LeetCode MySQL 1127. 用户购买平台 *
LeetCode MySQL 1127. 用户购买平台 *
Michael阿明
发布于 2021-02-19 10:43:25
发布于 2021-02-19 10:43:25
文章被收录于专栏：Michael阿明学习之路
文章目录
1. 题目
支出表: Spending
+-------------+---------+
| 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 仅 使用手机端用户、仅 使用桌面端用户、 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：
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。来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/user-purchase-platform 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。
2. 解题
- 先造出表的各种组合
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{"headers": ["spend_date", "platform"],
"values": [
["2019-07-01", "desktop"],
["2019-07-02", "desktop"],
["2019-07-01", "mobile"],
["2019-07-02", "mobile"],
["2019-07-01", "both"],
["2019-07-02", "both"]]}- 计算每天，某类属下的总金额、人数
select spend_date,
if(count(distinct platform)=1, platform, 'both') plat,
sum(amount) total_am,
count(distinct user_id) total_u # 1 total_u 这么写也对，就1个人
from Spending
group by spend_date, user_id{"headers": ["spend_date", "plat", "total_am", "total_u"],
"values": [
["2019-07-01", "both", 200, 1],
["2019-07-01", "mobile", 100, 1],
["2019-07-01", "desktop", 100, 1],
["2019-07-02", "mobile", 100, 1],
["2019-07-02", "desktop", 100, 1]]}- 上面2表连接
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{"headers": ["spend_date", "platform", "total_am", "total_u"], 。
"values": [
["2019-07-01", "desktop", 100, 1],
["2019-07-02", "desktop", 100, 1],
["2019-07-01", "mobile", 100, 1],
["2019-07-02", "mobile", 100, 1],
["2019-07-01", "both", 200, 1],
["2019-07-02", "both", null, null]]}- 对连接后的表，求和
# Write your MySQL query statement below
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本文参与 腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。
原始发表：2020/08/01 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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