书籍表 Books:
+----------------+---------+
| Column Name | Type |
+----------------+---------+
| book_id | int |
| name | varchar |
| available_from | date |
+----------------+---------+
book_id 是这个表的主键。
订单表 Orders:
+----------------+---------+
| 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 的 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 。
下面是样例输出结果:
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" |
+-----------+--------------------+
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/unpopular-books 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
# Write your MySQL query statement below
select b.book_id, name
from Books b left join
(
select book_id, sum(quantity) sell
from Orders
where datediff('2019-06-23', dispatch_date) <= 365
group by book_id
) t
using(book_id)
where available_from <= '2019-05-23'
and ifnull(t.sell, 0) < 10