LeetCode：Database 57.过去30天的用户活动 II

Posted 2021-06-24 Xiao Miao

要求：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。

Activity 表的结构

+---------------+---------+
| 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 表:

+---------+------------+---------------+---------------+
| 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语句：

select ifnull(round(sum(c)/count(*),2),0)as average_sessions_per_user
from(
select count(distinct user_id,session_id) as c
from activity 
where activity_date between adddate('2019-07-27',-29) and '2019-07-27'
group by user_id)a;