前言

一、题目：1141. 查询近30天活跃用户数

活动记录表：Activity

+---------------+---------+
| 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）。

以任意顺序返回结果表。

查询结果示例如下。

示例 1:

输入：
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   |
+---------+------------+---------------+---------------+
输出：
+------------+--------------+ 
| day        | active_users |
+------------+--------------+ 
| 2019-07-20 | 2            |
| 2019-07-21 | 2            |
+------------+--------------+ 
解释：注意非活跃用户的记录不需要展示。

二、解题

1.正确示范①

提交SQL

select activity_date day,count(distinct user_id) active_users
from Activity
where activity_date>'2019-06-27' and activity_date<='2019-07-27'
# activity_date>='2019-06-28' and activity_date<='2019-07-27'
group by activity_date

或者

select activity_date day,count(distinct user_id) active_users
from Activity
where activity_date>DATE_SUB('2019-07-27',INTERVAL 30 DAY)
and activity_date<='2019-07-27'
group by activity_date

或者

select activity_date day,count(distinct user_id) active_users
from Activity
where DATEDIFF('2019-07-27',activity_date)>=0 
and DATEDIFF('2019-07-27',activity_date) <30 
group by activity_date

运行结果

2.正确示范②

提交SQL

select activity_date day,count(distinct user_id) active_users
from Activity
where activity_date between '2019-06-28' and '2019-07-27'
# activity_date between DATE_SUB('2019-07-27',INTERVAL 29 DAY) and '2019-07-27'
group by activity_date

运行结果

3.正确示范③

提交SQL

select activity_date day,count(1) active_users
from(
select distinct activity_date,user_id
from Activity
where activity_date between '2019-06-28' and '2019-07-27'
) u
group by activity_date

运行结果

4.其他

总结

正确示范①思路：
先按活跃日期筛选
activity_date>‘2019-06-27’ and activity_date<=‘2019-07-27’
或者 activity_date>DATE_SUB(‘2019-07-27’,INTERVAL 30 DAY) and activity_date<=‘2019-07-27’
或者 DATEDIFF(‘2019-07-27’,activity_date)>=0 and DATEDIFF(‘2019-07-27’,activity_date) <30 ，
再按活跃日期分组 group by activity_date，
再取去重用户数 count(distinct user_id)；
正确示范②思路：
将正确示范①中活跃日期筛选改为
activity_date between ‘2019-06-28’ and ‘2019-07-27’ ，其他不变；
正确示范③思路：
先按活跃日期筛选，将数据按distinct activity_date,user_id去重后，
再按活跃日期分组 group by activity_date，
再取用户数 count(1)。