简单

SQL架构

给定一个 salary 表,如下所示,有 m = 男性 和 f = 女性 的值。交换所有的 f 和 m 值(例如,将所有 f 值更改为 m,反之亦然)。要求只使用一个更新(Update)语句,并且没有中间的临时表。

注意,您必只能写一个 Update 语句,请不要编写任何 Select 语句。

例如:

id

name

sex

salary

1

A

m

2500

2

B

f

1500

3

C

m

5500

4

D

f

500

运行你所编写的更新语句之后,将会得到以下表:

id

name

sex

salary

1

A

f

2500

2

B

m

1500

3

C

f

5500

4

D

m

500

sql

UPDATE salary
SET
    sex = CASE sex
        WHEN 'm' THEN 'f'
        ELSE 'm'
    END;

Update 和set的使用

1045. 买下所有产品的客户

难度中等

SQL架构

Customer 表:

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| customer_id | int     |
| product_key | int     |
+-------------+---------+
product_key 是 Customer 表的外键。

Product 表:

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| product_key | int     |
+-------------+---------+
product_key 是这张表的主键。

写一条 SQL 查询语句,从 Customer 表中查询购买了 Product 表中所有产品的客户的 id。

示例:

Code

Customer 表:
+-------------+-------------+
| customer_id | product_key |
+-------------+-------------+
| 1           | 5           |
| 2           | 6           |
| 3           | 5           |
| 3           | 6           |
| 1           | 6           |
+-------------+-------------+

Product 表:
+-------------+
| product_key |
+-------------+
| 5           |
| 6           |
+-------------+

Result 表:
+-------------+
| customer_id |
+-------------+
| 1           |
| 3           |
+-------------+
购买了所有产品(5 和 6)的客户的 id 是 1 和 3 。

sql

select customer_id
from Customer
group by customer_id
having count(distinct product_key)=(
select count(*) cnt
from Product)

1050. 合作过至少三次的演员和导演

难度简单

SQL架构

ActorDirector 表:

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| actor_id    | int     |
| director_id | int     |
| timestamp   | int     |
+-------------+---------+
timestamp 是这张表的主键.

写一条SQL查询语句获取合作过至少三次的演员和导演的 id 对 (actor_id, director_id)

示例:

Code

ActorDirector 表:
+-------------+-------------+-------------+
| actor_id    | director_id | timestamp   |
+-------------+-------------+-------------+
| 1           | 1           | 0           |
| 1           | 1           | 1           |
| 1           | 1           | 2           |
| 1           | 2           | 3           |
| 1           | 2           | 4           |
| 2           | 1           | 5           |
| 2           | 1           | 6           |
+-------------+-------------+-------------+

Result 表:
+-------------+-------------+
| actor_id    | director_id |
+-------------+-------------+
| 1           | 1           |
+-------------+-------------+
唯一的 id 对是 (1, 1),他们恰好合作了 3 次。

sql

select actor_id,director_id
from ActorDirector
group by actor_id,director_id
having count(*)>=3

1068. 产品销售分析 I

难度简单

SQL架构

销售表 Sales

Code

+-------------+-------+
| Column Name | Type  |
+-------------+-------+
| sale_id     | int   |
| product_id  | int   |
| year        | int   |
| quantity    | int   |
| price       | int   |
+-------------+-------+
(sale_id, year) 是销售表 Sales 的主键.
product_id 是产品表 Product 的外键.
注意: price 表示每单位价格

产品表 Product

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
+--------------+---------+
product_id 是表的主键.

写一条SQL 查询语句获取产品表 Product 中所有的 产品名称 product name 以及 该产品在 Sales 表中相对应的 上市年份 year价格 price

示例:

Code

Sales 表:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+ 
| 1       | 100        | 2008 | 10       | 5000  |
| 2       | 100        | 2009 | 12       | 5000  |
| 7       | 200        | 2011 | 15       | 9000  |
+---------+------------+------+----------+-------+

Product 表:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100        | Nokia        |
| 200        | Apple        |
| 300        | Samsung      |
+------------+--------------+

Result 表:
+--------------+-------+-------+
| product_name | year  | price |
+--------------+-------+-------+
| Nokia        | 2008  | 5000  |
| Nokia        | 2009  | 5000  |
| Apple        | 2011  | 9000  |
+--------------+-------+-------+

sql

select product_name,year,price
from Sales s left join Product p
on s.product_id  = p.product_id

1069. 产品销售分析 II

难度简单

SQL架构

销售表:Sales

Code

+-------------+-------+
| Column Name | Type  |
+-------------+-------+
| sale_id     | int   |
| product_id  | int   |
| year        | int   |
| quantity    | int   |
| price       | int   |
+-------------+-------+
sale_id 是这个表的主键。
product_id 是 Product 表的外键。
请注意价格是每单位的。

产品表:Product

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
+--------------+---------+
product_id 是这个表的主键。

编写一个 SQL 查询,按产品 id product_id 来统计每个产品的销售总量。

查询结果格式如下面例子所示:

Code

Sales 表:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+ 
| 1       | 100        | 2008 | 10       | 5000  |
| 2       | 100        | 2009 | 12       | 5000  |
| 7       | 200        | 2011 | 15       | 9000  |
+---------+------------+------+----------+-------+

Product 表:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100        | Nokia        |
| 200        | Apple        |
| 300        | Samsung      |
+------------+--------------+

Result 表:
+--------------+----------------+
| product_id   | total_quantity |
+--------------+----------------+
| 100          | 22             |
| 200          | 15             |
+--------------+----------------+

sql

select s.product_id,sum(quantity) total_quantity
from Sales s left join Product p
on s.product_id  = p.product_id
group by s.product_id

1070. 产品销售分析 III

难度中等

SQL架构

销售表 Sales

Code

+-------------+-------+
| Column Name | Type  |
+-------------+-------+
| sale_id     | int   |
| product_id  | int   |
| year        | int   |
| quantity    | int   |
| price       | int   |
+-------------+-------+
sale_id 是此表的主键。
product_id 是产品表的外键。
请注意,价格是按每单位计的。

产品表 Product

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
+--------------+---------+
product_id 是此表的主键。

编写一个 SQL 查询,选出每个销售产品的 第一年产品 id年份数量价格

查询结果格式如下:

Code

Sales table:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+ 
| 1       | 100        | 2008 | 10       | 5000  |
| 2       | 100        | 2009 | 12       | 5000  |
| 7       | 200        | 2011 | 15       | 9000  |
+---------+------------+------+----------+-------+

Product table:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100        | Nokia        |
| 200        | Apple        |
| 300        | Samsung      |
+------------+--------------+

Result table:
+------------+------------+----------+-------+
| product_id | first_year | quantity | price |
+------------+------------+----------+-------+ 
| 100        | 2008       | 10       | 5000  |
| 200        | 2011       | 15       | 9000  |
+------------+------------+----------+-------+

sql

select product_id,year first_year,quantity,price
from (
select s.product_id,year,quantity,price,
rank() over(partition by product_id order by year) rk
from Sales s left join Product p
on s.product_id  = p.product_id
)t1
where rk = 1

1075. 项目员工 I

难度简单

SQL架构

项目表 Project

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
主键为 (project_id, employee_id)。
employee_id 是员工表 Employee 表的外键。

员工表 Employee

Code

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
主键是 employee_id。

请写一个 SQL 语句,查询每一个项目中员工的 平均 工作年限,精确到小数点后两位

查询结果的格式如下:

Code

Project 表:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+

Employee 表:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+

Result 表:
+-------------+---------------+
| project_id  | average_years |
+-------------+---------------+
| 1           | 2.00          |
| 2           | 2.50          |
+-------------+---------------+
第一个项目中,员工的平均工作年限是 (3 + 2 + 1) / 3 = 2.00;第二个项目中,员工的平均工作年限是 (3 + 2) / 2 = 2.50

sql

select project_id,round(avg(experience_years),2) average_years
from Project p join Employee e
on p.employee_id=e.employee_id 
group by project_id

1076. 项目员工II

难度简单

SQL架构

Table: Project

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
主键为 (project_id, employee_id)。
employee_id 是员工表 Employee 表的外键。

Table: Employee

Code

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
主键是 employee_id。

编写一个SQL查询,报告所有雇员最多的项目。

查询结果格式如下所示:

Code

Project table:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+

Employee table:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+

Result table:
+-------------+
| project_id  |
+-------------+
| 1           |
+-------------+
第一个项目有3名员工,第二个项目有2名员工。

sql

select project_id 
from Project 
group by project_id 
having count(*) = 
(select count(*) `num` from Project group by project_id order by count(*) desc limit 1);

开窗

sql

select project_id from
(select project_id,rank()over(order by count(employee_id) desc) ranking from Project group by project_id) temp where ranking=1

1077. 项目员工 III

难度中等

SQL架构

项目表 Project

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
(project_id, employee_id) 是这个表的主键
employee_id 是员工表 Employee 的外键

员工表 Employee

Code

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
employee_id 是这个表的主键

写 一个 SQL 查询语句,报告在每一个项目中经验最丰富的雇员是谁。如果出现经验年数相同的情况,请报告所有具有最大经验年数的员工。

查询结果格式在以下示例中:

Code

Project 表:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+

Employee 表:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 3                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+

Result 表:
+-------------+---------------+
| project_id  | employee_id   |
+-------------+---------------+
| 1           | 1             |
| 1           | 3             |
| 2           | 1             |
+-------------+---------------+
employee_id 为 1 和 3 的员工在 project_id 为 1 的项目中拥有最丰富的经验。在 project_id 为 2 的项目中,employee_id 为 1 的员工拥有最丰富的经验。

sql

select project_id ,employee_id
from(
select project_id,e.employee_id,
rank()over(partition by project_id order by experience_years desc) rk
from Project p join Employee e
on p.employee_id=e.employee_id 
)t1
where rk=1

1082. 销售分析 I

难度简单22

SQL架构

产品表:Product

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
| unit_price   | int     |
+--------------+---------+
product_id 是这个表的主键.

销售表:Sales

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| seller_id   | int     |
| product_id  | int     |
| buyer_id    | int     |
| sale_date   | date    |
| quantity    | int     |
| price       | int     |
+------ ------+---------+
这个表没有主键,它可以有重复的行.
product_id 是 Product 表的外键.

编写一个 SQL 查询,查询总销售额最高的销售者,如果有并列的,就都展示出来。

查询结果格式如下所示:

Code

Product 表:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1          | S8           | 1000       |
| 2          | G4           | 800        |
| 3          | iPhone       | 1400       |
+------------+--------------+------------+

Sales 表:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
| 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
| 2         | 2          | 3        | 2019-06-02 | 1        | 800   |
| 3         | 3          | 4        | 2019-05-13 | 2        | 2800  |
+-----------+------------+----------+------------+----------+-------+

Result 表:
+-------------+
| seller_id   |
+-------------+
| 1           |
| 3           |
+-------------+
Id 为 1 和 3 的销售者,销售总金额都为最高的 2800。

sql

select seller_id
from (
select seller_id ,sum(price) tp,rank() over(order by sum(price) desc) rk
from Sales
group by seller_id
)t1
where rk =1

1083. 销售分析 II

难度简单13

SQL架构

Table: Product

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
| unit_price   | int     |
+--------------+---------+
product_id 是这张表的主键

Table: Sales

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| seller_id   | int     |
| product_id  | int     |
| buyer_id    | int     |
| sale_date   | date    |
| quantity    | int     |
| price       | int     |
+------ ------+---------+
这个表没有主键,它可以有重复的行.
product_id 是 Product 表的外键.

编写一个 SQL 查询,查询购买了 S8 手机却没有购买 iPhone 的买家。注意这里 S8 和 iPhone 是 Product 表中的产品。

查询结果格式如下图表示:

Code

Product table:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1          | S8           | 1000       |
| 2          | G4           | 800        |
| 3          | iPhone       | 1400       |
+------------+--------------+------------+

Sales table:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
| 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
| 2         | 1          | 3        | 2019-06-02 | 1        | 800   |
| 3         | 3          | 3        | 2019-05-13 | 2        | 2800  |
+-----------+------------+----------+------------+----------+-------+

Result table:
+-------------+
| buyer_id    |
+-------------+
| 1           |
+-------------+
id 为 1 的买家购买了一部 S8,但是却没有购买 iPhone,而 id 为 3 的买家却同时购买了这 2 部手机。

sql

select t.buyer_id from(
select s.buyer_id, p.product_name
from sales s
inner join
product p
on s.product_id=p.product_id and (p.product_name='S8' or p.product_name='iPhone')
group by s.buyer_id
having count(distinct p.product_name) = 1
) t
where t.product_name='S8'

效率低

sql

select s.buyer_id 
from sales as s left join product as p 
on s.product_id=p.product_id
group by buyer_id
having sum(p.product_name='S8')>0 and sum(p.product_name='iPhone')=0

1084. 销售分析III

难度简单13

SQL架构

Table: Product

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
| unit_price   | int     |
+--------------+---------+
product_id 是这个表的主键

Table: Sales

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| seller_id   | int     |
| product_id  | int     |
| buyer_id    | int     |
| sale_date   | date    |
| quantity    | int     |
| price       | int     |
+------ ------+---------+
这个表没有主键,它可以有重复的行.
product_id 是 Product 表的外键.

编写一个SQL查询,报告2019年春季才售出的产品。即2019-01-012019-03-31(含)之间出售的商品。

查询结果格式如下所示:

Code

Product table:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1          | S8           | 1000       |
| 2          | G4           | 800        |
| 3          | iPhone       | 1400       |
+------------+--------------+------------+

Sales table:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
| 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
| 2         | 2          | 3        | 2019-06-02 | 1        | 800   |
| 3         | 3          | 4        | 2019-05-13 | 2        | 2800  |
+-----------+------------+----------+------------+----------+-------+

Result table:
+-------------+--------------+
| product_id  | product_name |
+-------------+--------------+
| 1           | S8           |
+-------------+--------------+
id为1的产品仅在2019年春季销售,其他两个产品在之后销售。

876ms

正常解法

sql

select product_id, product_name  
from product 
where product_id not in (
select distinct product_id from sales 
where sale_date > '2019-03-31' or sale_date < '2019-01-01')

867ms

sum = count

sql

select product_id, product_name
from Sales join Product
using(product_id)
group by product_id
having sum(sale_date between "2019-01-01" and "2019-03-31") = count(sale_date)

987ms

sum 为0 类似于第一种解法,计算了sum效率就低了

sql

select p.product_id, p.product_name
from sales s, product p
where s.product_id = p.product_id
group by s.product_id
having sum(s.sale_date> '2019-03-31')=0 and sum(s.sale_date < '2019-01-01') = 0

上一解法sum换成max

859ms

sql

select p.product_id, p.product_name
from sales s, product p
where s.product_id = p.product_id
group by s.product_id
having min(s.sale_date)>='2019-01-01' and max(s.sale_date) <= '2019-03-31'

1097. 游戏玩法分析 V

难度困难

SQL架构

Activity 活动记录表

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| player_id    | int     |
| device_id    | int     |
| event_date   | date    |
| games_played | int     |
+--------------+---------+
(player_id,event_date)是此表的主键
这张表显示了某些游戏的玩家的活动情况
每一行是一个玩家的记录,他在某一天使用某个设备注销之前登录并玩了很多游戏(可能是 0)

我们将玩家的安装日期定义为该玩家的第一个登录日。

我们还将某个日期 X 的第 1 天留存时间定义为安装日期为 X 的玩家的数量,他们在 X 之后的一天重新登录,除以安装日期为 X 的玩家的数量,四舍五入到小数点后两位。

编写一个 SQL 查询,报告每个安装日期、当天安装游戏的玩家数量和第一天的留存时间。

查询结果格式如下所示:

Code

Activity 表:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1         | 2         | 2016-03-01 | 5            |
| 1         | 2         | 2016-03-02 | 6            |
| 2         | 3         | 2017-06-25 | 1            |
| 3         | 1         | 2016-03-01 | 0            |
| 3         | 4         | 2016-07-03 | 5            |
+-----------+-----------+------------+--------------+

Result 表:
+------------+----------+----------------+
| install_dt | installs | Day1_retention |
+------------+----------+----------------+
| 2016-03-01 | 2        | 0.50           |
| 2017-06-25 | 1        | 0.00           |
+------------+----------+----------------+
玩家 1 和 3 在 2016-03-01 安装了游戏,但只有玩家 1 在 2016-03-02 重新登录,所以 2016-03-01 的第一天留存时间是 1/2=0.50
玩家 2 在 2017-06-25 安装了游戏,但在 2017-06-26 没有重新登录,因此 2017-06-25 的第一天留存时间为 0/1=0.00

sql

select install_dt,count(distinct player_id)installs,
       round(sum(if(datediff(event_date,install_dt)=1,1,0))/count(distinct player_id),2) Day1_retention 
from 
(
    select *,min(event_date) over(partition by player_id) install_dt
    from Activity
)t1   
group by install_dt

1098. 小众书籍

难度中等

SQL架构

书籍表 Books

Code

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| book_id        | int     |
| name           | varchar |
| available_from | date    |
+----------------+---------+
book_id 是这个表的主键。

订单表 Orders

Code

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| order_id       | int     |
| book_id        | int     |
| quantity       | int     |
| dispatch_date  | date    |
+----------------+---------+
order_id 是这个表的主键。
book_id  是 Books 表的外键。

你需要写一段 SQL 命令,筛选出过去一年中订单总量 少于10本书籍

注意:不考虑 上架(available from)距今 不满一个月 的书籍。并且 假设今天是 2019-06-23

Write an SQL query that reports the books that have sold less than 10 copies in the last year, excluding books that have been available for less than 1 month from today. Assume today is 2019-06-23.

下面是样例输出结果:

Code

Books 表:
+---------+--------------------+----------------+
| book_id | name               | available_from |
+---------+--------------------+----------------+
| 1       | "Kalila And Demna" | 2010-01-01     |
| 2       | "28 Letters"       | 2012-05-12     |
| 3       | "The Hobbit"       | 2019-06-10     |
| 4       | "13 Reasons Why"   | 2019-06-01     |
| 5       | "The Hunger Games" | 2008-09-21     |
+---------+--------------------+----------------+

Orders 表:
+----------+---------+----------+---------------+
| order_id | book_id | quantity | dispatch_date |
+----------+---------+----------+---------------+
| 1        | 1       | 2        | 2018-07-26    |
| 2        | 1       | 1        | 2018-11-05    |
| 3        | 3       | 8        | 2019-06-11    |
| 4        | 4       | 6        | 2019-06-05    |
| 5        | 4       | 5        | 2019-06-20    |
| 6        | 5       | 9        | 2009-02-02    |
| 7        | 5       | 8        | 2010-04-13    |
+----------+---------+----------+---------------+

Result 表:
+-----------+--------------------+
| book_id   | name               |
+-----------+--------------------+
| 1         | "Kalila And Demna" |
| 2         | "28 Letters"       |
| 5         | "The Hunger Games" |
+-----------+--------------------+

这题中英文 都有歧义,看结果进行分析

sql

SELECT a.book_id, a.name 
FROM books a LEFT JOIN orders b ON a.book_id=b.book_id 
AND dispatch_date BETWEEN DATE_ADD('2019-06-23', INTERVAL -1 YEAR) AND '2019-06-23' 
WHERE a.available_from <= DATE_ADD('2019-06-23',INTERVAL -1 MONTH) 
GROUP BY a.book_id, a.name 
HAVING SUM(IFNULL(b.quantity,0)) < 10 
ORDER BY a.book_id;

此题有坑,首先订单近一年,再者quantity为null,还有近一月出版,sum不是订单是本数 还有就是date_add这个函数在hive里和mysql语法上有点小区别

1107. 每日新用户统计

难度中等

SQL架构

Traffic 表:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| activity      | enum    |
| activity_date | date    |
+---------------+---------+
该表没有主键,它可能有重复的行。
activity 列是 ENUM 类型,可能取 ('login', 'logout', 'jobs', 'groups', 'homepage') 几个值之一。

编写一个 SQL 查询,以查询从今天起最多 90 天内,每个日期该日期首次登录的用户数。假设今天是 2019-06-30.

查询结果格式如下例所示:

Code

Traffic 表:
+---------+----------+---------------+
| user_id | activity | activity_date |
+---------+----------+---------------+
| 1       | login    | 2019-05-01    |
| 1       | homepage | 2019-05-01    |
| 1       | logout   | 2019-05-01    |
| 2       | login    | 2019-06-21    |
| 2       | logout   | 2019-06-21    |
| 3       | login    | 2019-01-01    |
| 3       | jobs     | 2019-01-01    |
| 3       | logout   | 2019-01-01    |
| 4       | login    | 2019-06-21    |
| 4       | groups   | 2019-06-21    |
| 4       | logout   | 2019-06-21    |
| 5       | login    | 2019-03-01    |
| 5       | logout   | 2019-03-01    |
| 5       | login    | 2019-06-21    |
| 5       | logout   | 2019-06-21    |
+---------+----------+---------------+

Result 表:
+------------+-------------+
| login_date | user_count  |
+------------+-------------+
| 2019-05-01 | 1           |
| 2019-06-21 | 2           |
+------------+-------------+
请注意,我们只关心用户数非零的日期.
ID 为 5 的用户第一次登陆于 2019-03-01,因此他不算在 2019-06-21 的的统计内。

sql

select login_date,count(*) user_count
from(
    select user_id,min(activity_date) login_date
    from Traffic
    where activity = 'login'
    group by user_id
    having login_date>= DATE_ADD('2019-06-30',INTERVAL -90 DAY) 
)t1
group by login_date

1112. 每位学生的最高成绩

难度中等

SQL架构

表:Enrollments

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| student_id    | int     |
| course_id     | int     |
| grade         | int     |
+---------------+---------+
(student_id, course_id) 是该表的主键。

编写一个 SQL 查询,查询每位学生获得的最高成绩和它所对应的科目,若科目成绩并列,取 course_id 最小的一门。查询结果需按 student_id 增序进行排序。

查询结果格式如下所示:

Code

Enrollments 表:
+------------+-------------------+
| student_id | course_id | grade |
+------------+-----------+-------+
| 2          | 2         | 95    |
| 2          | 3         | 95    |
| 1          | 1         | 90    |
| 1          | 2         | 99    |
| 3          | 1         | 80    |
| 3          | 2         | 75    |
| 3          | 3         | 82    |
+------------+-----------+-------+

Result 表:
+------------+-------------------+
| student_id | course_id | grade |
+------------+-----------+-------+
| 1          | 2         | 99    |
| 2          | 2         | 95    |
| 3          | 3         | 82    |
+------------+-----------+-------+

sql

select student_id,course_id ,grade
from (
select student_id,course_id ,grade,
rank()over(partition by student_id order by grade desc,course_id) rK
from Enrollments
)t1
where rk =1

1113. 报告的记录

难度简单

SQL架构

动作表:Actions

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| post_id       | int     |
| action_date   | date    | 
| action        | enum    |
| extra         | varchar |
+---------------+---------+
此表没有主键,所以可能会有重复的行。
action 字段是 ENUM 类型的,包含:('view', 'like', 'reaction', 'comment', 'report', 'share')
extra 字段是可选的信息(可能为 null),其中的信息例如有:1.报告理由(a reason for report) 2.反应类型(a type of reaction)

编写一条SQL,查询每种 报告理由(report reason)在昨天的报告数量。假设今天是 2019-07-05

查询及结果的格式示例:

Code

Actions table:
+---------+---------+-------------+--------+--------+
| user_id | post_id | action_date | action | extra  |
+---------+---------+-------------+--------+--------+
| 1       | 1       | 2019-07-01  | view   | null   |
| 1       | 1       | 2019-07-01  | like   | null   |
| 1       | 1       | 2019-07-01  | share  | null   |
| 2       | 4       | 2019-07-04  | view   | null   |
| 2       | 4       | 2019-07-04  | report | spam   |
| 3       | 4       | 2019-07-04  | view   | null   |
| 3       | 4       | 2019-07-04  | report | spam   |
| 4       | 3       | 2019-07-02  | view   | null   |
| 4       | 3       | 2019-07-02  | report | spam   |
| 5       | 2       | 2019-07-04  | view   | null   |
| 5       | 2       | 2019-07-04  | report | racism |
| 5       | 5       | 2019-07-04  | view   | null   |
| 5       | 5       | 2019-07-04  | report | racism |
+---------+---------+-------------+--------+--------+

Result table:
+---------------+--------------+
| report_reason | report_count |
+---------------+--------------+
| spam          | 1            |
| racism        | 2            |
+---------------+--------------+ 
注意,我们只关心报告数量非零的结果。

sql

SELECT
	extra AS report_reason,
	COUNT(distinct post_id) AS report_count
FROM
	Actions
WHERE
	`action` = 'report' AND action_date = date_add('2019-07-05',Interval -1 day)
GROUP BY
	extra;

1132. 报告的记录 II

难度中等

SQL架构

动作表: Actions

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| post_id       | int     |
| action_date   | date    |
| action        | enum    |
| extra         | varchar |
+---------------+---------+
这张表没有主键,并有可能存在重复的行。
action 列的类型是 ENUM,可能的值为 ('view', 'like', 'reaction', 'comment', 'report', 'share')。
extra 列拥有一些可选信息,例如:报告理由(a reason for report)或反应类型(a type of reaction)等。

移除表: Removals

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| post_id       | int     |
| remove_date   | date    | 
+---------------+---------+
这张表的主键是 post_id。
这张表的每一行表示一个被移除的帖子,原因可能是由于被举报或被管理员审查。

编写一段 SQL 来查找:在被报告为垃圾广告的帖子中,被移除的帖子的每日平均占比,四舍五入到小数点后 2 位

查询结果的格式如下:

Code

Actions table:
+---------+---------+-------------+--------+--------+
| user_id | post_id | action_date | action | extra  |
+---------+---------+-------------+--------+--------+
| 1       | 1       | 2019-07-01  | view   | null   |
| 1       | 1       | 2019-07-01  | like   | null   |
| 1       | 1       | 2019-07-01  | share  | null   |
| 2       | 2       | 2019-07-04  | view   | null   |
| 2       | 2       | 2019-07-04  | report | spam   |
| 3       | 4       | 2019-07-04  | view   | null   |
| 3       | 4       | 2019-07-04  | report | spam   |
| 4       | 3       | 2019-07-02  | view   | null   |
| 4       | 3       | 2019-07-02  | report | spam   |
| 5       | 2       | 2019-07-03  | view   | null   |
| 5       | 2       | 2019-07-03  | report | racism |
| 5       | 5       | 2019-07-03  | view   | null   |
| 5       | 5       | 2019-07-03  | report | racism |
+---------+---------+-------------+--------+--------+

Removals table:
+---------+-------------+
| post_id | remove_date |
+---------+-------------+
| 2       | 2019-07-20  |
| 3       | 2019-07-18  |
+---------+-------------+

Result table:
+-----------------------+
| average_daily_percent |
+-----------------------+
| 75.00                 |
+-----------------------+
2019-07-04 的垃圾广告移除率是 50%,因为有两张帖子被报告为垃圾广告,但只有一个得到移除。
2019-07-02 的垃圾广告移除率是 100%,因为有一张帖子被举报为垃圾广告并得到移除。
其余几天没有收到垃圾广告的举报,因此平均值为:(50 + 100) / 2 = 75%
注意,输出仅需要一个平均值即可,我们并不关注移除操作的日期。

sql

SELECT ROUND(AVG(proportion) * 100, 2) AS average_daily_percent  
FROM (
    SELECT actions.action_date, COUNT(DISTINCT removals.post_id)/COUNT(DISTINCT actions.post_id) AS proportion
    FROM actions
    LEFT JOIN removals
    ON actions.post_id = removals.post_id
    WHERE extra = 'spam' 
    GROUP BY actions.action_date
) a

要理解spam的含义

1126. 查询活跃业务

难度中等

SQL架构

事件表:Events

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| business_id   | int     |
| event_type    | varchar |
| occurences    | int     | 
+---------------+---------+
此表的主键是 (business_id, event_type)。
表中的每一行记录了某种类型的事件在某些业务中多次发生的信息。

写一段 SQL 来查询所有活跃的业务。

如果一个业务的某个事件类型的发生次数大于此事件类型在所有业务中的平均发生次数,并且该业务至少有两个这样的事件类型,那么该业务就可被看做是活跃业务。

查询结果格式如下所示:

Code

Events table:
+-------------+------------+------------+
| business_id | event_type | occurences |
+-------------+------------+------------+
| 1           | reviews    | 7          |
| 3           | reviews    | 3          |
| 1           | ads        | 11         |
| 2           | ads        | 7          |
| 3           | ads        | 6          |
| 1           | page views | 3          |
| 2           | page views | 12         |
+-------------+------------+------------+

结果表
+-------------+
| business_id |
+-------------+
| 1           |
+-------------+ 
'reviews'、 'ads' 和 'page views' 的总平均发生次数分别是 (7+3)/2=5, (11+7+6)/3=8, (3+12)/2=7.5。
id 为 1 的业务有 7 个 'reviews' 事件(大于 5)和 11 个 'ads' 事件(大于 8),所以它是活跃业务。

sql

select business_id
from (
    select business_id,occurences,
    avg(occurences) over(partition by event_type) avo
    from Events
)t1
where  occurences > avo
group by business_id
having count(*)>=2

1127. 用户购买平台

难度困难

SQL架构

支出表: Spending

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| user_id     | int     |
| spend_date  | date    |
| platform    | enum    | 
| amount      | int     |
+-------------+---------+
这张表记录了用户在一个在线购物网站的支出历史,该在线购物平台同时拥有桌面端('desktop')和手机端('mobile')的应用程序。
这张表的主键是 (user_id, spend_date, platform)。
平台列 platform 是一种 ENUM ,类型为('desktop', 'mobile')。

写一段 SQL 来查找每天 使用手机端用户、 使用桌面端用户和 同时 使用桌面端和手机端的用户人数和总支出金额。

查询结果格式如下例所示:

Code

Spending table:
+---------+------------+----------+--------+
| user_id | spend_date | platform | amount |
+---------+------------+----------+--------+
| 1       | 2019-07-01 | mobile   | 100    |
| 1       | 2019-07-01 | desktop  | 100    |
| 2       | 2019-07-01 | mobile   | 100    |
| 2       | 2019-07-02 | mobile   | 100    |
| 3       | 2019-07-01 | desktop  | 100    |
| 3       | 2019-07-02 | desktop  | 100    |
+---------+------------+----------+--------+

Result table:
+------------+----------+--------------+-------------+
| spend_date | platform | total_amount | total_users |
+------------+----------+--------------+-------------+
| 2019-07-01 | desktop  | 100          | 1           |
| 2019-07-01 | mobile   | 100          | 1           |
| 2019-07-01 | both     | 200          | 1           |
| 2019-07-02 | desktop  | 100          | 1           |
| 2019-07-02 | mobile   | 100          | 1           |
| 2019-07-02 | both     | 0            | 0           |
+------------+----------+--------------+-------------+ 
在 2019-07-01, 用户1 同时 使用桌面端和手机端购买, 用户2 仅 使用了手机端购买,而用户3 仅 使用了桌面端购买。
在 2019-07-02, 用户2 仅 使用了手机端购买, 用户3 仅 使用了桌面端购买,且没有用户 同时 使用桌面端和手机端购买。

sql

select
    spend_date, platform,
    ifnull(sum(total_am),0) total_amount,
    ifnull(sum(total_u),0) total_users
from
(
    select p.spend_date, p.platform, t.total_am, t.total_u
    from
    (
        select distinct spend_date, "desktop" platform from Spending
        union
        select distinct spend_date, "mobile" platform from Spending
        union
        select distinct spend_date, "both" platform from Spending
    ) p
    left join
    (
        select spend_date, 
            if(count(distinct platform)=1, platform, 'both') plat,
            sum(amount) total_am,
            count(distinct user_id) total_u
        from Spending
        group by spend_date, user_id
    ) t
    on p.platform = t.plat and p.spend_date = t.spend_date
) temp
group by spend_date, platform

必须保证 desktop mobile both的顺序 ,所以 先列出三个字段

1141. 查询近30天活跃用户数

难度简单

SQL架构

活动记录表:Activity

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| session_id    | int     |
| activity_date | date    |
| activity_type | enum    |
+---------------+---------+
该表是用户在社交网站的活动记录。
该表没有主键,可能包含重复数据。
activity_type 字段为以下四种值 ('open_session', 'end_session', 'scroll_down', 'send_message')。
每个 session_id 只属于一个用户。

请写SQL查询出截至 2019-07-27(包含2019-07-27),近 30天的每日活跃用户数(当天只要有一条活动记录,即为活跃用户)。

查询结果示例如下:

Code

Activity table:
+---------+------------+---------------+---------------+
| user_id | session_id | activity_date | activity_type |
+---------+------------+---------------+---------------+
| 1       | 1          | 2019-07-20    | open_session  |
| 1       | 1          | 2019-07-20    | scroll_down   |
| 1       | 1          | 2019-07-20    | end_session   |
| 2       | 4          | 2019-07-20    | open_session  |
| 2       | 4          | 2019-07-21    | send_message  |
| 2       | 4          | 2019-07-21    | end_session   |
| 3       | 2          | 2019-07-21    | open_session  |
| 3       | 2          | 2019-07-21    | send_message  |
| 3       | 2          | 2019-07-21    | end_session   |
| 4       | 3          | 2019-06-25    | open_session  |
| 4       | 3          | 2019-06-25    | end_session   |
+---------+------------+---------------+---------------+

Result table:
+------------+--------------+ 
| day        | active_users |
+------------+--------------+ 
| 2019-07-20 | 2            |
| 2019-07-21 | 2            |
+------------+--------------+ 
非活跃用户的记录不需要展示。

sql

select activity_date day,count(distinct user_id) active_users
from Activity
where activity_date > date_add('2019-07-27',INTERVAL -1 MONTH)
group by activity_date

1142. 过去30天的用户活动 II

难度简单7收藏分享切换为英文关注反馈

SQL架构

Table: Activity

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| session_id    | int     |
| activity_date | date    |
| activity_type | enum    |
+---------------+---------+
该表没有主键,它可能有重复的行。
activity_type列是一种类型的ENUM(“ open_session”,“ end_session”,“ scroll_down”,“ send_message”)。
该表显示了社交媒体网站的用户活动。
请注意,每个会话完全属于一个用户。

编写SQL查询以查找截至2019年7月27日(含)的30天内每个用户的平均会话数,四舍五入到小数点后两位。我们只统计那些会话期间用户至少进行一项活动的有效会话。

查询结果格式如下例所示:

Code

Activity table:
+---------+------------+---------------+---------------+
| user_id | session_id | activity_date | activity_type |
+---------+------------+---------------+---------------+
| 1       | 1          | 2019-07-20    | open_session  |
| 1       | 1          | 2019-07-20    | scroll_down   |
| 1       | 1          | 2019-07-20    | end_session   |
| 2       | 4          | 2019-07-20    | open_session  |
| 2       | 4          | 2019-07-21    | send_message  |
| 2       | 4          | 2019-07-21    | end_session   |
| 3       | 2          | 2019-07-21    | open_session  |
| 3       | 2          | 2019-07-21    | send_message  |
| 3       | 2          | 2019-07-21    | end_session   |
| 3       | 5          | 2019-07-21    | open_session  |
| 3       | 5          | 2019-07-21    | scroll_down   |
| 3       | 5          | 2019-07-21    | end_session   |
| 4       | 3          | 2019-06-25    | open_session  |
| 4       | 3          | 2019-06-25    | end_session   |
+---------+------------+---------------+---------------+

Result table:
+---------------------------+ 
| average_sessions_per_user |
+---------------------------+ 
| 1.33                      |
+---------------------------+ 
User 1 和 2 在过去30天内各自进行了1次会话,而用户3进行了2次会话,因此平均值为(1 +1 + 2)/ 3 = 1.33。

sql

SELECT IFNULL(ROUND(COUNT(DISTINCT session_id) / COUNT(DISTINCT user_id), 2), 0) AS average_sessions_per_user
from Activity
where activity_date > date_add('2019-07-27',INTERVAL -1 MONTH)

1个session id 代表一次会话

1148. 文章浏览 I

难度简单3收藏分享切换为英文关注反馈

SQL架构

Views 表:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| article_id    | int     |
| author_id     | int     |
| viewer_id     | int     |
| view_date     | date    |
+---------------+---------+
此表无主键,因此可能会存在重复行。
此表的每一行都表示某人在某天浏览了某位作者的某篇文章。
请注意,同一人的 author_id 和 viewer_id 是相同的。

请编写一条 SQL 查询以找出所有浏览过自己文章的作者,结果按照 id 升序排列。

查询结果的格式如下所示:

Code

Views 表:
+------------+-----------+-----------+------------+
| article_id | author_id | viewer_id | view_date  |
+------------+-----------+-----------+------------+
| 1          | 3         | 5         | 2019-08-01 |
| 1          | 3         | 6         | 2019-08-02 |
| 2          | 7         | 7         | 2019-08-01 |
| 2          | 7         | 6         | 2019-08-02 |
| 4          | 7         | 1         | 2019-07-22 |
| 3          | 4         | 4         | 2019-07-21 |
| 3          | 4         | 4         | 2019-07-21 |
+------------+-----------+-----------+------------+

结果表:
+------+
| id   |
+------+
| 4    |
| 7    |
+------+

sql

select distinct author_id id
from Views
where author_id=  viewer_id
order by id

1149. 文章浏览 II

难度中等4收藏分享切换为英文关注反馈

SQL架构

Table: Views

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| article_id    | int     |
| author_id     | int     |
| viewer_id     | int     |
| view_date     | date    |
+---------------+---------+
此表无主键,因此可能会存在重复行。此表的每一行都表示某人在某天浏览了某位作者的某篇文章。 请注意,同一人的 author_id 和 viewer_id 是相同的。

编写一条 SQL 查询来找出在同一天阅读至少两篇文章的人,结果按照 id 升序排序。

查询结果的格式如下:

Code

Views table:
+------------+-----------+-----------+------------+
| article_id | author_id | viewer_id | view_date  |
+------------+-----------+-----------+------------+
| 1          | 3         | 5         | 2019-08-01 |
| 3          | 4         | 5         | 2019-08-01 |
| 1          | 3         | 6         | 2019-08-02 |
| 2          | 7         | 7         | 2019-08-01 |
| 2          | 7         | 6         | 2019-08-02 |
| 4          | 7         | 1         | 2019-07-22 |
| 3          | 4         | 4         | 2019-07-21 |
| 3          | 4         | 4         | 2019-07-21 |
+------------+-----------+-----------+------------+

Result table:
+------+
| id   |
+------+
| 5    |
| 6    |
+------+

sql

SELECT DISTINCT viewer_id AS id
FROM Views
GROUP BY view_date, viewer_id
HAVING COUNT(DISTINCT article_id) >= 2
ORDER BY viewer_id

1158. 市场分析 I

难度中等

SQL架构

Table: Users

Code

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| user_id        | int     |
| join_date      | date    |
| favorite_brand | varchar |
+----------------+---------+
此表主键是 user_id,表中描述了购物网站的用户信息,用户可以在此网站上进行商品买卖。

Table: Orders

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| item_id       | int     |
| buyer_id      | int     |
| seller_id     | int     |
+---------------+---------+
此表主键是 order_id,外键是 item_id 和(buyer_id,seller_id)。

Table: Item

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| item_id       | int     |
| item_brand    | varchar |
+---------------+---------+
此表主键是 item_id。

请写出一条SQL语句以查询每个用户的注册日期和在 2019 年作为买家的订单总数。

查询结果格式如下:

Code

Users table:
+---------+------------+----------------+
| user_id | join_date  | favorite_brand |
+---------+------------+----------------+
| 1       | 2018-01-01 | Lenovo         |
| 2       | 2018-02-09 | Samsung        |
| 3       | 2018-01-19 | LG             |
| 4       | 2018-05-21 | HP             |
+---------+------------+----------------+

Orders table:
+----------+------------+---------+----------+-----------+
| order_id | order_date | item_id | buyer_id | seller_id |
+----------+------------+---------+----------+-----------+
| 1        | 2019-08-01 | 4       | 1        | 2         |
| 2        | 2018-08-02 | 2       | 1        | 3         |
| 3        | 2019-08-03 | 3       | 2        | 3         |
| 4        | 2018-08-04 | 1       | 4        | 2         |
| 5        | 2018-08-04 | 1       | 3        | 4         |
| 6        | 2019-08-05 | 2       | 2        | 4         |
+----------+------------+---------+----------+-----------+

Items table:
+---------+------------+
| item_id | item_brand |
+---------+------------+
| 1       | Samsung    |
| 2       | Lenovo     |
| 3       | LG         |
| 4       | HP         |
+---------+------------+

Result table:
+-----------+------------+----------------+
| buyer_id  | join_date  | orders_in_2019 |
+-----------+------------+----------------+
| 1         | 2018-01-01 | 1              |
| 2         | 2018-02-09 | 2              |
| 3         | 2018-01-19 | 0              |
| 4         | 2018-05-21 | 0              |
+-----------+------------+----------------+

sql

select user_id buyer_id,join_date, ifnull(cnt,0)orders_in_2019
from Users u left join 
(select buyer_id,count(*) cnt
from Orders
where year(order_date) = 2019
group by  buyer_id
)t1
on u.user_id =t1.buyer_id

1159. 市场分析 II

难度困难

SQL架构

表: Users

Code

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| user_id        | int     |
| join_date      | date    |
| favorite_brand | varchar |
+----------------+---------+
user_id 是该表的主键
表中包含一位在线购物网站用户的个人信息,用户可以在该网站出售和购买商品。

表: Orders

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| item_id       | int     |
| buyer_id      | int     |
| seller_id     | int     |
+---------------+---------+
order_id 是该表的主键
item_id 是 Items 表的外键
buyer_id 和 seller_id 是 Users 表的外键

表: Items

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| item_id       | int     |
| item_brand    | varchar |
+---------------+---------+
item_id 是该表的主键

写一个 SQL 查询确定每一个用户按日期顺序卖出的第二件商品的品牌是否是他们最喜爱的品牌。如果一个用户卖出少于两件商品,查询的结果是 no

题目保证没有一个用户在一天中卖出超过一件商品

下面是查询结果格式的例子:

Code

Users table:
+---------+------------+----------------+
| user_id | join_date  | favorite_brand |
+---------+------------+----------------+
| 1       | 2019-01-01 | Lenovo         |
| 2       | 2019-02-09 | Samsung        |
| 3       | 2019-01-19 | LG             |
| 4       | 2019-05-21 | HP             |
+---------+------------+----------------+

Orders table:
+----------+------------+---------+----------+-----------+
| order_id | order_date | item_id | buyer_id | seller_id |
+----------+------------+---------+----------+-----------+
| 1        | 2019-08-01 | 4       | 1        | 2         |
| 2        | 2019-08-02 | 2       | 1        | 3         |
| 3        | 2019-08-03 | 3       | 2        | 3         |
| 4        | 2019-08-04 | 1       | 4        | 2         |
| 5        | 2019-08-04 | 1       | 3        | 4         |
| 6        | 2019-08-05 | 2       | 2        | 4         |
+----------+------------+---------+----------+-----------+

Items table:
+---------+------------+
| item_id | item_brand |
+---------+------------+
| 1       | Samsung    |
| 2       | Lenovo     |
| 3       | LG         |
| 4       | HP         |
+---------+------------+

Result table:
+-----------+--------------------+
| seller_id | 2nd_item_fav_brand |
+-----------+--------------------+
| 1         | no                 |
| 2         | yes                |
| 3         | yes                |
| 4         | no                 |
+-----------+--------------------+

id 为 1 的用户的查询结果是 no,因为他什么也没有卖出
id为 2 和 3 的用户的查询结果是 yes,因为他们卖出的第二件商品的品牌是他们自己最喜爱的品牌
id为 4 的用户的查询结果是 no,因为他卖出的第二件商品的品牌不是他最喜爱的品牌

sql

select user_id seller_id, if(item_brand=favorite_brand,'yes','no') 2nd_item_fav_brand 
from Users u
left join 
(
    select i.item_id,seller_id,item_brand
    from Items i
    join 
    (
        select seller_id,item_id,rank() over(partition by seller_id  order by order_date ) rk
        from Orders 
    )t1 
    on i.item_id = t1.item_id 
    where rk  =2
)t2
on u.user_id = t2.seller_id

1164. 指定日期的产品价格

难度中等

SQL架构

产品数据表: Products

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| new_price     | int     |
| change_date   | date    |
+---------------+---------+
这张表的主键是 (product_id, change_date)。
这张表的每一行分别记录了 某产品 在某个日期 更改后 的新价格。

写一段 SQL来查找在 2019-08-16 时全部产品的价格,假设所有产品在修改前的价格都是 10。

查询结果格式如下例所示:

Code

Products table:
+------------+-----------+-------------+
| product_id | new_price | change_date |
+------------+-----------+-------------+
| 1          | 20        | 2019-08-14  |
| 2          | 50        | 2019-08-14  |
| 1          | 30        | 2019-08-15  |
| 1          | 35        | 2019-08-16  |
| 2          | 65        | 2019-08-17  |
| 3          | 20        | 2019-08-18  |
+------------+-----------+-------------+

Result table:
+------------+-------+
| product_id | price |
+------------+-------+
| 2          | 50    |
| 1          | 35    |
| 3          | 10    |
+------------+-------+

sql

select distinct p.product_id,ifnull(t1.new_price,10) price
from Products p
left join 
(
    select product_id,new_price
    from (
        select product_id,new_price,change_date,Max(change_date) over(partition by product_id  ) md
        from  Products
        where change_date<='2019-08-16'
    )tmp
    where change_date = md
)t1
on p. product_id = t1.product_id
order by price desc

1173. 即时食物配送 I

难度简单

SQL架构

配送表: Delivery

Code

+-----------------------------+---------+
| Column Name                 | Type    |
+-----------------------------+---------+
| delivery_id                 | int     |
| customer_id                 | int     |
| order_date                  | date    |
| customer_pref_delivery_date | date    |
+-----------------------------+---------+
delivery_id 是表的主键。
该表保存着顾客的食物配送信息,顾客在某个日期下了订单,并指定了一个期望的配送日期(和下单日期相同或者在那之后)。

如果顾客期望的配送日期和下单日期相同,则该订单称为 「即时订单」,否则称为「计划订单」。

写一条 SQL 查询语句获取即时订单所占的百分比, 保留两位小数。

查询结果如下所示:

Code

Delivery 表:
+-------------+-------------+------------+-----------------------------+
| delivery_id | customer_id | order_date | customer_pref_delivery_date |
+-------------+-------------+------------+-----------------------------+
| 1           | 1           | 2019-08-01 | 2019-08-02                  |
| 2           | 5           | 2019-08-02 | 2019-08-02                  |
| 3           | 1           | 2019-08-11 | 2019-08-11                  |
| 4           | 3           | 2019-08-24 | 2019-08-26                  |
| 5           | 4           | 2019-08-21 | 2019-08-22                  |
| 6           | 2           | 2019-08-11 | 2019-08-13                  |
+-------------+-------------+------------+-----------------------------+

Result 表:
+----------------------+
| immediate_percentage |
+----------------------+
| 33.33                |
+----------------------+
2 和 3 号订单为即时订单,其他的为计划订单。

sql

select round(sum(if(order_date=customer_pref_delivery_date,1,0))/count(*)*100,2) immediate_percentage
from Delivery

1174. 即时食物配送 II

难度中等

SQL架构

配送表: Delivery

Code

+-----------------------------+---------+
| Column Name                 | Type    |
+-----------------------------+---------+
| delivery_id                 | int     |
| customer_id                 | int     |
| order_date                  | date    |
| customer_pref_delivery_date | date    |
+-----------------------------+---------+
delivery_id 是表的主键。
该表保存着顾客的食物配送信息,顾客在某个日期下了订单,并指定了一个期望的配送日期(和下单日期相同或者在那之后)。

如果顾客期望的配送日期和下单日期相同,则该订单称为 「即时订单」,否则称为「计划订单」。

「首次订单」是顾客最早创建的订单。我们保证一个顾客只会有一个「首次订单」。

写一条 SQL 查询语句获取即时订单在所有用户的首次订单中的比例。保留两位小数。

查询结果如下所示:

Code

Delivery 表:
+-------------+-------------+------------+-----------------------------+
| delivery_id | customer_id | order_date | customer_pref_delivery_date |
+-------------+-------------+------------+-----------------------------+
| 1           | 1           | 2019-08-01 | 2019-08-02                  |
| 2           | 2           | 2019-08-02 | 2019-08-02                  |
| 3           | 1           | 2019-08-11 | 2019-08-12                  |
| 4           | 3           | 2019-08-24 | 2019-08-24                  |
| 5           | 3           | 2019-08-21 | 2019-08-22                  |
| 6           | 2           | 2019-08-11 | 2019-08-13                  |
| 7           | 4           | 2019-08-09 | 2019-08-09                  |
+-------------+-------------+------------+-----------------------------+

Result 表:
+----------------------+
| immediate_percentage |
+----------------------+
| 50.00                |
+----------------------+
1 号顾客的 1 号订单是首次订单,并且是计划订单。
2 号顾客的 2 号订单是首次订单,并且是即时订单。
3 号顾客的 5 号订单是首次订单,并且是计划订单。
4 号顾客的 7 号订单是首次订单,并且是即时订单。
因此,一半顾客的首次订单是即时的。

开窗

sql

select round(sum(if(order_date = fo && fo=d,1,0))/count(distinct customer_id)*100,2) immediate_percentage 
from 
(
select customer_id,order_date,min(order_date)over ( partition by customer_id) fo,
    if(order_date=customer_pref_delivery_date,order_date ,null) d
from Delivery
)t1

另一种思路

sql

select round (
    sum(order_date = customer_pref_delivery_date) * 100 /
    count(*),
    2
) as immediate_percentage
from Delivery
where (customer_id, order_date) in (
    select customer_id, min(order_date)
    from delivery
    group by customer_id
)

1179. 重新格式化部门表

难度

SQL架构

部门表 Department

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| revenue       | int     |
| month         | varchar |
+---------------+---------+
(id, month) 是表的联合主键。
这个表格有关于每个部门每月收入的信息。
月份(month)可以取下列值 ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]。

编写一个 SQL 查询来重新格式化表,使得新的表中有一个部门 id 列和一些对应 每个月 的收入(revenue)列。

查询结果格式如下面的示例所示:

Code

Department 表:
+------+---------+-------+
| id   | revenue | month |
+------+---------+-------+
| 1    | 8000    | Jan   |
| 2    | 9000    | Jan   |
| 3    | 10000   | Feb   |
| 1    | 7000    | Feb   |
| 1    | 6000    | Mar   |
+------+---------+-------+

查询得到的结果表:
+------+-------------+-------------+-------------+-----+-------------+
| id   | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
+------+-------------+-------------+-------------+-----+-------------+
| 1    | 8000        | 7000        | 6000        | ... | null        |
| 2    | 9000        | null        | null        | ... | null        |
| 3    | null        | 10000       | null        | ... | null        |
+------+-------------+-------------+-------------+-----+-------------+

注意,结果表有 13 列 (1个部门 id 列 + 12个月份的收入列)。

sql

SELECT id,
SUM(CASE `month` WHEN 'Jan' THEN revenue END) Jan_Revenue,
SUM(CASE `month` WHEN 'Feb' THEN revenue END) Feb_Revenue,
SUM(CASE `month` WHEN 'Mar' THEN revenue END) Mar_Revenue,
SUM(CASE `month` WHEN 'Apr' THEN revenue END) Apr_Revenue,
SUM(CASE `month` WHEN 'May' THEN revenue END) May_Revenue,
SUM(CASE `month` WHEN 'Jun' THEN revenue END) Jun_Revenue,
SUM(CASE `month` WHEN 'Jul' THEN revenue END) Jul_Revenue,
SUM(CASE `month` WHEN 'Aug' THEN revenue END) Aug_Revenue,
SUM(CASE `month` WHEN 'Sep' THEN revenue END) Sep_Revenue,
SUM(CASE `month` WHEN 'Oct' THEN revenue END) Oct_Revenue,
SUM(CASE `month` WHEN 'Nov' THEN revenue END) Nov_Revenue,
SUM(CASE `month` WHEN 'Dec' THEN revenue END) Dec_Revenue
FROM Department
GROUP BY id;

sql

1193. 每月交易 I
SQL架构
Table: Transactions

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| country       | varchar |
| state         | enum    |
| amount        | int     |
| trans_date    | date    |
+---------------+---------+
id 是这个表的主键。
该表包含有关传入事务的信息。
state 列类型为 “[”批准“,”拒绝“] 之一。
 

编写一个 sql 查询来查找每个月和每个国家/地区的事务数及其总金额、已批准的事务数及其总金额。

查询结果格式如下所示:

Transactions table:
+------+---------+----------+--------+------------+
| id   | country | state    | amount | trans_date |
+------+---------+----------+--------+------------+
| 121  | US      | approved | 1000   | 2018-12-18 |
| 122  | US      | declined | 2000   | 2018-12-19 |
| 123  | US      | approved | 2000   | 2019-01-01 |
| 124  | DE      | approved | 2000   | 2019-01-07 |
+------+---------+----------+--------+------------+

Result table:
+----------+---------+-------------+----------------+--------------------+-----------------------+
| month    | country | trans_count | approved_count | trans_total_amount | approved_total_amount |
+----------+---------+-------------+----------------+--------------------+-----------------------+
| 2018-12  | US      | 2           | 1              | 3000               | 1000                  |
| 2019-01  | US      | 1           | 1              | 2000               | 2000                  |
| 2019-01  | DE      | 1           | 1              | 2000               | 2000                  |
+----------+---------+-------------+----------------+--------------------+-----------------------+

sql

SELECT DATE_FORMAT(trans_date, '%Y-%m') AS month,
    country,
    COUNT(*) AS trans_count,
    COUNT(IF(state = 'approved', 1, NULL)) AS approved_count,
    SUM(amount) AS trans_total_amount,
    SUM(IF(state = 'approved', amount, 0)) AS approved_total_amount
FROM Transactions
GROUP BY month, country

1193. 每月交易 I

难度中等

SQL架构

Table: Transactions

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| country       | varchar |
| state         | enum    |
| amount        | int     |
| trans_date    | date    |
+---------------+---------+
id 是这个表的主键。
该表包含有关传入事务的信息。
state 列类型为 “[”批准“,”拒绝“] 之一。

编写一个 sql 查询来查找每个月和每个国家/地区的事务数及其总金额、已批准的事务数及其总金额。

查询结果格式如下所示:

Code

Transactions table:
+------+---------+----------+--------+------------+
| id   | country | state    | amount | trans_date |
+------+---------+----------+--------+------------+
| 121  | US      | approved | 1000   | 2018-12-18 |
| 122  | US      | declined | 2000   | 2018-12-19 |
| 123  | US      | approved | 2000   | 2019-01-01 |
| 124  | DE      | approved | 2000   | 2019-01-07 |
+------+---------+----------+--------+------------+

Result table:
+----------+---------+-------------+----------------+--------------------+-----------------------+
| month    | country | trans_count | approved_count | trans_total_amount | approved_total_amount |
+----------+---------+-------------+----------------+--------------------+-----------------------+
| 2018-12  | US      | 2           | 1              | 3000               | 1000                  |
| 2019-01  | US      | 1           | 1              | 2000               | 2000                  |
| 2019-01  | DE      | 1           | 1              | 2000               | 2000                  |
+----------+---------+-------------+----------------+--------------------+-----------------------+

sql

SELECT DATE_FORMAT(trans_date, '%Y-%m') AS month,
    country,
    COUNT(*) AS trans_count,
    COUNT(IF(state = 'approved', 1, NULL)) AS approved_count,
    SUM(amount) AS trans_total_amount,
    SUM(IF(state = 'approved', amount, 0)) AS approved_total_amount
FROM Transactions
GROUP BY month, country

1205. 每月交易II

难度中等

SQL架构

Transactions 记录表

Code

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| id             | int     |
| country        | varchar |
| state          | enum    |
| amount         | int     |
| trans_date     | date    |
+----------------+---------+
id 是这个表的主键。
该表包含有关传入事务的信息。
状态列是类型为 [approved(已批准)、declined(已拒绝)] 的枚举。

Chargebacks

Code

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| trans_id       | int     |
| charge_date    | date    |
+----------------+---------+
退单包含有关放置在事务表中的某些事务的传入退单的基本信息。
trans_id 是 transactions 表的 id 列的外键。
每项退单都对应于之前进行的交易,即使未经批准。

编写一个 SQL 查询,以查找每个月和每个国家/地区的已批准交易的数量及其总金额、退单的数量及其总金额。

注意:在您的查询中,给定月份和国家,忽略所有为零的行。

查询结果格式如下所示:

Code

Transactions 表:
+------+---------+----------+--------+------------+
| id   | country | state    | amount | trans_date |
+------+---------+----------+--------+------------+
| 101  | US      | approved | 1000   | 2019-05-18 |
| 102  | US      | declined | 2000   | 2019-05-19 |
| 103  | US      | approved | 3000   | 2019-06-10 |
| 104  | US      | declined | 4000   | 2019-06-13 |
| 105  | US      | approved | 5000   | 2019-06-15 |
+------+---------+----------+--------+------------+

Chargebacks 表:
+------------+------------+
| trans_id   | trans_date |
+------------+------------+
| 102        | 2019-05-29 |
| 101        | 2019-06-30 |
| 105        | 2019-09-18 |
+------------+------------+

Result 表:
+----------+---------+----------------+-----------------+-------------------+--------------------+
| month    | country | approved_count | approved_amount | chargeback_count  | chargeback_amount  |
+----------+---------+----------------+-----------------+-------------------+--------------------+
| 2019-05  | US      | 1              | 1000            | 1                 | 2000               |
| 2019-06  | US      | 2              | 8000            | 1                 | 1000               |
| 2019-09  | US      | 0              | 0               | 1                 | 5000               |
+----------+---------+----------------+-----------------+-------------------+--------------------+

sql

select 
    date_format(trans_date, '%Y-%m') month, 
    country,
    sum(state = 'approved') approved_count,
    sum(if(state = 'approved', amount, 0)) approved_amount,
    sum(state = 'chargeback') chargeback_count,
    sum(if(state = 'chargeback', amount, 0)) chargeback_amount
from (
    select * from transactions
    union all
    select id, country, 'chargeback' state, amount, c.trans_date
    from chargebacks c left join transactions t 
    on c.trans_id = t.id
) tmp 
group by month, country
having approved_amount or chargeback_amount

1194. 锦标赛优胜者

难度困难

SQL架构

Players 玩家表

Code

+-------------+-------+
| Column Name | Type  |
+-------------+-------+
| player_id   | int   |
| group_id    | int   |
+-------------+-------+
玩家 ID 是此表的主键。
此表的每一行表示每个玩家的组。

Matches 赛事表

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| match_id      | int     |
| first_player  | int     |
| second_player | int     | 
| first_score   | int     |
| second_score  | int     |
+---------------+---------+
match_id 是此表的主键。
每一行是一场比赛的记录,第一名和第二名球员包含每场比赛的球员 ID。
第一个玩家和第二个玩家的分数分别包含第一个玩家和第二个玩家的分数。
你可以假设,在每一场比赛中,球员都属于同一组。

每组的获胜者是在组内得分最高的选手。如果平局,player_id 最小 的选手获胜。

编写一个 SQL 查询来查找每组中的获胜者。

查询结果格式如下所示

Code

Players 表:
+-----------+------------+
| player_id | group_id   |
+-----------+------------+
| 15        | 1          |
| 25        | 1          |
| 30        | 1          |
| 45        | 1          |
| 10        | 2          |
| 35        | 2          |
| 50        | 2          |
| 20        | 3          |
| 40        | 3          |
+-----------+------------+

Matches 表:
+------------+--------------+---------------+-------------+--------------+
| match_id   | first_player | second_player | first_score | second_score |
+------------+--------------+---------------+-------------+--------------+
| 1          | 15           | 45            | 3           | 0            |
| 2          | 30           | 25            | 1           | 2            |
| 3          | 30           | 15            | 2           | 0            |
| 4          | 40           | 20            | 5           | 2            |
| 5          | 35           | 50            | 1           | 1            |
+------------+--------------+---------------+-------------+--------------+

Result 表:
+-----------+------------+
| group_id  | player_id  |
+-----------+------------+ 
| 1         | 15         |
| 2         | 35         |
| 3         | 40         |
+-----------+------------+

union all

sql

SELECT group_id, player_id
FROM (
    SELECT group_id, player_id, SUM(score) AS score
    FROM (
        -- 每个用户总的 first_score
        SELECT Players.group_id, Players.player_id, SUM(Matches.first_score) AS score
        FROM Players JOIN Matches ON Players.player_id = Matches.first_player
        GROUP BY Players.player_id

        UNION ALL

        -- 每个用户总的 second_score
        SELECT Players.group_id, Players.player_id, SUM(Matches.second_score) AS score
        FROM Players JOIN Matches ON Players.player_id = Matches.second_player
        GROUP BY Players.player_id
    ) s
    GROUP BY player_id
    ORDER BY score DESC, player_id
) result
GROUP BY group_id

sql

select group_id, player_id
from (
    select players.*, sum(if(player_id = first_player, first_score, second_score)) score
    from players join matches
    on player_id = first_player or player_id = second_player
    group by player_id
    order by score desc, player_id
) tmp
group by group_id

1204. 最后一个能进入电梯的人

难度中等

SQL架构

表: Queue

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| person_id   | int     |
| person_name | varchar |
| weight      | int     |
| turn        | int     |
+-------------+---------+
person_id 是这个表的主键。
该表展示了所有等待电梯的人的信息。
表中 person_id 和 turn 列将包含从 1 到 n 的所有数字,其中 n 是表中的行数。

电梯最大载重量为 1000

写一条 SQL 查询语句查找最后一个能进入电梯且不超过重量限制的 person_name 。题目确保队列中第一位的人可以进入电梯 。

查询结果如下所示 :

Code

Queue 表
+-----------+-------------------+--------+------+
| person_id | person_name       | weight | turn |
+-----------+-------------------+--------+------+
| 5         | George Washington | 250    | 1    |
| 3         | John Adams        | 350    | 2    |
| 6         | Thomas Jefferson  | 400    | 3    |
| 2         | Will Johnliams    | 200    | 4    |
| 4         | Thomas Jefferson  | 175    | 5    |
| 1         | James Elephant    | 500    | 6    |
+-----------+-------------------+--------+------+

Result 表
+-------------------+
| person_name       |
+-------------------+
| Thomas Jefferson  |
+-------------------+

为了简化,Queue 表按 turn 列由小到大排序。
上例中 George Washington(id 5), John Adams(id 3) 和 Thomas Jefferson(id 6) 将可以进入电梯,因为他们的体重和为 250 + 350 + 400 = 1000。
Thomas Jefferson(id 6) 是最后一个体重合适并进入电梯的人。

sql

select person_name
from(
    select person_name  ,sum(weight) over(order by turn) t
    from Queue
)t1
where t<=1000
order by t desc 
limit 1

1211. 查询结果的质量和占比

难度简单

SQL架构

查询表 Queries

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| query_name  | varchar |
| result      | varchar |
| position    | int     |
| rating      | int     |
+-------------+---------+
此表没有主键,并可能有重复的行。
此表包含了一些从数据库中收集的查询信息。
“位置”(position)列的值为 1 到 500 。
“评分”(rating)列的值为 1 到 5 。评分小于 3 的查询被定义为质量很差的查询。

将查询结果的质量 quality 定义为:

各查询结果的评分与其位置之间比率的平均值。

将劣质查询百分比 poor_query_percentage 为:

评分小于 3 的查询结果占全部查询结果的百分比。

编写一组 SQL 来查找每次查询的名称(query_name)、质量(quality) 和 劣质查询百分比(poor_query_percentage)。

质量(quality) 和劣质查询百分比(poor_query_percentage) 都应四舍五入到小数点后两位。

查询结果格式如下所示:

Code

Queries table:
+------------+-------------------+----------+--------+
| query_name | result            | position | rating |
+------------+-------------------+----------+--------+
| Dog        | Golden Retriever  | 1        | 5      |
| Dog        | German Shepherd   | 2        | 5      |
| Dog        | Mule              | 200      | 1      |
| Cat        | Shirazi           | 5        | 2      |
| Cat        | Siamese           | 3        | 3      |
| Cat        | Sphynx            | 7        | 4      |
+------------+-------------------+----------+--------+

Result table:
+------------+---------+-----------------------+
| query_name | quality | poor_query_percentage |
+------------+---------+-----------------------+
| Dog        | 2.50    | 33.33                 |
| Cat        | 0.66    | 33.33                 |
+------------+---------+-----------------------+

Dog 查询结果的质量为 ((5 / 1) + (5 / 2) + (1 / 200)) / 3 = 2.50
Dog 查询结果的劣质查询百分比为 (1 / 3) * 100 = 33.33

Cat 查询结果的质量为 ((2 / 5) + (3 / 3) + (4 / 7)) / 3 = 0.66
Cat 查询结果的劣质查询百分比为 (1 / 3) * 100 = 33.33

sql

select query_name,
    round(avg(rating/position),2) quality,
    round(sum(if(rating<3,1,0))/count(*)*100,2) poor_query_percentage
from Queries
group by query_name

1212. 查询球队积分

难度中等

SQL架构

Table: Teams

Code

+---------------+----------+
| Column Name   | Type     |
+---------------+----------+
| team_id       | int      |
| team_name     | varchar  |
+---------------+----------+
此表的主键是 team_id,表中的每一行都代表一支独立足球队。

Table: Matches

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| match_id      | int     |
| host_team     | int     |
| guest_team    | int     | 
| host_goals    | int     |
| guest_goals   | int     |
+---------------+---------+
此表的主键是 match_id,表中的每一行都代表一场已结束的比赛,比赛的主客队分别由它们自己的 id 表示,他们的进球由 host_goals 和 guest_goals 分别表示。

积分规则如下:

  • 赢一场得三分;
  • 平一场得一分;
  • 输一场不得分。

写出一条SQL语句以查询每个队的 team_idteam_namenum_points。结果根据 num_points 降序排序,如果有两队积分相同,那么这两队按 team_id 升序排序

查询结果格式如下:

Code

Teams table:
+-----------+--------------+
| team_id   | team_name    |
+-----------+--------------+
| 10        | Leetcode FC  |
| 20        | NewYork FC   |
| 30        | Atlanta FC   |
| 40        | Chicago FC   |
| 50        | Toronto FC   |
+-----------+--------------+

Matches table:
+------------+--------------+---------------+-------------+--------------+
| match_id   | host_team    | guest_team    | host_goals  | guest_goals  |
+------------+--------------+---------------+-------------+--------------+
| 1          | 10           | 20            | 3           | 0            |
| 2          | 30           | 10            | 2           | 2            |
| 3          | 10           | 50            | 5           | 1            |
| 4          | 20           | 30            | 1           | 0            |
| 5          | 50           | 30            | 1           | 0            |
+------------+--------------+---------------+-------------+--------------+

Result table:
+------------+--------------+---------------+
| team_id    | team_name    | num_points    |
+------------+--------------+---------------+
| 10         | Leetcode FC  | 7             |
| 20         | NewYork FC   | 3             |
| 50         | Toronto FC   | 3             |
| 30         | Atlanta FC   | 1             |
| 40         | Chicago FC   | 0             |
+------------+--------------+---------------+

sql

select team_id , team_name ,sum(
if(team_id = host_team && host_goals>guest_goals ,3,0)+
if(team_id = host_team && host_goals=guest_goals,1,0)+
if(team_id = guest_team && host_goals=guest_goals,1,0)+
if(team_id = guest_team && host_goals<guest_goals ,3,0)
)num_points
from Teams t left join Matches m
on t.team_id =m.host_team  or t.team_id =m.guest_team  
group by team_id ,team_name
order by num_points desc,team_id

1225. 报告系统状态的连续日期

难度困难

SQL架构

Table: Failed

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| fail_date    | date    |
+--------------+---------+
该表主键为 fail_date。
该表包含失败任务的天数.

Table: Succeeded

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| success_date | date    |
+--------------+---------+
该表主键为 success_date。
该表包含成功任务的天数.

系统 每天 运行一个任务。每个任务都独立于先前的任务。任务的状态可以是失败或是成功。

编写一个 SQL 查询 2019-01-012019-12-31 期间任务连续同状态 period_state 的起止日期(start_dateend_date)。即如果任务失败了,就是失败状态的起止日期,如果任务成功了,就是成功状态的起止日期。

最后结果按照起始日期 start_date 排序

查询结果样例如下所示:

Code

Failed table:
+-------------------+
| fail_date         |
+-------------------+
| 2018-12-28        |
| 2018-12-29        |
| 2019-01-04        |
| 2019-01-05        |
+-------------------+

Succeeded table:
+-------------------+
| success_date      |
+-------------------+
| 2018-12-30        |
| 2018-12-31        |
| 2019-01-01        |
| 2019-01-02        |
| 2019-01-03        |
| 2019-01-06        |
+-------------------+


Result table:
+--------------+--------------+--------------+
| period_state | start_date   | end_date     |
+--------------+--------------+--------------+
| succeeded    | 2019-01-01   | 2019-01-03   |
| failed       | 2019-01-04   | 2019-01-05   |
| succeeded    | 2019-01-06   | 2019-01-06   |
+--------------+--------------+--------------+

结果忽略了 2018 年的记录,因为我们只关心从 2019-01-01 到 2019-12-31 的记录
从 2019-01-01 到 2019-01-03 所有任务成功,系统状态为 "succeeded"。
从 2019-01-04 到 2019-01-05 所有任务失败,系统状态为 "failed"。
从 2019-01-06 到 2019-01-06 所有任务成功,系统状态为 "succeeded"。

开窗函数

sql

select type  period_state, min(date) start_date, max(date) as end_date
from
(
    select type, date, subdate(date,row_number()over(partition by type order by date)) as diff
    from
    (
        select 'failed' as type, fail_date as date from Failed
        where fail_date between '2019-01-01' and '2019-12-31'
        union all
        select 'succeeded' as type, success_date as date from Succeeded
        where success_date between '2019-01-01' and '2019-12-31'
    ) a
)b
group by type,diff
order by start_date

1241. 每个帖子的评论数

难度简单

SQL架构

Submissions 结构如下:

Code

+---------------+----------+
| 列名           | 类型     |
+---------------+----------+
| sub_id        | int      |
| parent_id     | int      |
+---------------+----------+
上表没有主键, 所以可能会出现重复的行。
每行可以是一个帖子或对该帖子的评论。
如果是帖子的话,parent_id 就是 null。
对于评论来说,parent_id 就是表中对应帖子的 sub_id。

编写 SQL 语句以查找每个帖子的评论数。

结果表应包含帖子的 post_id 和对应的评论数 number_of_comments 并且按 post_id 升序排列。

Submissions 可能包含重复的评论。您应该计算每个帖子的唯一评论数。

Submissions 可能包含重复的帖子。您应该将它们视为一个帖子。

查询结果格式如下例所示:

Code

Submissions table:
+---------+------------+
| sub_id  | parent_id  |
+---------+------------+
| 1       | Null       |
| 2       | Null       |
| 1       | Null       |
| 12      | Null       |
| 3       | 1          |
| 5       | 2          |
| 3       | 1          |
| 4       | 1          |
| 9       | 1          |
| 10      | 2          |
| 6       | 7          |
+---------+------------+

结果表:
+---------+--------------------+
| post_id | number_of_comments |
+---------+--------------------+
| 1       | 3                  |
| 2       | 2                  |
| 12      | 0                  |
+---------+--------------------+

表中 ID 为 1 的帖子有 ID 为 3、4 和 9 的三个评论。表中 ID 为 3 的评论重复出现了,所以我们只对它进行了一次计数。
表中 ID 为 2 的帖子有 ID 为 5 和 10 的两个评论。
ID 为 12 的帖子在表中没有评论。
表中 ID 为 6 的评论是对 ID 为 7 的已删除帖子的评论,因此我们将其忽略。

后join

sql

select s.sub_id post_id,ifnull(number_of_comments,0) number_of_comments
from Submissions s
left join (
  select parent_id ,count(distinct sub_id) number_of_comments
  from Submissions 
  group by parent_id
) t1
on s.sub_id = t1.parent_id
where s.parent_id  is null
group by post_id,number_of_comments
order by sub_id

先join

sql

SELECT post_id, COUNT(sub_id) AS number_of_comments
FROM (
    SELECT DISTINCT post.sub_id AS post_id, sub.sub_id AS sub_id
    FROM Submissions post
    LEFT JOIN Submissions sub
    ON post.sub_id = sub.parent_id
    WHERE post.parent_id is null
) T
GROUP BY post_id
ORDER BY post_id ASC

1251. 平均售价

难度简单

SQL架构

Table: Prices

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| start_date    | date    |
| end_date      | date    |
| price         | int     |
+---------------+---------+
(product_id,start_date,end_date) 是 Prices 表的主键。
Prices 表的每一行表示的是某个产品在一段时期内的价格。
每个产品的对应时间段是不会重叠的,这也意味着同一个产品的价格时段不会出现交叉。

Table: UnitsSold

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| purchase_date | date    |
| units         | int     |
+---------------+---------+
UnitsSold 表没有主键,它可能包含重复项。
UnitsSold 表的每一行表示的是每种产品的出售日期,单位和产品 id。

编写SQL查询以查找每种产品的平均售价。 average_price 应该四舍五入到小数点后两位。 查询结果格式如下例所示:

Code

Prices table:
+------------+------------+------------+--------+
| product_id | start_date | end_date   | price  |
+------------+------------+------------+--------+
| 1          | 2019-02-17 | 2019-02-28 | 5      |
| 1          | 2019-03-01 | 2019-03-22 | 20     |
| 2          | 2019-02-01 | 2019-02-20 | 15     |
| 2          | 2019-02-21 | 2019-03-31 | 30     |
+------------+------------+------------+--------+
 
UnitsSold table:
+------------+---------------+-------+
| product_id | purchase_date | units |
+------------+---------------+-------+
| 1          | 2019-02-25    | 100   |
| 1          | 2019-03-01    | 15    |
| 2          | 2019-02-10    | 200   |
| 2          | 2019-03-22    | 30    |
+------------+---------------+-------+

Result table:
+------------+---------------+
| product_id | average_price |
+------------+---------------+
| 1          | 6.96          |
| 2          | 16.96         |
+------------+---------------+
平均售价 = 产品总价 / 销售的产品数量。
产品 1 的平均售价 = ((100 * 5)+(15 * 20) )/ 115 = 6.96
产品 2 的平均售价 = ((200 * 15)+(30 * 30) )/ 230 = 16.96

sql

SELECT
    product_id,
    Round(SUM(sales) / SUM(units), 2) AS average_price
FROM (
    SELECT
        Prices.product_id AS product_id,
        Prices.price * UnitsSold.units AS sales,
        UnitsSold.units AS units
    FROM Prices 
    JOIN UnitsSold ON Prices.product_id = UnitsSold.product_id
    WHERE UnitsSold.purchase_date BETWEEN Prices.start_date AND Prices.end_date
) T
GROUP BY product_id

2表关联的时候直接把日期做过滤

1264. 页面推荐

难度中等

SQL架构

朋友关系列表: Friendship

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user1_id      | int     |
| user2_id      | int     |
+---------------+---------+
这张表的主键是 (user1_id, user2_id)。
这张表的每一行代表着 user1_id 和 user2_id 之间存在着朋友关系。

喜欢列表: Likes

Code

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| user_id     | int     |
| page_id     | int     |
+-------------+---------+
这张表的主键是 (user_id, page_id)。
这张表的每一行代表着 user_id 喜欢 page_id。

写一段 SQL 向user_id = 1 的用户,推荐其朋友们喜欢的页面。不要推荐该用户已经喜欢的页面。

你返回的结果中不应当包含重复项。

返回结果的格式如下例所示:

Code

Friendship table:
+----------+----------+
| user1_id | user2_id |
+----------+----------+
| 1        | 2        |
| 1        | 3        |
| 1        | 4        |
| 2        | 3        |
| 2        | 4        |
| 2        | 5        |
| 6        | 1        |
+----------+----------+
 
Likes table:
+---------+---------+
| user_id | page_id |
+---------+---------+
| 1       | 88      |
| 2       | 23      |
| 3       | 24      |
| 4       | 56      |
| 5       | 11      |
| 6       | 33      |
| 2       | 77      |
| 3       | 77      |
| 6       | 88      |
+---------+---------+

Result table:
+------------------+
| recommended_page |
+------------------+
| 23               |
| 24               |
| 56               |
| 33               |
| 77               |
+------------------+
用户1 同 用户2, 3, 4, 6 是朋友关系。
推荐页面为: 页面23 来自于 用户2, 页面24 来自于 用户3, 页面56 来自于 用户3 以及 页面33 来自于 用户6。
页面77 同时被 用户2 和 用户3 推荐。
页面88 没有被推荐,因为 用户1 已经喜欢了它。

sql

select distinct page_id  recommended_page
from Likes
where user_id in(
select if(user1_id=1,user2_id,user1_id) user_id
from  Friendship
where  user1_id =1 or user2_id =1
)
and page_id not in (select page_id from Likes where user_id=1)

1270. 向公司CEO汇报工作的所有人

难度中等

SQL架构

员工表:Employees

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| employee_id   | int     |
| employee_name | varchar |
| manager_id    | int     |
+---------------+---------+
employee_id 是这个表的主键。
这个表中每一行中,employee_id 表示职工的 ID,employee_name 表示职工的名字,manager_id 表示该职工汇报工作的直线经理。
这个公司 CEO 是 employee_id = 1 的人。

用 SQL 查询出所有直接或间接向公司 CEO 汇报工作的职工的 employee_id 。

由于公司规模较小,经理之间的间接关系不超过 3 个经理。

可以以任何顺序返回的结果,不需要去重。

查询结果示例如下:

Code

Employees table:
+-------------+---------------+------------+
| employee_id | employee_name | manager_id |
+-------------+---------------+------------+
| 1           | Boss          | 1          |
| 3           | Alice         | 3          |
| 2           | Bob           | 1          |
| 4           | Daniel        | 2          |
| 7           | Luis          | 4          |
| 8           | Jhon          | 3          |
| 9           | Angela        | 8          |
| 77          | Robert        | 1          |
+-------------+---------------+------------+

Result table:
+-------------+
| employee_id |
+-------------+
| 2           |
| 77          |
| 4           |
| 7           |
+-------------+

公司 CEO 的 employee_id 是 1.
employee_id 是 2 和 77 的职员直接汇报给公司 CEO。
employee_id 是 4 的职员间接汇报给公司 CEO 4 --> 2 --> 1 。
employee_id 是 7 的职员间接汇报给公司 CEO 7 --> 4 --> 2 --> 1 。
employee_id 是 3, 8 ,9 的职员不会直接或间接的汇报给公司 CEO。

sql

select employee_id
from 
(
select a.employee_id
from Employees a 
left join Employees b on a.manager_id = b.employee_id
left join Employees c on b.manager_id = c.employee_id
where a.manager_id=1 or b.manager_id=1  or c.manager_id=1
)t1
where employee_id!=1

1280. 学生们参加各科测试的次数

难度简单

SQL架构

学生表: Students

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| student_id    | int     |
| student_name  | varchar |
+---------------+---------+
主键为 student_id(学生ID),该表内的每一行都记录有学校一名学生的信息。

科目表: Subjects

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| subject_name | varchar |
+--------------+---------+
主键为 subject_name(科目名称),每一行记录学校的一门科目名称。

考试表: Examinations

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| student_id   | int     |
| subject_name | varchar |
+--------------+---------+
这张表压根没有主键,可能会有重复行。
学生表里的一个学生修读科目表里的每一门科目,而这张考试表的每一行记录就表示学生表里的某个学生参加了一次科目表里某门科目的测试。

sql

SELECT a.student_id, a.student_name, b.subject_name, COUNT(e.subject_name) AS attended_exams
FROM Students a CROSS JOIN Subjects b
LEFT JOIN Examinations e ON a.student_id = e.student_id AND b.subject_name = e.subject_name
GROUP BY a.student_id, b.subject_name
ORDER BY a.student_id, b.subject_name

CROSS JOIN Mysql中没有full outer join hive中可以用

1285. 找到连续区间的开始和结束数字

难度中等

SQL架构

表:Logs

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| log_id        | int     |
+---------------+---------+
id 是上表的主键。
上表的每一行包含日志表中的一个 ID。

后来一些 ID 从 Logs 表中删除。编写一个 SQL 查询得到 Logs 表中的连续区间的开始数字和结束数字。

将查询表按照 start_id 排序。

查询结果格式如下面的例子:

Code

Logs 表:
+------------+
| log_id     |
+------------+
| 1          |
| 2          |
| 3          |
| 7          |
| 8          |
| 10         |
+------------+

结果表:
+------------+--------------+
| start_id   | end_id       |
+------------+--------------+
| 1          | 3            |
| 7          | 8            |
| 10         | 10           |
+------------+--------------+
结果表应包含 Logs 表中的所有区间。
从 1 到 3 在表中。
从 4 到 6 不在表中。
从 7 到 8 在表中。
9 不在表中。
10 在表中。

sql

SELECT
    MIN(log_id) start_id,
    MAX(log_id) end_id
FROM
    (SELECT
        log_id, 
        log_id - row_number() OVER(ORDER BY log_id) as diff
    FROM Logs) t
GROUP BY diff

相似1225题

1294. 不同国家的天气类型

难度简单

SQL架构

国家表:Countries

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| country_id    | int     |
| country_name  | varchar |
+---------------+---------+
country_id 是这张表的主键。
该表的每行有 country_id 和 country_name 两列。

天气表:Weather

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| country_id    | int     |
| weather_state | varchar |
| day           | date    |
+---------------+---------+
(country_id, day) 是该表的复合主键。
该表的每一行记录了某个国家某一天的天气情况。

写一段 SQL 来找到表中每个国家在 2019 年 11 月的天气类型。

天气类型的定义如下:当 weather_state 的平均值小于或等于15返回 Cold,当 weather_state 的平均值大于或等于 25 返回 Hot,否则返回 Warm

你可以以任意顺序返回你的查询结果。

查询结果格式如下所示:

Code

Countries table:
+------------+--------------+
| country_id | country_name |
+------------+--------------+
| 2          | USA          |
| 3          | Australia    |
| 7          | Peru         |
| 5          | China        |
| 8          | Morocco      |
| 9          | Spain        |
+------------+--------------+
Weather table:
+------------+---------------+------------+
| country_id | weather_state | day        |
+------------+---------------+------------+
| 2          | 15            | 2019-11-01 |
| 2          | 12            | 2019-10-28 |
| 2          | 12            | 2019-10-27 |
| 3          | -2            | 2019-11-10 |
| 3          | 0             | 2019-11-11 |
| 3          | 3             | 2019-11-12 |
| 5          | 16            | 2019-11-07 |
| 5          | 18            | 2019-11-09 |
| 5          | 21            | 2019-11-23 |
| 7          | 25            | 2019-11-28 |
| 7          | 22            | 2019-12-01 |
| 7          | 20            | 2019-12-02 |
| 8          | 25            | 2019-11-05 |
| 8          | 27            | 2019-11-15 |
| 8          | 31            | 2019-11-25 |
| 9          | 7             | 2019-10-23 |
| 9          | 3             | 2019-12-23 |
+------------+---------------+------------+
Result table:
+--------------+--------------+
| country_name | weather_type |
+--------------+--------------+
| USA          | Cold         |
| Austraila    | Cold         |
| Peru         | Hot          |
| China        | Warm         |
| Morocco      | Hot          |
+--------------+--------------+
USA 11 月的平均 weather_state 为 (15) / 1 = 15 所以天气类型为 Cold。
Australia 11 月的平均 weather_state 为 (-2 + 0 + 3) / 3 = 0.333 所以天气类型为 Cold。
Peru 11 月的平均 weather_state 为 (25) / 1 = 25 所以天气类型为 Hot。
China 11 月的平均 weather_state 为 (16 + 18 + 21) / 3 = 18.333 所以天气类型为 Warm。
Morocco 11 月的平均 weather_state 为 (25 + 27 + 31) / 3 = 27.667 所以天气类型为 Hot。
我们并不知道 Spain 在 11 月的 weather_state 情况所以无需将他包含在结果中。

sql

select country_name,( case when avg(weather_state)<=15 then 'Cold'
                        when avg(weather_state)>=25 then 'Hot'
                        else 'Warm' end ) weather_type
from Countries c join Weather w on c.country_id = w.country_id
where date_format(day,"%Y-%m")='2019-11'
group by country_name

1303. 求团队人数

难度简单

SQL架构

员工表:Employee

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| employee_id   | int     |
| team_id       | int     |
+---------------+---------+
employee_id 字段是这张表的主键,表中的每一行都包含每个员工的 ID 和他们所属的团队。

编写一个 SQL 查询,以求得每个员工所在团队的总人数。

查询结果中的顺序无特定要求。

查询结果格式示例如下:

Code

Employee Table:
+-------------+------------+
| employee_id | team_id    |
+-------------+------------+
|     1       |     8      |
|     2       |     8      |
|     3       |     8      |
|     4       |     7      |
|     5       |     9      |
|     6       |     9      |
+-------------+------------+
Result table:
+-------------+------------+
| employee_id | team_size  |
+-------------+------------+
|     1       |     3      |
|     2       |     3      |
|     3       |     3      |
|     4       |     1      |
|     5       |     2      |
|     6       |     2      |
+-------------+------------+
ID 为 1、2、3 的员工是 team_id 为 8 的团队的成员,
ID 为 4 的员工是 team_id 为 7 的团队的成员,
ID 为 5、6 的员工是 team_id 为 9 的团队的成员。

sql

select employee_id,count(*) over(partition by team_id) team_size 
from Employee

1308. 不同性别每日分数总计

难度中等

SQL架构

表: Scores

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| player_name   | varchar |
| gender        | varchar |
| day           | date    |
| score_points  | int     |
+---------------+---------+
(gender, day)是该表的主键
一场比赛是在女队和男队之间举行的
该表的每一行表示一个名叫 (player_name) 性别为 (gender) 的参赛者在某一天获得了 (score_points) 的分数
如果参赛者是女性,那么 gender 列为 'F',如果参赛者是男性,那么 gender 列为 'M'

写一条SQL语句查询每种性别在每一天的总分,并按性别和日期对查询结果排序

下面是查询结果格式的例子:

Code

Scores表:
+-------------+--------+------------+--------------+
| player_name | gender | day        | score_points |
+-------------+--------+------------+--------------+
| Aron        | F      | 2020-01-01 | 17           |
| Alice       | F      | 2020-01-07 | 23           |
| Bajrang     | M      | 2020-01-07 | 7            |
| Khali       | M      | 2019-12-25 | 11           |
| Slaman      | M      | 2019-12-30 | 13           |
| Joe         | M      | 2019-12-31 | 3            |
| Jose        | M      | 2019-12-18 | 2            |
| Priya       | F      | 2019-12-31 | 23           |
| Priyanka    | F      | 2019-12-30 | 17           |
+-------------+--------+------------+--------------+
结果表:
+--------+------------+-------+
| gender | day        | total |
+--------+------------+-------+
| F      | 2019-12-30 | 17    |
| F      | 2019-12-31 | 40    |
| F      | 2020-01-01 | 57    |
| F      | 2020-01-07 | 80    |
| M      | 2019-12-18 | 2     |
| M      | 2019-12-25 | 13    |
| M      | 2019-12-30 | 26    |
| M      | 2019-12-31 | 29    |
| M      | 2020-01-07 | 36    |
+--------+------------+-------+
女性队伍:
第一天是 2019-12-30,Priyanka 获得 17 分,队伍的总分是 17 分
第二天是 2019-12-31, Priya 获得 23 分,队伍的总分是 40 分
第三天是 2020-01-01, Aron 获得 17 分,队伍的总分是 57 分
第四天是 2020-01-07, Alice 获得 23 分,队伍的总分是 80 分
男性队伍:
第一天是 2019-12-18, Jose 获得 2 分,队伍的总分是 2 分
第二天是 2019-12-25, Khali 获得 11 分,队伍的总分是 13 分
第三天是 2019-12-30, Slaman 获得 13 分,队伍的总分是 26 分
第四天是 2019-12-31, Joe 获得 3 分,队伍的总分是 29 分
第五天是 2020-01-07, Bajrang 获得 7 分,队伍的总分是 36 分

sql

select gender , day  ,sum( score_points) over(partition by gender order by day)  total
from Scores

1321. 餐馆营业额变化增长

难度中等

SQL架构

表: Customer

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
| visited_on    | date    |
| amount        | int     |
+---------------+---------+
(customer_id, visited_on) 是该表的主键
该表包含一家餐馆的顾客交易数据
visited_on 表示 (customer_id) 的顾客在 visited_on 那天访问了餐馆
amount 是一个顾客某一天的消费总额

你是餐馆的老板,现在你想分析一下可能的营业额变化增长(每天至少有一位顾客)

写一条 SQL 查询计算以 7 天(某日期 + 该日期前的 6 天)为一个时间段的顾客消费平均值

查询结果格式的例子如下:

  • 查询结果按 visited_on 排序
  • average_amount保留两位小数,日期数据的格式为 (‘YYYY-MM-DD’)

Code

Customer 表:
+-------------+--------------+--------------+-------------+
| customer_id | name         | visited_on   | amount      |
+-------------+--------------+--------------+-------------+
| 1           | Jhon         | 2019-01-01   | 100         |
| 2           | Daniel       | 2019-01-02   | 110         |
| 3           | Jade         | 2019-01-03   | 120         |
| 4           | Khaled       | 2019-01-04   | 130         |
| 5           | Winston      | 2019-01-05   | 110         | 
| 6           | Elvis        | 2019-01-06   | 140         | 
| 7           | Anna         | 2019-01-07   | 150         |
| 8           | Maria        | 2019-01-08   | 80          |
| 9           | Jaze         | 2019-01-09   | 110         | 
| 1           | Jhon         | 2019-01-10   | 130         | 
| 3           | Jade         | 2019-01-10   | 150         | 
+-------------+--------------+--------------+-------------+

结果表:
+--------------+--------------+----------------+
| visited_on   | amount       | average_amount |
+--------------+--------------+----------------+
| 2019-01-07   | 860          | 122.86         |
| 2019-01-08   | 840          | 120            |
| 2019-01-09   | 840          | 120            |
| 2019-01-10   | 1000         | 142.86         |
+--------------+--------------+----------------+

第一个七天消费平均值从 2019-01-01 到 2019-01-07 是 (100 + 110 + 120 + 130 + 110 + 140 + 150)/7 = 122.86
第二个七天消费平均值从 2019-01-02 到 2019-01-08 是 (110 + 120 + 130 + 110 + 140 + 150 + 80)/7 = 120
第三个七天消费平均值从 2019-01-03 到 2019-01-09 是 (120 + 130 + 110 + 140 + 150 + 80 + 110)/7 = 120
第四个七天消费平均值从 2019-01-04 到 2019-01-10 是 (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150)/7 = 142.86

sql

select  visited_on,amount,round(amount/7,2) average_amount
from (
    select visited_on,ant,lag(visited_on,6,null) over(order by visited_on) lg,
            sum(ant) over(order by visited_on rows between 6 PRECEDING and current row) amount
    from(
        select  visited_on  ,sum(amount) ant
        from Customer
        group by visited_on
    )t1
)t2
where lg is not null

1322. 广告效果

难度简单8收藏分享切换为英文关注反馈

SQL架构

表: Ads

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| ad_id         | int     |
| user_id       | int     |
| action        | enum    |
+---------------+---------+
(ad_id, user_id) 是该表的主键
该表的每一行包含一条广告的 ID(ad_id),用户的 ID(user_id) 和用户对广告采取的行为 (action)
action 列是一个枚举类型 ('Clicked', 'Viewed', 'Ignored') 。

一家公司正在运营这些广告并想计算每条广告的效果。

广告效果用点击通过率(Click-Through Rate:CTR)来衡量,公式如下:

写一条SQL语句来查询每一条广告的 ctr

ctr 要保留两位小数。结果需要按 ctr 降序、按 ad_id 升序 进行排序。

查询结果示例如下:

Code

Ads 表:
+-------+---------+---------+
| ad_id | user_id | action  |
+-------+---------+---------+
| 1     | 1       | Clicked |
| 2     | 2       | Clicked |
| 3     | 3       | Viewed  |
| 5     | 5       | Ignored |
| 1     | 7       | Ignored |
| 2     | 7       | Viewed  |
| 3     | 5       | Clicked |
| 1     | 4       | Viewed  |
| 2     | 11      | Viewed  |
| 1     | 2       | Clicked |
+-------+---------+---------+
结果表:
+-------+-------+
| ad_id | ctr   |
+-------+-------+
| 1     | 66.67 |
| 3     | 50.00 |
| 2     | 33.33 |
| 5     | 0.00  |
+-------+-------+
对于 ad_id = 1, ctr = (2/(2+1)) * 100 = 66.67
对于 ad_id = 2, ctr = (1/(1+2)) * 100 = 33.33
对于 ad_id = 3, ctr = (1/(1+1)) * 100 = 50.00
对于 ad_id = 5, ctr = 0.00, 注意 ad_id = 5 没有被点击 (Clicked) 或查看 (Viewed) 过
注意我们不关心 action 为 Ingnored 的广告
结果按 ctr(降序),ad_id(升序)排序

精简

sql

SELECT ad_id,
    ROUND(IFNULL(SUM(action = 'Clicked') /
        (SUM(action = 'Clicked') + SUM(action = 'Viewed')) * 100, 0), 2) AS ctr
FROM Ads
GROUP BY ad_id
ORDER BY ctr DESC, ad_id ASC;

笨方法

sql

select a.ad_id,ifnull(ctr,0) ctr
from Ads a
left join (
    select ad_id,round(sum(if(action='Clicked',1,0))/count(*)*100,2) ctr 
    from Ads 
    where action !='Ignored'
    group by ad_id
)t1 
on a.ad_id= t1.ad_id
group by ad_id,ctr
order by ctr desc,ad_id

1327. 列出指定时间段内所有的下单产品

难度简单

SQL架构

表: Products

Code

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| product_id       | int     |
| product_name     | varchar |
| product_category | varchar |
+------------------+---------+
product_id 是该表主键。
该表包含该公司产品的数据。

表: Orders

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| order_date    | date    |
| unit          | int     |
+---------------+---------+
该表无主键,可能包含重复行。
product_id 是表单 Products 的外键。
unit 是在日期 order_date 内下单产品的数目。

写一个 SQL 语句,要求获取在 2020 年 2 月份下单的数量不少于 100 的产品的名字和数目。

返回结果表单的顺序无要求。

查询结果的格式如下:

Code

Products 表:
+-------------+-----------------------+------------------+
| product_id  | product_name          | product_category |
+-------------+-----------------------+------------------+
| 1           | Leetcode Solutions    | Book             |
| 2           | Jewels of Stringology | Book             |
| 3           | HP                    | Laptop           |
| 4           | Lenovo                | Laptop           |
| 5           | Leetcode Kit          | T-shirt          |
+-------------+-----------------------+------------------+

Orders 表:
+--------------+--------------+----------+
| product_id   | order_date   | unit     |
+--------------+--------------+----------+
| 1            | 2020-02-05   | 60       |
| 1            | 2020-02-10   | 70       |
| 2            | 2020-01-18   | 30       |
| 2            | 2020-02-11   | 80       |
| 3            | 2020-02-17   | 2        |
| 3            | 2020-02-24   | 3        |
| 4            | 2020-03-01   | 20       |
| 4            | 2020-03-04   | 30       |
| 4            | 2020-03-04   | 60       |
| 5            | 2020-02-25   | 50       |
| 5            | 2020-02-27   | 50       |
| 5            | 2020-03-01   | 50       |
+--------------+--------------+----------+

Result 表:
+--------------------+---------+
| product_name       | unit    |
+--------------------+---------+
| Leetcode Solutions | 130     |
| Leetcode Kit       | 100     |
+--------------------+---------+

2020 年 2 月份下单 product_id = 1 的产品的数目总和为 (60 + 70) = 130 。
2020 年 2 月份下单 product_id = 2 的产品的数目总和为 80 。
2020 年 2 月份下单 product_id = 3 的产品的数目总和为 (2 + 3) = 5 。
2020 年 2 月份 product_id = 4 的产品并没有下单。
2020 年 2 月份下单 product_id = 5 的产品的数目总和为 (50 + 50) = 100 。

sql

select product_name,sum(unit) unit 
from Orders o left join Products p
on o.product_id=p.product_id
where date_format(order_date,'%Y-%m')='2020-02' 
group by product_name
having unit>=100

1336. 每次访问的交易次数

难度困难

SQL架构

表: Visits

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| visit_date    | date    |
+---------------+---------+
(user_id, visit_date) 是该表的主键
该表的每行表示 user_id 在 visit_date 访问了银行

表: Transactions

Code

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| user_id          | int     |
| transaction_date | date    |
| amount           | int     |
+------------------+---------+
该表没有主键,所以可能有重复行
该表的每一行表示 user_id 在 transaction_date 完成了一笔 amount 数额的交易
可以保证用户 (user) 在 transaction_date 访问了银行 (也就是说 Visits 表包含 (user_id, transaction_date) 行)

银行想要得到银行客户在一次访问时的交易次数和相应的在一次访问时该交易次数的客户数量的图表

写一条 SQL 查询多少客户访问了银行但没有进行任何交易,多少客户访问了银行进行了一次交易等等

结果包含两列:

  • transactions_count: 客户在一次访问中的交易次数
  • visits_count:transactions_count 交易次数下相应的一次访问时的客户数量

Code

transactions_count` 的值从 `0` 到所有用户一次访问中的 `max(transactions_count)

transactions_count 排序

下面是查询结果格式的例子:

Code

Visits 表:
+---------+------------+
| user_id | visit_date |
+---------+------------+
| 1       | 2020-01-01 |
| 2       | 2020-01-02 |
| 12      | 2020-01-01 |
| 19      | 2020-01-03 |
| 1       | 2020-01-02 |
| 2       | 2020-01-03 |
| 1       | 2020-01-04 |
| 7       | 2020-01-11 |
| 9       | 2020-01-25 |
| 8       | 2020-01-28 |
+---------+------------+
Transactions 表:
+---------+------------------+--------+
| user_id | transaction_date | amount |
+---------+------------------+--------+
| 1       | 2020-01-02       | 120    |
| 2       | 2020-01-03       | 22     |
| 7       | 2020-01-11       | 232    |
| 1       | 2020-01-04       | 7      |
| 9       | 2020-01-25       | 33     |
| 9       | 2020-01-25       | 66     |
| 8       | 2020-01-28       | 1      |
| 9       | 2020-01-25       | 99     |
+---------+------------------+--------+
结果表:
+--------------------+--------------+
| transactions_count | visits_count |
+--------------------+--------------+
| 0                  | 4            |
| 1                  | 5            |
| 2                  | 0            |
| 3                  | 1            |
+--------------------+--------------+
* 对于 transactions_count = 0, visits 中 (1, "2020-01-01"), (2, "2020-01-02"), (12, "2020-01-01") 和 (19, "2020-01-03") 没有进行交易,所以 visits_count = 4 。
* 对于 transactions_count = 1, visits 中 (2, "2020-01-03"), (7, "2020-01-11"), (8, "2020-01-28"), (1, "2020-01-02") 和 (1, "2020-01-04") 进行了一次交易,所以 visits_count = 5 。
* 对于 transactions_count = 2, 没有客户访问银行进行了两次交易,所以 visits_count = 0 。
* 对于 transactions_count = 3, visits 中 (9, "2020-01-25") 进行了三次交易,所以 visits_count = 1 。
* 对于 transactions_count >= 4, 没有客户访问银行进行了超过3次交易,所以我们停止在 transactions_count = 3 。

如下是这个例子的图表:

sql

SELECT *
FROM
(
    SELECT t5.rnb AS transactions_count, IFNULL(visits_count, 0) AS visits_count
    FROM
    (
        SELECT 0 AS rnb
        UNION
        SELECT ROW_NUMBER() OVER () AS rnb
        FROM Transactions
    ) t5
    LEFT JOIN
    (
        SELECT
            cnt AS transactions_count
            ,COUNT(user_id) AS visits_count
        FROM
        (
            SELECT t1.user_id, COUNT(t2.amount) AS cnt
            FROM Visits t1
            LEFT JOIN Transactions t2
            ON t1.user_id = t2.user_id AND t1.visit_date = t2.transaction_date
            GROUP BY user_id, visit_date
        ) t3
        GROUP BY cnt
    ) t4
    ON t5.rnb = t4.transactions_count
) t6
WHERE transactions_count <= (
    SELECT COUNT(t2.amount) AS cnt
    FROM Visits t1
    LEFT JOIN Transactions t2
    ON t1.user_id = t2.user_id AND t1.visit_date = t2.transaction_date
    GROUP BY t1.user_id, visit_date
    ORDER BY cnt DESC
    LIMIT 1)

难点 从0自增序列,2交易的人数为0

sql

select pcnt transactions_count,count(*) visits_count
from (
select visit_date,
        sum(if(amount is  null,0,1)) over(partition by transaction_date ) pcnt,
        count(*) over(partition by  visit_date ) tcnt
from Visits v left join Transactions t
on v.user_id= t.user_id and v.visit_date=t.transaction_date
)t1
group by pcnt

这个得出结果是[0, 4], [1, 5], [3, 3] 少了[2,0] 还没想到什么好办法能把[2,0]加进去。。。

1341. 电影评分

难度中等

SQL架构

表:Movies

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| title         | varchar |
+---------------+---------+
movie_id 是这个表的主键。
title 是电影的名字。

表:Users

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| name          | varchar |
+---------------+---------+
user_id 是表的主键。

表:Movie_Rating

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| user_id       | int     |
| rating        | int     |
| created_at    | date    |
+---------------+---------+
(movie_id, user_id) 是这个表的主键。
这个表包含用户在其评论中对电影的评分 rating 。
created_at 是用户的点评日期。

请你编写一组 SQL 查询:

  • 查找评论电影数量最多的用户名。 如果出现平局,返回字典序较小的用户名。
  • 查找在

2020 年 2 月 平均评分最高

的电影名称。

如果出现平局,返回字典序较小的电影名称。

查询分两行返回,查询结果格式如下例所示:

Code

Movies 表:
+-------------+--------------+
| movie_id    |  title       |
+-------------+--------------+
| 1           | Avengers     |
| 2           | Frozen 2     |
| 3           | Joker        |
+-------------+--------------+

Users 表:
+-------------+--------------+
| user_id     |  name        |
+-------------+--------------+
| 1           | Daniel       |
| 2           | Monica       |
| 3           | Maria        |
| 4           | James        |
+-------------+--------------+

Movie_Rating 表:
+-------------+--------------+--------------+-------------+
| movie_id    | user_id      | rating       | created_at  |
+-------------+--------------+--------------+-------------+
| 1           | 1            | 3            | 2020-01-12  |
| 1           | 2            | 4            | 2020-02-11  |
| 1           | 3            | 2            | 2020-02-12  |
| 1           | 4            | 1            | 2020-01-01  |
| 2           | 1            | 5            | 2020-02-17  | 
| 2           | 2            | 2            | 2020-02-01  | 
| 2           | 3            | 2            | 2020-03-01  |
| 3           | 1            | 3            | 2020-02-22  | 
| 3           | 2            | 4            | 2020-02-25  | 
+-------------+--------------+--------------+-------------+

Result 表:
+--------------+
| results      |
+--------------+
| Daniel       |
| Frozen 2     |
+--------------+

Daniel 和 Monica 都点评了 3 部电影("Avengers", "Frozen 2" 和 "Joker") 但是 Daniel 字典序比较小。
Frozen 2 和 Joker 在 2 月的评分都是 3.5,但是 Frozen 2 的字典序比较小。

sql

select name results 
from
(
    select m.user_id ,u.name 
    from Movie_Rating m left join Users u 
    on m.user_id = u.user_id
    group by user_id
    order by count(*) desc,name
    limit 1
)t1
union 
(
select title results
from Movie_Rating r left join Movies m
on r.movie_id =m.movie_id 
where date_format(created_at,'%Y-%m')='2020-02'
group by r.movie_id
order by avg(rating) desc,title 
limit 1
)

1350. 院系无效的学生

难度简单

SQL架构

院系表: Departments

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
+---------------+---------+
id 是该表的主键
该表包含一所大学每个院系的 id 信息

学生表: Students

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
| department_id | int     |
+---------------+---------+
id 是该表的主键
该表包含一所大学每个学生的 id 和他/她就读的院系信息

写一条 SQL 语句以查询那些所在院系不存在的学生的 id 和姓名

可以以任何顺序返回结果

下面是返回结果格式的例子

Code

Departments 表:
+------+--------------------------+
| id   | name                     |
+------+--------------------------+
| 1    | Electrical Engineering   |
| 7    | Computer Engineering     |
| 13   | Bussiness Administration |
+------+--------------------------+

Students 表:
+------+----------+---------------+
| id   | name     | department_id |
+------+----------+---------------+
| 23   | Alice    | 1             |
| 1    | Bob      | 7             |
| 5    | Jennifer | 13            |
| 2    | John     | 14            |
| 4    | Jasmine  | 77            |
| 3    | Steve    | 74            |
| 6    | Luis     | 1             |
| 8    | Jonathan | 7             |
| 7    | Daiana   | 33            |
| 11   | Madelynn | 1             |
+------+----------+---------------+

结果表:
+------+----------+
| id   | name     |
+------+----------+
| 2    | John     |
| 7    | Daiana   |
| 4    | Jasmine  |
| 3    | Steve    |
+------+----------+

John, Daiana, Steve 和 Jasmine 所在的院系分别是 14, 33, 74 和 77, 其中 14, 33, 74 和 77 并不存在于院系表

sql

select id,name  
from Students 
where department_id not in 
(
    select id 
    from Departments
)

1355. 活动参与者

难度中等

SQL架构

表: Friends

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
| activity      | varchar |
+---------------+---------+
id 是朋友的 id 和该表的主键
name 是朋友的名字
activity 是朋友参加的活动的名字

表: Activities

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
+---------------+---------+
id 是该表的主键
name 是活动的名字

写一条 SQL 查询那些既没有最多,也没有最少参与者的活动的名字

可以以任何顺序返回结果,Activities 表的每项活动的参与者都来自 Friends 表

下面是查询结果格式的例子:

Code

Friends 表:
+------+--------------+---------------+
| id   | name         | activity      |
+------+--------------+---------------+
| 1    | Jonathan D.  | Eating        |
| 2    | Jade W.      | Singing       |
| 3    | Victor J.    | Singing       |
| 4    | Elvis Q.     | Eating        |
| 5    | Daniel A.    | Eating        |
| 6    | Bob B.       | Horse Riding  |
+------+--------------+---------------+

Activities 表:
+------+--------------+
| id   | name         |
+------+--------------+
| 1    | Eating       |
| 2    | Singing      |
| 3    | Horse Riding |
+------+--------------+

Result 表:
+--------------+
| activity     |
+--------------+
| Singing      |
+--------------+

Eating 活动有三个人参加, 是最多人参加的活动 (Jonathan D. , Elvis Q. and Daniel A.)
Horse Riding 活动有一个人参加, 是最少人参加的活动 (Bob B.)
Singing 活动有两个人参加 (Victor J. and Jade W.)

sql

select activity
from (
    select activity,
    rank()over(order by cnt) rk1,
    rank()over(order by cnt desc) rk2
    from
    (
        select  activity  ,count(*) cnt
        from  Friends
        group by  activity 
    )t1
)t2
where rk1 !=1 and rk2 != 1

不需要关联 Activities表,因为 至少有一人参加

1364. 顾客的可信联系人数量

难度中等

SQL架构

顾客表:Customers

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| customer_name | varchar |
| email         | varchar |
+---------------+---------+
customer_id 是这张表的主键。
此表的每一行包含了某在线商店顾客的姓名和电子邮件。

联系方式表:Contacts

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | id      |
| contact_name  | varchar |
| contact_email | varchar |
+---------------+---------+
(user_id, contact_email) 是这张表的主键。
此表的每一行表示编号为 user_id 的顾客的某位联系人的姓名和电子邮件。
此表包含每位顾客的联系人信息,但顾客的联系人不一定存在于顾客表中。

发票表:Invoices

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| invoice_id   | int     |
| price        | int     |
| user_id      | int     |
+--------------+---------+
invoice_id 是这张表的主键。
此表的每一行分别表示编号为 user_id 的顾客拥有有一张编号为 invoice_id、价格为 price 的发票。

为每张发票 invoice_id 编写一个SQL查询以查找以下内容:

  • customer_name:与发票相关的顾客名称。
  • price:发票的价格。
  • contacts_cnt:该顾客的联系人数量。
  • trusted_contacts_cnt:可信联系人的数量:既是该顾客的联系人又是商店顾客的联系人数量(即:可信联系人的电子邮件存在于客户表中)。

将查询的结果按照 invoice_id 排序。

查询结果的格式如下例所示:

Code

Customers table:
+-------------+---------------+--------------------+
| customer_id | customer_name | email              |
+-------------+---------------+--------------------+
| 1           | Alice         | alice@leetcode.com |
| 2           | Bob           | bob@leetcode.com   |
| 13          | John          | john@leetcode.com  |
| 6           | Alex          | alex@leetcode.com  |
+-------------+---------------+--------------------+
Contacts table:
+-------------+--------------+--------------------+
| user_id     | contact_name | contact_email      |
+-------------+--------------+--------------------+
| 1           | Bob          | bob@leetcode.com   |
| 1           | John         | john@leetcode.com  |
| 1           | Jal          | jal@leetcode.com   |
| 2           | Omar         | omar@leetcode.com  |
| 2           | Meir         | meir@leetcode.com  |
| 6           | Alice        | alice@leetcode.com |
+-------------+--------------+--------------------+
Invoices table:
+------------+-------+---------+
| invoice_id | price | user_id |
+------------+-------+---------+
| 77         | 100   | 1       |
| 88         | 200   | 1       |
| 99         | 300   | 2       |
| 66         | 400   | 2       |
| 55         | 500   | 13      |
| 44         | 60    | 6       |
+------------+-------+---------+
Result table:
+------------+---------------+-------+--------------+----------------------+
| invoice_id | customer_name | price | contacts_cnt | trusted_contacts_cnt |
+------------+---------------+-------+--------------+----------------------+
| 44         | Alex          | 60    | 1            | 1                    |
| 55         | John          | 500   | 0            | 0                    |
| 66         | Bob           | 400   | 2            | 0                    |
| 77         | Alice         | 100   | 3            | 2                    |
| 88         | Alice         | 200   | 3            | 2                    |
| 99         | Bob           | 300   | 2            | 0                    |
+------------+---------------+-------+--------------+----------------------+
Alice 有三位联系人,其中两位(Bob 和 John)是可信联系人。
Bob 有两位联系人, 他们中的任何一位都不是可信联系人。
Alex 只有一位联系人(Alice),并是一位可信联系人。
John 没有任何联系人。

sql

select invoice_id ,customer_name,price,ifnull(cnt,0) contacts_cnt,ifnull(bc,0) trusted_contacts_cnt 
from Invoices i
left join (
select user_id ,count(*) cnt
from Contacts
group by user_id
) t1
on i.user_id=t1.user_id
left join (
select  user_id ,count(*) bc
from Contacts
    where contact_name in
    (
        select customer_name
        from Customers
    )
group by user_id 
)t2
on i.user_id = t2.user_id
left join Customers c
on i.user_id= c.customer_id
order by invoice_id

就是麻烦点 各种join

1369. 获取最近第二次的活动

难度困难

SQL架构

表: UserActivity

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| username      | varchar |
| activity      | varchar |
| startDate     | Date    |
| endDate       | Date    |
+---------------+---------+
该表不包含主键
该表包含每个用户在一段时间内进行的活动的信息
名为 username 的用户在 startDate 到 endDate 日内有一次活动

写一条SQL查询展示每一位用户 最近第二次 的活动

如果用户仅有一次活动,返回该活动

一个用户不能同时进行超过一项活动,以 任意 顺序返回结果

下面是查询结果格式的例子:

Code

UserActivity 表:
+------------+--------------+-------------+-------------+
| username   | activity     | startDate   | endDate     |
+------------+--------------+-------------+-------------+
| Alice      | Travel       | 2020-02-12  | 2020-02-20  |
| Alice      | Dancing      | 2020-02-21  | 2020-02-23  |
| Alice      | Travel       | 2020-02-24  | 2020-02-28  |
| Bob        | Travel       | 2020-02-11  | 2020-02-18  |
+------------+--------------+-------------+-------------+

Result 表:
+------------+--------------+-------------+-------------+
| username   | activity     | startDate   | endDate     |
+------------+--------------+-------------+-------------+
| Alice      | Dancing      | 2020-02-21  | 2020-02-23  |
| Bob        | Travel       | 2020-02-11  | 2020-02-18  |
+------------+--------------+-------------+-------------+

Alice 最近第二次的活动是从 2020-02-24 到 2020-02-28 的旅行, 在此之前的 2020-02-21 到 2020-02-23 她进行了舞蹈
Bob 只有一条记录,我们就取这条记录

sql

select username, activity ,startDate,endDate 
from 
(
select username, activity ,startDate,endDate ,
    rank()over(partition by username order by startDate desc) rk,
    lag( startDate ,1,null)over(partition by username order by startDate ) lg
from UserActivity
)t1
where rk=2 or  (rk = 1 &&  lg is null)

1378. 使用唯一标识码替换员工ID

难度简单

SQL架构

Employees 表:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
+---------------+---------+
id 是这张表的主键。
这张表的每一行分别代表了某公司其中一位员工的名字和 ID 。

EmployeeUNI 表:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| unique_id     | int     |
+---------------+---------+
(id, unique_id) 是这张表的主键。
这张表的每一行包含了该公司某位员工的 ID 和他的唯一标识码(unique ID)。

写一段SQL查询来展示每位用户的 唯一标识码(unique ID );如果某位员工没有唯一标识码,使用 null 填充即可。

你可以以 任意 顺序返回结果表。

查询结果的格式如下例所示:

Code

Employees table:
+----+----------+
| id | name     |
+----+----------+
| 1  | Alice    |
| 7  | Bob      |
| 11 | Meir     |
| 90 | Winston  |
| 3  | Jonathan |
+----+----------+

EmployeeUNI table:
+----+-----------+
| id | unique_id |
+----+-----------+
| 3  | 1         |
| 11 | 2         |
| 90 | 3         |
+----+-----------+

EmployeeUNI table:
+-----------+----------+
| unique_id | name     |
+-----------+----------+
| null      | Alice    |
| null      | Bob      |
| 2         | Meir     |
| 3         | Winston  |
| 1         | Jonathan |
+-----------+----------+

Alice and Bob 没有唯一标识码, 因此我们使用 null 替代。
Meir 的唯一标识码是 2 。
Winston 的唯一标识码是 3 。
Jonathan 唯一标识码是 1 。

sql

select unique_id,e.name
from Employees  e left join EmployeeUNI u
on e.id = u.id

1384. 按年度列出销售总额

难度困难

SQL架构

Product 表:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| product_name  | varchar |
+---------------+---------+
product_id 是这张表的主键。
product_name 是产品的名称。

Sales 表:

Code

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| product_id          | int     |
| period_start        | varchar |
| period_end          | date    |
| average_daily_sales | int     |
+---------------------+---------+
product_id 是这张表的主键。
period_start 和 period_end 是该产品销售期的起始日期和结束日期,且这两个日期包含在销售期内。
average_daily_sales 列存储销售期内该产品的日平均销售额。

编写一段SQL查询每个产品每年的总销售额,并包含 product_id, product_name 以及 report_year 等信息。

销售年份的日期介于 2018 年到 2020 年之间。你返回的结果需要按 product_id 和 report_year 排序

查询结果格式如下例所示:

Code

Product table:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 1          | LC Phone     |
| 2          | LC T-Shirt   |
| 3          | LC Keychain  |
+------------+--------------+

Sales table:
+------------+--------------+-------------+---------------------+
| product_id | period_start | period_end  | average_daily_sales |
+------------+--------------+-------------+---------------------+
| 1          | 2019-01-25   | 2019-02-28  | 100                 |
| 2          | 2018-12-01   | 2020-01-01  | 10                  |
| 3          | 2019-12-01   | 2020-01-31  | 1                   |
+------------+--------------+-------------+---------------------+

Result table:
+------------+--------------+-------------+--------------+
| product_id | product_name | report_year | total_amount |
+------------+--------------+-------------+--------------+
| 1          | LC Phone     |    2019     | 3500         |
| 2          | LC T-Shirt   |    2018     | 310          |
| 2          | LC T-Shirt   |    2019     | 3650         |
| 2          | LC T-Shirt   |    2020     | 10           |
| 3          | LC Keychain  |    2019     | 31           |
| 3          | LC Keychain  |    2020     | 31           |
+------------+--------------+-------------+--------------+
LC Phone 在 2019-01-25 至 2019-02-28 期间销售,该产品销售时间总计35天。销售总额 35*100 = 3500。
LC T-shirt 在 2018-12-01 至 2020-01-01 期间销售,该产品在2018年、2019年、2020年的销售时间分别是31天、365天、1天,2018年、2019年、2020年的销售总额分别是31*10=310、365*10=3650、1*10=10。
LC Keychain 在 2019-12-01 至 2020-01-31 期间销售,该产品在2019年、2020年的销售时间分别是:31天、31天,2019年、2020年的销售总额分别是31*1=31、31*1=31。

sql

(
 select Sales.product_id, product_name, '2018' as 'report_year', if(period_start<'2019-01-01', (datediff(if(period_end<'2019-01-01', period_end, date('2018-12-31')), if(period_start>='2018-01-01', period_start, date('2018-01-01')))+1)*average_daily_sales, 0) as total_amount
from Sales  
join Product on Sales.product_id = Product.product_id 
having  total_amount>0
)
union(
select Sales.product_id, product_name, '2019' as 'report_year', if( period_start<'2020-01-01', (datediff(if(period_end<'2020-01-01', period_end, date('2019-12-31')), if(period_start>='2019-01-01', period_start, date('2019-01-01')))+1)*average_daily_sales , 0) as total_amount
from Sales  
join Product on (Sales.product_id = Product.product_id )
having  total_amount>0
)
union(
select Sales.product_id, product_name, '2020' as 'report_year', (datediff(if(period_end<'2021-01-01', period_end, date('2020-12-31')), if(period_start>='2020-01-01', period_start, date('2020-01-01')))+1)*average_daily_sales as total_amount
from Sales  
join Product  on (Sales.product_id = Product.product_id)
having total_amount>0
) 
order by product_id, report_year

各个年份进行union,就是年份判断的时候麻烦些

1393. 股票的资本损益

难度中等

SQL架构

Stocks 表:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| stock_name    | varchar |
| operation     | enum    |
| operation_day | int     |
| price         | int     |
+---------------+---------+
(stock_name, day) 是这张表的主键
operation 列使用的是一种枚举类型,包括:('Sell','Buy')
此表的每一行代表了名为 stock_name 的某支股票在 operation_day 这一天的操作价格。
保证股票的每次'Sell'操作前,都有相应的'Buy'操作。

编写一个SQL查询来报告每支股票的资本损益。

股票的资本损益是一次或多次买卖股票后的全部收益或损失。

以任意顺序返回结果即可。

SQL查询结果的格式如下例所示:

Code

Stocks 表:
+---------------+-----------+---------------+--------+
| stock_name    | operation | operation_day | price  |
+---------------+-----------+---------------+--------+
| Leetcode      | Buy       | 1             | 1000   |
| Corona Masks  | Buy       | 2             | 10     |
| Leetcode      | Sell      | 5             | 9000   |
| Handbags      | Buy       | 17            | 30000  |
| Corona Masks  | Sell      | 3             | 1010   |
| Corona Masks  | Buy       | 4             | 1000   |
| Corona Masks  | Sell      | 5             | 500    |
| Corona Masks  | Buy       | 6             | 1000   |
| Handbags      | Sell      | 29            | 7000   |
| Corona Masks  | Sell      | 10            | 10000  |
+---------------+-----------+---------------+--------+

Result 表:
+---------------+-------------------+
| stock_name    | capital_gain_loss |
+---------------+-------------------+
| Corona Masks  | 9500              |
| Leetcode      | 8000              |
| Handbags      | -23000            |
+---------------+-------------------+
Leetcode 股票在第一天以1000美元的价格买入,在第五天以9000美元的价格卖出。资本收益=9000-1000=8000美元。
Handbags 股票在第17天以30000美元的价格买入,在第29天以7000美元的价格卖出。资本损失=7000-30000=-23000美元。
Corona Masks 股票在第1天以10美元的价格买入,在第3天以1010美元的价格卖出。在第4天以1000美元的价格再次购买,在第5天以500美元的价格出售。最后,它在第6天以1000美元的价格被买走,在第10天以10000美元的价格被卖掉。资本损益是每次(’Buy'->'Sell')操作资本收益或损失的和=(1010-10)+(500-1000)+(10000-1000)=1000-500+9000=9500美元。

sql

select stock_name,sell-buy capital_gain_loss
from(
select stock_name ,
       sum(if(operation='Buy', price,0))over(partition by stock_name ) buy,
       sum(if(operation='Sell',price,0))over(partition by stock_name) sell
from Stocks s
)t1
group by stock_name,buy,sell

1398. 购买了产品A和产品B却没有购买产品C的顾客

难度中等

SQL架构

Customers 表:

Code

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| customer_id         | int     |
| customer_name       | varchar |
+---------------------+---------+
customer_id 是这张表的主键。
customer_name 是顾客的名称。

Orders 表:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| customer_id   | int     |
| product_name  | varchar |
+---------------+---------+
order_id 是这张表的主键。
customer_id 是购买了名为 "product_name" 产品顾客的id。

请你设计 SQL 查询来报告购买了产品 A 和产品 B 却没有购买产品 C 的顾客的 ID 和姓名( customer_idcustomer_name ),我们将基于此结果为他们推荐产品 C 。 您返回的查询结果需要按照 customer_id 排序

查询结果如下例所示。

Code

Customers table:
+-------------+---------------+
| customer_id | customer_name |
+-------------+---------------+
| 1           | Daniel        |
| 2           | Diana         |
| 3           | Elizabeth     |
| 4           | Jhon          |
+-------------+---------------+

Orders table:
+------------+--------------+---------------+
| order_id   | customer_id  | product_name  |
+------------+--------------+---------------+
| 10         |     1        |     A         |
| 20         |     1        |     B         |
| 30         |     1        |     D         |
| 40         |     1        |     C         |
| 50         |     2        |     A         |
| 60         |     3        |     A         |
| 70         |     3        |     B         |
| 80         |     3        |     D         |
| 90         |     4        |     C         |
+------------+--------------+---------------+

Result table:
+-------------+---------------+
| customer_id | customer_name |
+-------------+---------------+
| 3           | Elizabeth     |
+-------------+---------------+
只有 customer_id 为 3 的顾客购买了产品 A 和产品 B ,却没有购买产品 C 。

sql

select  o.customer_id, customer_name 
from Orders o left join  Customers c
on o.customer_id=c.customer_id
group by customer_id
having sum(product_name ='A')>=1 and sum(product_name='B')>=1 and sum(product_name='C')=0

1407. 排名靠前的旅行者

难度简单

SQL架构

表单: Users

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
+---------------+---------+
id 是该表单主键.
name 是用户名字.

表单: Rides

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| user_id       | int     |
| distance      | int     |
+---------------+---------+
id 是该表单主键.
user_id 是本次行程的用户的 id, 而该用户此次行程距离为 distance.

写一段 SQL , 报告每个用户的旅行距离.

返回的结果表单, 以 travelled_distance 降序排列, 如果有两个或者更多的用户旅行了相同的距离, 那么再以 name 升序排列.

查询结果格式, 如下例所示.

Code

Users 表单:
+------+-----------+
| id   | name      |
+------+-----------+
| 1    | Alice     |
| 2    | Bob       |
| 3    | Alex      |
| 4    | Donald    |
| 7    | Lee       |
| 13   | Jonathan  |
| 19   | Elvis     |
+------+-----------+

Rides 表单:
+------+----------+----------+
| id   | user_id  | distance |
+------+----------+----------+
| 1    | 1        | 120      |
| 2    | 2        | 317      |
| 3    | 3        | 222      |
| 4    | 7        | 100      |
| 5    | 13       | 312      |
| 6    | 19       | 50       |
| 7    | 7        | 120      |
| 8    | 19       | 400      |
| 9    | 7        | 230      |
+------+----------+----------+

Result 表单:
+----------+--------------------+
| name     | travelled_distance |
+----------+--------------------+
| Elvis    | 450                |
| Lee      | 450                |
| Bob      | 317                |
| Jonathan | 312                |
| Alex     | 222                |
| Alice    | 120                |
| Donald   | 0                  |
+----------+--------------------+
Elvis 和 Lee 旅行了 450 英里, Elvis 是排名靠前的旅行者, 因为他的名字在字母表上的排序比 Lee 更小.
Bob, Jonathan, Alex 和 Alice 只有一次行程, 我们只按此次行程的全部距离对他们排序.
Donald 没有任何行程, 他的旅行距离为 0.

sql

select name,sum(ifnull(distance,0)) travelled_distance 
from Users u left join Rides r
on u.id = r.user_id
group by name
order by travelled_distance  desc, name

1412. 查找成绩处于中游的学生

难度困难

SQL架构

表: Student

Code

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| student_id          | int     |
| student_name        | varchar |
+---------------------+---------+
student_id 是该表主键.
student_name 学生名字.

表: Exam

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| exam_id       | int     |
| student_id    | int     |
| score         | int     |
+---------------+---------+
(exam_id, student_id) 是该表主键.
学生 student_id 在测验 exam_id 中得分为 score.

成绩处于中游的学生是指至少参加了一次测验, 且得分既不是最高分也不是最低分的学生。

写一个 SQL 语句,找出在所有测验中都处于中游的学生 (student_id, student_name)

不要返回从来没有参加过测验的学生。返回结果表按照 student_id 排序。

查询结果格式如下。

Code

Student 表:
+-------------+---------------+
| student_id  | student_name  |
+-------------+---------------+
| 1           | Daniel        |
| 2           | Jade          |
| 3           | Stella        |
| 4           | Jonathan      |
| 5           | Will          |
+-------------+---------------+

Exam 表:
+------------+--------------+-----------+
| exam_id    | student_id   | score     |
+------------+--------------+-----------+
| 10         |     1        |    70     |
| 10         |     2        |    80     |
| 10         |     3        |    90     |
| 20         |     1        |    80     |
| 30         |     1        |    70     |
| 30         |     3        |    80     |
| 30         |     4        |    90     |
| 40         |     1        |    60     |
| 40         |     2        |    70     |
| 40         |     4        |    80     |
+------------+--------------+-----------+

Result 表:
+-------------+---------------+
| student_id  | student_name  |
+-------------+---------------+
| 2           | Jade          |
+-------------+---------------+

对于测验 1: 学生 1 和 3 分别获得了最低分和最高分。
对于测验 2: 学生 1 既获得了最高分, 也获得了最低分。
对于测验 3 和 4: 学生 1 和 4 分别获得了最低分和最高分。
学生 2 和 5 没有在任一场测验中获得了最高分或者最低分。
因为学生 5 从来没有参加过任何测验, 所以他被排除于结果表。
由此, 我们仅仅返回学生 2 的信息。

sql

select e.student_id,student_name
from Exam e left join Student s
on e.student_id=s.student_id
where e.student_id not in(
    select student_id
    from(
        select student_id,rank() over(partition by exam_id order by score desc) rkmax, rank() over(partition by exam_id order by score ) rkmin
        from Exam 
    )t1
    where rkmax = 1 or rkmin =1
)
group by e.student_id,student_name
order by e.student_id

1421. 净现值查询

难度中等

SQL架构

表: NPV

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| year          | int     |
| npv           | int     |
+---------------+---------+
(id, year) 是该表主键.
该表有每一笔存货的年份, id 和对应净现值的信息.

表: Queries

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| year          | int     |
+---------------+---------+
(id, year) 是该表主键.
该表有每一次查询所对应存货的 id 和年份的信息.

写一个 SQL, 找到 Queries 表中每一次查询的净现值.

结果表没有顺序要求.

查询结果的格式如下所示:

Code

NPV 表:
+------+--------+--------+
| id   | year   | npv    |
+------+--------+--------+
| 1    | 2018   | 100    |
| 7    | 2020   | 30     |
| 13   | 2019   | 40     |
| 1    | 2019   | 113    |
| 2    | 2008   | 121    |
| 3    | 2009   | 12     |
| 11   | 2020   | 99     |
| 7    | 2019   | 0      |
+------+--------+--------+

Queries 表:
+------+--------+
| id   | year   |
+------+--------+
| 1    | 2019   |
| 2    | 2008   |
| 3    | 2009   |
| 7    | 2018   |
| 7    | 2019   |
| 7    | 2020   |
| 13   | 2019   |
+------+--------+

结果表:
+------+--------+--------+
| id   | year   | npv    |
+------+--------+--------+
| 1    | 2019   | 113    |
| 2    | 2008   | 121    |
| 3    | 2009   | 12     |
| 7    | 2018   | 0      |
| 7    | 2019   | 0      |
| 7    | 2020   | 30     |
| 13   | 2019   | 40     |
+------+--------+--------+

(7, 2018)的净现值不在 NPV 表中, 我们把它看作是 0.
所有其它查询的净现值都能在 NPV 表中找到.

sql

select q.id,q.year,ifnull(npv,0) npv
from Queries q left join NPV n
on q.id = n.id and q.year = n.year

npv 净现值概念 了解下

1435. 制作会话柱状图

难度简单

SQL架构

表:Sessions

Code

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| session_id          | int     |
| duration            | int     |
+---------------------+---------+
session_id 是该表主键
duration 是用户访问应用的时间, 以秒为单位

你想知道用户在你的 app 上的访问时长情况。因此决定统计访问时长区间分别为 “[0-5>”, “[5-10>”, “[10-15>” 和 “15 or more” (单位:分钟)的会话数量,并以此绘制柱状图。

写一个SQL查询来报告(访问时长区间,会话总数)。结果可用任何顺序呈现。

下方为查询的输出格式:

Code

Sessions 表:
+-------------+---------------+
| session_id  | duration      |
+-------------+---------------+
| 1           | 30            |
| 2           | 199           |
| 3           | 299           |
| 4           | 580           |
| 5           | 1000          |
+-------------+---------------+

Result 表:
+--------------+--------------+
| bin          | total        |
+--------------+--------------+
| [0-5>        | 3            |
| [5-10>       | 1            |
| [10-15>      | 0            |
| 15 or more   | 1            |
+--------------+--------------+

对于 session_id 1,2 和 3 ,它们的访问时间大于等于 0 分钟且小于 5 分钟。
对于 session_id 4,它的访问时间大于等于 5 分钟且小于 10 分钟。
没有会话的访问时间大于等于 10 分钟且小于 15 分钟。
对于 session_id 5, 它的访问时间大于等于 15 分钟。

Union

sql

select '[0-5>' as bin, count(*) as total from Sessions where duration/60>=0 and duration/60<5
union
select '[5-10>' as bin, count(*) as total from Sessions where duration/60>=5 and duration/60<10
union
select '[10-15>' as bin, count(*) as total from Sessions where duration/60>=10 and duration/60<15
union
select '15 or more'as bin, count(*) as total from Sessions where duration/60>=15

还有很多其他解法

sql

select a.bin, count(b.bin) as total
from
(
    select '[0-5>' as bin union select '[5-10>' as bin union select '[10-15>' as bin union select '15 or more' as bin 
)a
left join 
(
    select case
        when duration < 300 then '[0-5>'
        when duration >= 300 and duration < 600 then '[5-10>'
        when duration >= 600 and duration < 900 then '[10-15>'
        else '15 or more'
        end bin
    from Sessions 
)b
on a.bin = b.bin
group by a.bin

1440. 计算布尔表达式的值

难度中等

SQL架构

Variables:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| name          | varchar |
| value         | int     |
+---------------+---------+
name 是该表主键.
该表包含了存储的变量及其对应的值.

Expressions:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| left_operand  | varchar |
| operator      | enum    |
| right_operand | varchar |
+---------------+---------+
(left_operand, operator, right_operand) 是该表主键.
该表包含了需要计算的布尔表达式.
operator 是枚举类型, 取值于('<', '>', '=')
left_operand 和 right_operand 的值保证存在于 Variables 表单中.

写一个 SQL 查询, 以计算表 Expressions 中的布尔表达式.

返回的结果表没有顺序要求.

查询结果格式如下例所示.

Code

Variables 表:
+------+-------+
| name | value |
+------+-------+
| x    | 66    |
| y    | 77    |
+------+-------+

Expressions 表:
+--------------+----------+---------------+
| left_operand | operator | right_operand |
+--------------+----------+---------------+
| x            | >        | y             |
| x            | <        | y             |
| x            | =        | y             |
| y            | >        | x             |
| y            | <        | x             |
| x            | =        | x             |
+--------------+----------+---------------+

Result 表:
+--------------+----------+---------------+-------+
| left_operand | operator | right_operand | value |
+--------------+----------+---------------+-------+
| x            | >        | y             | false |
| x            | <        | y             | true  |
| x            | =        | y             | false |
| y            | >        | x             | true  |
| y            | <        | x             | false |
| x            | =        | x             | true  |
+--------------+----------+---------------+-------+
如上所示, 你需要通过使用 Variables 表来找到 Expressions 表中的每一个布尔表达式的值.

sql

select e.left_operand,e.operator,e.right_operand,
case e.operator
    when '>' then if(v1.value>v2.value,'true','false')
    when '<' then if(v1.value<v2.value,'true','false')
    else  if(v1.value=v2.value,'true','false')
end value
from Expressions e
left join Variables v1 on v1.name = e.left_operand 
left join Variables v2 on v2.name = e.right_operand

1445. 苹果和桔子

难度中等

SQL架构

表: Sales

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| sale_date     | date    |
| fruit         | enum    | 
| sold_num      | int     | 
+---------------+---------+
(sale_date,fruit) 是该表主键.
该表包含了每一天中"苹果" 和 "桔子"的销售情况.

写一个 SQL 查询, 报告每一天 苹果桔子 销售的数目的差异.

返回的结果表, 按照格式为 (‘YYYY-MM-DD’) 的 sale_date 排序.

查询结果表如下例所示:

Code

Sales 表:
+------------+------------+-------------+
| sale_date  | fruit      | sold_num    |
+------------+------------+-------------+
| 2020-05-01 | apples     | 10          |
| 2020-05-01 | oranges    | 8           |
| 2020-05-02 | apples     | 15          |
| 2020-05-02 | oranges    | 15          |
| 2020-05-03 | apples     | 20          |
| 2020-05-03 | oranges    | 0           |
| 2020-05-04 | apples     | 15          |
| 2020-05-04 | oranges    | 16          |
+------------+------------+-------------+

Result 表:
+------------+--------------+
| sale_date  | diff         |
+------------+--------------+
| 2020-05-01 | 2            |
| 2020-05-02 | 0            |
| 2020-05-03 | 20           |
| 2020-05-04 | -1           |
+------------+--------------+

在 2020-05-01, 卖了 10 个苹果 和 8 个桔子 (差异为 10 - 8 = 2).
在 2020-05-02, 卖了 15 个苹果 和 15 个桔子 (差异为 15 - 15 = 0).
在 2020-05-03, 卖了 20 个苹果 和 0 个桔子 (差异为 20 - 0 = 20).
在 2020-05-04, 卖了 15 个苹果 和 16 个桔子 (差异为 15 - 16 = -1).

sql

select  sale_date,sold_num-ld diff
from 
(
select sale_date,sold_num , fruit ,lead(sold_num ,1,null) over(partition by  sale_date ) ld
from Sales
)t1
where fruit='apples'

1454. 活跃用户

难度中等

SQL架构

Accounts:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
+---------------+---------+
id 是该表主键.
该表包含账户 id 和账户的用户名.

Logins:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| login_date    | date    |
+---------------+---------+
该表无主键, 可能包含重复项.
该表包含登录用户的账户 id 和登录日期. 用户也许一天内登录多次.

写一个 SQL 查询, 找到活跃用户的 id 和 name.

活跃用户是指那些至少连续 5 天登录账户的用户.

返回的结果表按照 id 排序.

结果表格式如下例所示:

Code

Accounts 表:
+----+----------+
| id | name     |
+----+----------+
| 1  | Winston  |
| 7  | Jonathan |
+----+----------+

Logins 表:
+----+------------+
| id | login_date |
+----+------------+
| 7  | 2020-05-30 |
| 1  | 2020-05-30 |
| 7  | 2020-05-31 |
| 7  | 2020-06-01 |
| 7  | 2020-06-02 |
| 7  | 2020-06-02 |
| 7  | 2020-06-03 |
| 1  | 2020-06-07 |
| 7  | 2020-06-10 |
+----+------------+

Result 表:
+----+----------+
| id | name     |
+----+----------+
| 7  | Jonathan |
+----+----------+
id = 1 的用户 Winston 仅仅在不同的 2 天内登录了 2 次, 所以, Winston 不是活跃用户.
id = 7 的用户 Jonathon 在不同的 6 天内登录了 7 次, , 6 天中有 5 天是连续的, 所以, Jonathan 是活跃用户.

后续问题: 如果活跃用户是那些至少连续 n 天登录账户的用户, 你能否写出通用的解决方案?

sql

select t3.id,name
from 
(
    select distinct id
    from 
    (
        select id,login_date,lead(login_date,4,null) over(partition by id order by login_date) ld
        from 
        (
            select id,login_date 
            from Logins
            group by id,login_date
        )t1
    )t2
    where datediff(ld,login_date)=4
)t3
left join Accounts a
on t3.id = a.id

注意用户当天重复登入

1459. 矩形面积

难度中等

SQL架构

表: Points

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| x_value       | int     |
| y_value       | int     |
+---------------+---------+
id 是该表主键.
每个点都表示为二维空间 (x_value, y_value).

写一个 SQL 语句, 报告由表中任意两点可以形成的所有可能的矩形.

结果表中的每一行包含三列 (p1, p2, area) 如下:

  • p1p2 是矩形两个对角的 id 且 p1 < p2.
  • 矩形的面积由列 area 表示.

请按照面积大小降序排列,如果面积相同的话, 则按照 p1 和 p2 升序对结果表排序

Code

Points 表:
+----------+-------------+-------------+
| id       | x_value     | y_value     |
+----------+-------------+-------------+
| 1        | 2           | 8           |
| 2        | 4           | 7           |
| 3        | 2           | 10          |
+----------+-------------+-------------+

Result 表:
+----------+-------------+-------------+
| p1       | p2          | area        |
+----------+-------------+-------------+
| 2        | 3           | 6           |
| 1        | 2           | 2           |
+----------+-------------+-------------+

p1 应该小于 p2 并且面积大于 0.
p1 = 1 且 p2 = 2 时, 面积等于 |2-4| * |8-7| = 2.
p1 = 2 且 p2 = 3 时, 面积等于 |4-2| * |7-10| = 6.
p1 = 1 且 p2 = 3 时, 是不可能为矩形的, 因为面积等于 0.

sql

select a.id P1,b.id P2,abs(a.x_value-b.x_value)*abs(a.y_value-b.y_value) as area
from Points a,Points b
where a.id<b.id and a.x_value != b.x_value and a.y_value != b.y_value
order by area desc,P1 ,P2

1468. 计算税后工资

难度中等

SQL架构

Salaries 表:

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| company_id    | int     |
| employee_id   | int     |
| employee_name | varchar |
| salary        | int     |
+---------------+---------+
(company_id, employee_id) 是这个表的主键
这个表包括员工的company id, id, name 和 salary

写一条查询 SQL 来查找每个员工的税后工资

每个公司的税率计算依照以下规则

  • 如果这个公司员工最高工资不到 1000 ,税率为 0%
  • 如果这个公司员工最高工资在 1000 到 10000 之间,税率为 24%
  • 如果这个公司员工最高工资大于 10000 ,税率为 49%

按任意顺序返回结果,税后工资结果取整

结果表格式如下例所示:

Code

Salaries 表:
+------------+-------------+---------------+--------+
| company_id | employee_id | employee_name | salary |
+------------+-------------+---------------+--------+
| 1          | 1           | Tony          | 2000   |
| 1          | 2           | Pronub        | 21300  |
| 1          | 3           | Tyrrox        | 10800  |
| 2          | 1           | Pam           | 300    |
| 2          | 7           | Bassem        | 450    |
| 2          | 9           | Hermione      | 700    |
| 3          | 7           | Bocaben       | 100    |
| 3          | 2           | Ognjen        | 2200   |
| 3          | 13          | Nyancat       | 3300   |
| 3          | 15          | Morninngcat   | 1866   |
+------------+-------------+---------------+--------+

Result 表:
+------------+-------------+---------------+--------+
| company_id | employee_id | employee_name | salary |
+------------+-------------+---------------+--------+
| 1          | 1           | Tony          | 1020   |
| 1          | 2           | Pronub        | 10863  |
| 1          | 3           | Tyrrox        | 5508   |
| 2          | 1           | Pam           | 300    |
| 2          | 7           | Bassem        | 450    |
| 2          | 9           | Hermione      | 700    |
| 3          | 7           | Bocaben       | 76     |
| 3          | 2           | Ognjen        | 1672   |
| 3          | 13          | Nyancat       | 2508   |
| 3          | 15          | Morninngcat   | 5911   |
+------------+-------------+---------------+--------+
对于公司 1 ,最高工资是 21300 ,其每个员工的税率为 49%
对于公司 2 ,最高工资是 700 ,其每个员工税率为 0%
对于公司 3 ,最高工资是 7777 ,其每个员工税率是 24%
税后工资计算 = 工资 - ( 税率 / 100)*工资
对于上述案例,Morninngcat 的税后工资 = 7777 - 7777 * ( 24 / 100) = 7777 - 1866.48 = 5910.52 ,取整为 5911

sql

select company_id,employee_id , employee_name,
round(case when maxsalary<1000 then salary
       when maxsalary<10000 then salary*(1-0.24)
       else salary*(1-0.49) end ,0)salary
from(
    select *,max(salary) over(partition by company_id ) maxsalary
    from Salaries 
)t1

1479. 周内每天的销售情况

难度困难

SQL架构

表:Orders

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| customer_id   | int     |
| order_date    | date    | 
| item_id       | varchar |
| quantity      | int     |
+---------------+---------+
(order_id, item_id) 是该表主键
该表包含了订单信息
order_date 是id为 item_id 的商品被id为 customer_id 的消费者订购的日期.

表:Items

Code

+---------------------+---------+
| Column Name         | Type    |
+---------------------+---------+
| item_id             | varchar |
| item_name           | varchar |
| item_category       | varchar |
+---------------------+---------+
item_id 是该表主键
item_name 是商品的名字
item_category 是商品的类别

你是企业主,想要获得分类商品和周内每天的销售报告。

写一个SQL语句,报告 周内每天 每个商品类别下订购了多少单位。

返回结果表单 按商品类别排序

查询结果格式如下例所示:

Code

Orders 表:
+------------+--------------+-------------+--------------+-------------+
| order_id   | customer_id  | order_date  | item_id      | quantity    |
+------------+--------------+-------------+--------------+-------------+
| 1          | 1            | 2020-06-01  | 1            | 10          |
| 2          | 1            | 2020-06-08  | 2            | 10          |
| 3          | 2            | 2020-06-02  | 1            | 5           |
| 4          | 3            | 2020-06-03  | 3            | 5           |
| 5          | 4            | 2020-06-04  | 4            | 1           |
| 6          | 4            | 2020-06-05  | 5            | 5           |
| 7          | 5            | 2020-06-05  | 1            | 10          |
| 8          | 5            | 2020-06-14  | 4            | 5           |
| 9          | 5            | 2020-06-21  | 3            | 5           |
+------------+--------------+-------------+--------------+-------------+

Items 表:
+------------+----------------+---------------+
| item_id    | item_name      | item_category |
+------------+----------------+---------------+
| 1          | LC Alg. Book   | Book          |
| 2          | LC DB. Book    | Book          |
| 3          | LC SmarthPhone | Phone         |
| 4          | LC Phone 2020  | Phone         |
| 5          | LC SmartGlass  | Glasses       |
| 6          | LC T-Shirt XL  | T-Shirt       |
+------------+----------------+---------------+

Result 表:
+------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
| Category   | Monday    | Tuesday   | Wednesday | Thursday  | Friday    | Saturday  | Sunday    |
+------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
| Book       | 20        | 5         | 0         | 0         | 10        | 0         | 0         |
| Glasses    | 0         | 0         | 0         | 0         | 5         | 0         | 0         |
| Phone      | 0         | 0         | 5         | 1         | 0         | 0         | 10        |
| T-Shirt    | 0         | 0         | 0         | 0         | 0         | 0         | 0         |
+------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
在周一(2020-06-01, 2020-06-08),Book分类(ids: 1, 2)下,总共销售了20个单位(10 + 10)
在周二(2020-06-02),Book分类(ids: 1, 2)下,总共销售了5个单位
在周三(2020-06-03),Phone分类(ids: 3, 4)下,总共销售了5个单位
在周四(2020-06-04),Phone分类(ids: 3, 4)下,总共销售了1个单位
在周五(2020-06-05),Book分类(ids: 1, 2)下,总共销售了10个单位,Glasses分类(ids: 5)下,总共销售了5个单位
在周六, 没有商品销售
在周天(2020-06-14, 2020-06-21),Phone分类(ids: 3, 4)下,总共销售了10个单位(5 + 5)
没有销售 T-Shirt 类别的商品

sql

select item_category as category, 
sum(case when num = 2 then quantity else 0 end) as Monday,
sum(case when num = 3 then quantity else 0 end) as Tuesday,
sum(case when num = 4 then quantity else 0 end) as Wednesday,
sum(case when num = 5 then quantity else 0 end) as Thursday,
sum(case when num = 6 then quantity else 0 end) as Friday,
sum(case when num = 7 then quantity else 0 end) as Saturday,
sum(case when num = 1 then quantity else 0 end) as Sunday
from
(select item_category, quantity,dayofweek(order_date) as num from 
items i left join orders o 
on i.item_id=o.item_id) t
group by item_category
order by item_category

1485. 按日期分组销售产品

难度简单

SQL架构

Activities

Code

+-------------+---------+
| 列名         | 类型    |
+-------------+---------+
| sell_date   | date    |
| product     | varchar |
+-------------+---------+
此表没有主键,它可能包含重复项。
此表的每一行都包含产品名称和在市场上销售的日期。

编写一个 SQL 查询来查找每个日期、销售的不同产品的数量及其名称。 每个日期的销售产品名称应按词典序排列。 返回按 sell_date 排序的结果表。

查询结果格式如下例所示。

Code

Activities 表:
+------------+-------------+
| sell_date  | product     |
+------------+-------------+
| 2020-05-30 | Headphone   |
| 2020-06-01 | Pencil      |
| 2020-06-02 | Mask        |
| 2020-05-30 | Basketball  |
| 2020-06-01 | Bible       |
| 2020-06-02 | Mask        |
| 2020-05-30 | T-Shirt     |
+------------+-------------+

Result 表:
+------------+----------+------------------------------+
| sell_date  | num_sold | products                     |
+------------+----------+------------------------------+
| 2020-05-30 | 3        | Basketball,Headphone,T-shirt |
| 2020-06-01 | 2        | Bible,Pencil                 |
| 2020-06-02 | 1        | Mask                         |
+------------+----------+------------------------------+
对于2020-05-30,出售的物品是 (Headphone, Basketball, T-shirt),按词典序排列,并用逗号 ',' 分隔。
对于2020-06-01,出售的物品是 (Pencil, Bible),按词典序排列,并用逗号分隔。
对于2020-06-02,出售的物品是 (Mask),只需返回该物品名。

sql

select sell_date, count(distinct product) num_sold, 
    group_concat(distinct product order by product) products
from Activities
group by sell_date

行转列

1495. 上月播放的儿童适宜电影

难度简单

SQL架构

表: TVProgram

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| program_date  | date    |
| content_id    | int     |
| channel       | varchar |
+---------------+---------+
(program_date, content_id) 是该表主键.
该表包含电视上的节目信息.
content_id 是电视一些频道上的节目的 id.

表: Content

Code

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| content_id       | varchar |
| title            | varchar |
| Kids_content     | enum    |
| content_type     | varchar |
+------------------+---------+
content_id 是该表主键.
Kids_content 是枚举类型, 取值为('Y', 'N'), 其中: 
'Y' 表示儿童适宜内容, 而'N'表示儿童不宜内容.
content_type 表示内容的类型, 比如电影, 电视剧等.

写一个 SQL 语句, 报告在 2020 年 6 月份播放的儿童适宜电影的去重电影名.

返回的结果表单没有顺序要求.

查询结果的格式如下例所示.

Code

TVProgram 表:
+--------------------+--------------+-------------+
| program_date       | content_id   | channel     |
+--------------------+--------------+-------------+
| 2020-06-10 08:00   | 1            | LC-Channel  |
| 2020-05-11 12:00   | 2            | LC-Channel  |
| 2020-05-12 12:00   | 3            | LC-Channel  |
| 2020-05-13 14:00   | 4            | Disney Ch   |
| 2020-06-18 14:00   | 4            | Disney Ch   |
| 2020-07-15 16:00   | 5            | Disney Ch   |
+--------------------+--------------+-------------+

Content 表:
+------------+----------------+---------------+---------------+
| content_id | title          | Kids_content  | content_type  |
+------------+----------------+---------------+---------------+
| 1          | Leetcode Movie | N             | Movies        |
| 2          | Alg. for Kids  | Y             | Series        |
| 3          | Database Sols  | N             | Series        |
| 4          | Aladdin        | Y             | Movies        |
| 5          | Cinderella     | Y             | Movies        |
+------------+----------------+---------------+---------------+

Result 表:
+--------------+
| title        |
+--------------+
| Aladdin      |
+--------------+
"Leetcode Movie" 是儿童不宜的电影.
"Alg. for Kids" 不是电影.
"Database Sols" 不是电影
"Alladin" 是电影, 儿童适宜, 并且在 2020 年 6 月份播放.
"Cinderella" 不在 2020 年 6 月份播放.

sql

select distinct title   
from TVProgram t left join Content c
on t.content_id  = c.content_id 
where  Kids_content ='Y' 
and date_format(program_date ,'%Y-%m')='2020-06'
and content_type='Movies'

LEFT()函数参见:https://www.begtut.com/sql/func-mysql-left.html REGEXP语法参见: https://www.cnblogs.com/timssd/p/5882742.html https://www.cnblogs.com/zhaopanpan/p/10133224.html DATE_FORMAT()函数参见:https://www.w3school.com.cn/sql/func_date_format.asp EXTRACT()函数参见:https://www.runoob.com/sql/func-extract.html DATEDIFF()函数参见:https://www.runoob.com/sql/func-datediff-mysql.html YEAR()函数参见:https://blog.csdn.net/moakun/article/details/82528829 MONTH()函数参见:https://www.yiibai.com/mysql/month.html

1501. 可以放心投资的国家

难度中等

SQL架构

Person:

Code

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| id             | int     |
| name           | varchar |
| phone_number   | varchar |
+----------------+---------+
id 是该表主键.
该表每一行包含一个人的名字和电话号码.
电话号码的格式是:'xxx-yyyyyyy', 其中xxx是国家码(3个字符), yyyyyyy是电话号码(7个字符), x和y都表示数字. 同时, 国家码和电话号码都可以包含前导0.

Country:

Code

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| name           | varchar |
| country_code   | varchar |
+----------------+---------+
country_code是该表主键.
该表每一行包含国家名和国家码. country_code的格式是'xxx', x是数字.

Calls:

Code

+-------------+------+
| Column Name | Type |
+-------------+------+
| caller_id   | int  |
| callee_id   | int  |
| duration    | int  |
+-------------+------+
该表无主键, 可能包含重复行.
每一行包含呼叫方id, 被呼叫方id和以分钟为单位的通话时长. caller_id != callee_id

一家电信公司想要投资新的国家. 该公司想要投资的国家是: 该国的平均通话时长要严格地大于全球平均通话时长.

写一段 SQL, 找到所有该公司可以投资的国家.

返回的结果表没有顺序要求.

查询的结果格式如下例所示.

Code

Person 表:
+----+----------+--------------+
| id | name     | phone_number |
+----+----------+--------------+
| 3  | Jonathan | 051-1234567  |
| 12 | Elvis    | 051-7654321  |
| 1  | Moncef   | 212-1234567  |
| 2  | Maroua   | 212-6523651  |
| 7  | Meir     | 972-1234567  |
| 9  | Rachel   | 972-0011100  |
+----+----------+--------------+

Country 表:
+----------+--------------+
| name     | country_code |
+----------+--------------+
| Peru     | 051          |
| Israel   | 972          |
| Morocco  | 212          |
| Germany  | 049          |
| Ethiopia | 251          |
+----------+--------------+

Calls 表:
+-----------+-----------+----------+
| caller_id | callee_id | duration |
+-----------+-----------+----------+
| 1         | 9         | 33       |
| 2         | 9         | 4        |
| 1         | 2         | 59       |
| 3         | 12        | 102      |
| 3         | 12        | 330      |
| 12        | 3         | 5        |
| 7         | 9         | 13       |
| 7         | 1         | 3        |
| 9         | 7         | 1        |
| 1         | 7         | 7        |
+-----------+-----------+----------+

Result 表:
+----------+
| country  |
+----------+
| Peru     |
+----------+
国家Peru的平均通话时长是 (102 + 102 + 330 + 330 + 5 + 5) / 6 = 145.666667
国家Israel的平均通话时长是 (33 + 4 + 13 + 13 + 3 + 1 + 1 + 7) / 8 = 9.37500
国家Morocco的平均通话时长是 (33 + 4 + 59 + 59 + 3 + 7) / 6 = 27.5000 
全球平均通话时长 = (2 * (33 + 3 + 59 + 102 + 330 + 5 + 13 + 3 + 1 + 7)) / 20 = 55.70000
所以, Peru是唯一的平均通话时长大于全球平均通话时长的国家, 也是唯一的推荐投资的国家.

笛卡尔积

sql

select c2.name as country 
from Calls c1,Person p,Country c2
where (p.id=c1.caller_id or p.id=c1.callee_id) and c2.country_code=left(p.phone_number,3)
group by c2.name 
having avg(duration)>(select avg(duration) from Calls)

思路更清晰

sql

with people_country as
(
    select id, c.name country
    from Person p left join Country c
    on left(p.phone_number,3) = c.country_code
)

select country
from
(
    select country, avg(duration) avgtime
    from
    (
        select caller_id id, duration
        from Calls
        union all
        select callee_id, duration
        from Calls
    ) t left join people_country
    using(id)
    group by country
) temp
where avgtime > 
    (
        select avg(duration) avgtime
        from
        (
            select caller_id, duration
            from Calls
            union all
            select callee_id, duration
            from Calls
        ) t
    )

1511. 消费者下单频率

难度简单

SQL架构

表: Customers

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
| country       | varchar |
+---------------+---------+
customer_id 是该表主键.
该表包含公司消费者的信息.

表: Product

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| description   | varchar |
| price         | int     |
+---------------+---------+
product_id 是该表主键.
该表包含公司产品的信息.
price 是本产品的花销.

表: Orders

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| customer_id   | int     |
| product_id    | int     |
| order_date    | date    |
| quantity      | int     |
+---------------+---------+
order_id 是该表主键.
该表包含消费者下单的信息.
customer_id 是买了数量为"quantity", id为"product_id"产品的消费者的 id.
Order_date 是订单发货的日期, 格式为('YYYY-MM-DD').

写一个 SQL 语句, 报告消费者的 id 和名字, 其中消费者在 2020 年 6 月和 7 月, 每月至少花费了$100.

结果表无顺序要求.

查询结果格式如下例所示.

Code

Customers
+--------------+-----------+-------------+
| customer_id  | name      | country     |
+--------------+-----------+-------------+
| 1            | Winston   | USA         |
| 2            | Jonathan  | Peru        |
| 3            | Moustafa  | Egypt       |
+--------------+-----------+-------------+

Product
+--------------+-------------+-------------+
| product_id   | description | price       |
+--------------+-------------+-------------+
| 10           | LC Phone    | 300         |
| 20           | LC T-Shirt  | 10          |
| 30           | LC Book     | 45          |
| 40           | LC Keychain | 2           |
+--------------+-------------+-------------+

Orders
+--------------+-------------+-------------+-------------+-----------+
| order_id     | customer_id | product_id  | order_date  | quantity  |
+--------------+-------------+-------------+-------------+-----------+
| 1            | 1           | 10          | 2020-06-10  | 1         |
| 2            | 1           | 20          | 2020-07-01  | 1         |
| 3            | 1           | 30          | 2020-07-08  | 2         |
| 4            | 2           | 10          | 2020-06-15  | 2         |
| 5            | 2           | 40          | 2020-07-01  | 10        |
| 6            | 3           | 20          | 2020-06-24  | 2         |
| 7            | 3           | 30          | 2020-06-25  | 2         |
| 9            | 3           | 30          | 2020-05-08  | 3         |
+--------------+-------------+-------------+-------------+-----------+

Result 表:
+--------------+------------+
| customer_id  | name       |  
+--------------+------------+
| 1            | Winston    |
+--------------+------------+ 
Winston 在2020年6月花费了$300(300 * 1), 在7月花费了$100(10 * 1 + 45 * 2).
Jonathan 在2020年6月花费了$600(300 * 2), 在7月花费了$20(2 * 10).
Moustafa 在2020年6月花费了$110 (10 * 2 + 45 * 2), 在7月花费了$0.

sql

select customer_id,name
from Customers
where customer_id in
(select customer_id
    from
        (select customer_id, month(order_date) as month , sum(quantity*price) as total
        from Orders o left join Product p on o.product_id = p.product_id
        where month(order_date) = 6 or month(order_date)=7
        group by customer_id,month(order_date)
        ) as t1
    where total >=100
    group by customer_id
    having count(*)>=2
)

1517. Find Users With Valid E-Mails

难度简单

SQL架构

Table: Users

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| name          | varchar |
| mail          | varchar |
+---------------+---------+
user_id is the primary key for this table.
This table contains information of the users signed up in a website. Some e-mails are invalid.

Write an SQL query to find the users who have valid emails.

A valid e-mail has a prefix name and a domain where:

  • The prefix name is a string that may contain letters (upper or lower case), digits, underscore '_', period '.' and/or dash '-'. The prefix name must start with a letter.
  • The domain is '@leetcode.com'.

Return the result table in any order.

The query result format is in the following example.

Code

Users
+---------+-----------+-------------------------+
| user_id | name      | mail                    |
+---------+-----------+-------------------------+
| 1       | Winston   | winston@leetcode.com    |
| 2       | Jonathan  | jonathanisgreat         |
| 3       | Annabelle | bella-@leetcode.com     |
| 4       | Sally     | sally.come@leetcode.com |
| 5       | Marwan    | quarz#2020@leetcode.com |
| 6       | David     | david69@gmail.com       |
| 7       | Shapiro   | .shapo@leetcode.com     |
+---------+-----------+-------------------------+

Result table:
+---------+-----------+-------------------------+
| user_id | name      | mail                    |
+---------+-----------+-------------------------+
| 1       | Winston   | winston@leetcode.com    |
| 3       | Annabelle | bella-@leetcode.com     |
| 4       | Sally     | sally.come@leetcode.com |
+---------+-----------+-------------------------+
The mail of user 2 doesn't have a domain.
The mail of user 5 has # sign which is not allowed.
The mail of user 6 doesn't have leetcode domain.
The mail of user 7 starts with a period.

考察正则表达式的使用

sql

SELECT *
FROM Users
WHERE mail REGEXP '^[a-zA-Z]+[\\w_\\.\\-]*@leetcode.com$'   
ORDER BY user_id;

sql

select * from Users
where mail regexp '^[a-zA-Z]+[a-zA-Z0-9_\\./\\-]{0,}@leetcode.com$'
order by user_id

坑点: 1、前缀可能是一个字母,比如“J@leetcode.com”,所以匹配非首字母外的前缀字符数量要用{0,}或*,不能用+。 2、题意要求:underscore ‘’, period ‘.’ and/or dash ‘-‘,/没加单引号,不留神可能写漏/。 3、后缀可能是“@leetcodeecom”,所以要对“.”加转义符号。 4、后缀可能是“@LEETCODE.COM”,默认是不区分大小写匹配,所以要加上“BINARY”区分大小写。 语法: 1、https://www.cnblogs.com/timssd/p/5882742.html 2、https://www.cnblogs.com/zhaopanpan/p/10133224.html 3、”双反斜杠+w”表示字母、数字、下划线,相对”a-zA-Z0-9”的写法更简洁。

1527. Patients With a Condition

难度简单

SQL架构

Table: Patients

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| patient_id   | int     |
| patient_name | varchar |
| conditions   | varchar |
+--------------+---------+
patient_id is the primary key for this table.
'conditions' contains 0 or more code separated by spaces. 
This table contains information of the patients in the hospital.

Write an SQL query to report the patient_id, patient_name all conditions of patients who have Type I Diabetes. Type I Diabetes always starts with DIAB1 prefix

Return the result table in any order.

The query result format is in the following example.

Code

Patients
+------------+--------------+--------------+
| patient_id | patient_name | conditions   |
+------------+--------------+--------------+
| 1          | Daniel       | YFEV COUGH   |
| 2          | Alice        |              |
| 3          | Bob          | DIAB100 MYOP |
| 4          | George       | ACNE DIAB100 |
| 5          | Alain        | DIAB201      |
+------------+--------------+--------------+

Result table:
+------------+--------------+--------------+
| patient_id | patient_name | conditions   |
+------------+--------------+--------------+
| 3          | Bob          | DIAB100 MYOP |
| 4          | George       | ACNE DIAB100 | 
+------------+--------------+--------------+
Bob and George both have a condition that starts with DIAB1.

sql

select  patient_id , patient_name ,conditions 
from Patients
where conditions like '%DIAB1%'

1532. The Most Recent Three Orders

难度中等

SQL架构

Table: Customers

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
+---------------+---------+
customer_id is the primary key for this table.
This table contains information about customers.

Table: Orders

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| customer_id   | int     |
| cost          | int     |
+---------------+---------+
order_id is the primary key for this table.
This table contains information about the orders made by customer_id.
Each customer has one order per day.

Write an SQL query to find the most recent 3 orders of each user. If a user ordered less than 3 orders return all of their orders.

Return the result table sorted by customer_name in ascending order and in case of a tie by the customer_id in ascending order. If there still a tie, order them by the order_date in descending order.

The query result format is in the following example:

Code

Customers
+-------------+-----------+
| customer_id | name      |
+-------------+-----------+
| 1           | Winston   |
| 2           | Jonathan  |
| 3           | Annabelle |
| 4           | Marwan    |
| 5           | Khaled    |
+-------------+-----------+

Orders
+----------+------------+-------------+------+
| order_id | order_date | customer_id | cost |
+----------+------------+-------------+------+
| 1        | 2020-07-31 | 1           | 30   |
| 2        | 2020-07-30 | 2           | 40   |
| 3        | 2020-07-31 | 3           | 70   |
| 4        | 2020-07-29 | 4           | 100  |
| 5        | 2020-06-10 | 1           | 1010 |
| 6        | 2020-08-01 | 2           | 102  |
| 7        | 2020-08-01 | 3           | 111  |
| 8        | 2020-08-03 | 1           | 99   |
| 9        | 2020-08-07 | 2           | 32   |
| 10       | 2020-07-15 | 1           | 2    |
+----------+------------+-------------+------+

Result table:
+---------------+-------------+----------+------------+
| customer_name | customer_id | order_id | order_date |
+---------------+-------------+----------+------------+
| Annabelle     | 3           | 7        | 2020-08-01 |
| Annabelle     | 3           | 3        | 2020-07-31 |
| Jonathan      | 2           | 9        | 2020-08-07 |
| Jonathan      | 2           | 6        | 2020-08-01 |
| Jonathan      | 2           | 2        | 2020-07-30 |
| Marwan        | 4           | 4        | 2020-07-29 |
| Winston       | 1           | 8        | 2020-08-03 |
| Winston       | 1           | 1        | 2020-07-31 |
| Winston       | 1           | 10       | 2020-07-15 |
+---------------+-------------+----------+------------+
Winston has 4 orders, we discard the order of "2020-06-10" because it is the oldest order.
Annabelle has only 2 orders, we return them.
Jonathan has exactly 3 orders.
Marwan ordered only one time.
We sort the result table by customer_name in ascending order, by customer_id in ascending order and by order_date in descending order in case of a tie.

Follow-up: Can you write a general solution for the most recent n orders?

sql

select name customer_name ,customer_id,order_id,order_date
from (
select  name ,o.customer_id,order_id,order_date ,rank()over(partition by o.customer_id order by order_date desc) rk
from Orders o left join Customers c
on o.customer_id=c.customer_id
)t1
where rk <=3
order by customer_name ,customer_id,order_date desc

1543. Fix Product Name Format

难度简单

SQL架构

Table: Sales

Code

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| sale_id      | int     |
| product_name | varchar |
| sale_date    | date    |
+--------------+---------+
sale_id is the primary key for this table.
Each row of this table contains the product name and the date it was sold.

Since table Sales was filled manually in the year 2000, product_name may contain leading and/or trailing white spaces, also they are case-insensitive.

Write an SQL query to report

  • product_name in lowercase without leading or trailing white spaces.
  • sale_date in the format ('YYYY-MM')
  • total the number of times the product was sold in this month.

Return the result table ordered by product_name in ascending order, in case of a tie order it by sale_date in ascending order.

The query result format is in the following example.

Code

Sales
+------------+------------------+--------------+
| sale_id    | product_name     | sale_date    |
+------------+------------------+--------------+
| 1          |      LCPHONE     | 2000-01-16   |
| 2          |    LCPhone       | 2000-01-17   |
| 3          |     LcPhOnE      | 2000-02-18   |
| 4          |      LCKeyCHAiN  | 2000-02-19   |
| 5          |   LCKeyChain     | 2000-02-28   |
| 6          | Matryoshka       | 2000-03-31   | 
+------------+------------------+--------------+

Result table:
+--------------+--------------+----------+
| product_name | sale_date    | total    |
+--------------+--------------+----------+
| lcphone      | 2000-01      | 2        |
| lckeychain   | 2000-02      | 2        | 
| lcphone      | 2000-02      | 1        | 
| matryoshka   | 2000-03      | 1        | 
+--------------+--------------+----------+

In January, 2 LcPhones were sold, please note that the product names are not case sensitive and may contain spaces.
In Februery, 2 LCKeychains and 1 LCPhone were sold.
In March, 1 matryoshka was sold.

sql

select trim(lower(product_name)) as product_name, 
        date_format(sale_date,'%Y-%m') as sale_date,
        count(*) as total 
from Sales 
group by trim(lower(product_name)), date_format(sale_date,'%Y-%m') 
order by product_name asc, sale_date asc

注意大小写、空格

1549. The Most Recent Orders for Each Product

难度中等

SQL架构

Table: Customers

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
+---------------+---------+
customer_id is the primary key for this table.
This table contains information about the customers.

Table: Orders

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| customer_id   | int     |
| product_id    | int     |
+---------------+---------+
order_id is the primary key for this table.
This table contains information about the orders made by customer_id.
There will be no product ordered by the same user more than once in one day.

Table: Products

Code

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| product_name  | varchar |
| price         | int     |
+---------------+---------+
product_id is the primary key for this table.
This table contains information about the Products.

Write an SQL query to find the most recent order(s) of each product.

Return the result table sorted by product_name in ascending order and in case of a tie by the product_id in ascending order. If there still a tie, order them by the order_id in ascending order.

The query result format is in the following example:

Code

Customers
+-------------+-----------+
| customer_id | name      |
+-------------+-----------+
| 1           | Winston   |
| 2           | Jonathan  |
| 3           | Annabelle |
| 4           | Marwan    |
| 5           | Khaled    |
+-------------+-----------+

Orders
+----------+------------+-------------+------------+
| order_id | order_date | customer_id | product_id |
+----------+------------+-------------+------------+
| 1        | 2020-07-31 | 1           | 1          |
| 2        | 2020-07-30 | 2           | 2          |
| 3        | 2020-08-29 | 3           | 3          |
| 4        | 2020-07-29 | 4           | 1          |
| 5        | 2020-06-10 | 1           | 2          |
| 6        | 2020-08-01 | 2           | 1          |
| 7        | 2020-08-01 | 3           | 1          |
| 8        | 2020-08-03 | 1           | 2          |
| 9        | 2020-08-07 | 2           | 3          |
| 10       | 2020-07-15 | 1           | 2          |
+----------+------------+-------------+------------+

Products
+------------+--------------+-------+
| product_id | product_name | price |
+------------+--------------+-------+
| 1          | keyboard     | 120   |
| 2          | mouse        | 80    |
| 3          | screen       | 600   |
| 4          | hard disk    | 450   |
+------------+--------------+-------+

Result table:
+--------------+------------+----------+------------+
| product_name | product_id | order_id | order_date |
+--------------+------------+----------+------------+
| keyboard     | 1          | 6        | 2020-08-01 |
| keyboard     | 1          | 7        | 2020-08-01 |
| mouse        | 2          | 8        | 2020-08-03 |
| screen       | 3          | 3        | 2020-08-29 |
+--------------+------------+----------+------------+
keyboard's most recent order is in 2020-08-01, it was ordered two times this day.
mouse's most recent order is in 2020-08-03, it was ordered only once this day.
screen's most recent order is in 2020-08-29, it was ordered only once this day.
The hard disk was never ordered and we don't include it in the result table.

sql

select product_name,product_id,order_id,order_date
from
(
select product_name ,o.product_id ,order_id,order_date ,
    rank() over(partition by o.product_id order by order_date desc) rk
from Orders o left join Products p
on o.product_id =p.product_id 
)t1
where rk =1
order by product_name,product_id,order_id

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