MySQL错题整理(1127. 用户购买平台)
题目描述: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:
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。
select t2.spend_date, t2.platform,
ifnull(sum(amount),0) total_amount, ifnull(count(user_id),0) total_users
from
(select distinct spend_date, "desktop" as platform from Spending
union
select distinct spend_date, "mobile" as platform from Spending
union
select distinct spend_date, "both" as platform from Spending
) t2
left join
(select spend_date, sum(amount) amount, user_id,
case when count(*) = 1 then platform else "both" end as platform
from Spending
group by spend_date, user_id) t1
on t1.spend_date = t2.spend_date
and t1.platform = t2. platform
group by t2.spend_date, t2.platform
注意事项:
1.注意union的使用
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