Leetcode1107. 每日新用户统计（中等）

题目
Traffic 表：

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| activity      | enum    |
| activity_date | date    |
+---------------+---------+

该表没有主键，它可能有重复的行。
activity 列是 ENUM 类型，可能取 ('login', 'logout', 'jobs', 'groups', 'homepage') 几个值之一。

编写一个 SQL 查询，以查询从今天起最多 90 天内，每个日期该日期首次登录的用户数。假设今天是 2019-06-30.

查询结果格式如下例所示：

Traffic 表：

+---------+----------+---------------+
| user_id | activity | activity_date |
+---------+----------+---------------+
| 1       | login    | 2019-05-01    |
| 1       | homepage | 2019-05-01    |
| 1       | logout   | 2019-05-01    |
| 2       | login    | 2019-06-21    |
| 2       | logout   | 2019-06-21    |
| 3       | login    | 2019-01-01    |
| 3       | jobs     | 2019-01-01    |
| 3       | logout   | 2019-01-01    |
| 4       | login    | 2019-06-21    |
| 4       | groups   | 2019-06-21    |
| 4       | logout   | 2019-06-21    |
| 5       | login    | 2019-03-01    |
| 5       | logout   | 2019-03-01    |
| 5       | login    | 2019-06-21    |
| 5       | logout   | 2019-06-21    |
+---------+----------+---------------+

Result 表：

+------------+-------------+
| login_date | user_count  |
+------------+-------------+
| 2019-05-01 | 1           |
| 2019-06-21 | 2           |
+------------+-------------+

请注意，我们只关心用户数非零的日期.
ID 为 5 的用户第一次登陆于 2019-03-01，因此他不算在 2019-06-21 的的统计内。

解答
先查询每个用户的首次登陆

select T.user_id, min(T.activity_date) as first_date
from Traffic as T
where T.activity = 'login'
group by T.user_id

再去掉超过90 天的情况

select *
from (select T.user_id, min(T.activity_date) as first_date
from Traffic as T
where T.activity = 'login'
group by T.user_id) as tmp
where datediff(tmp.first_date, '2019-06-30') <= 90

然后对activity_date分组 统计user_id的数量即可

select tmp.first_date as login_date, count(tmp.user_id) as user_count
from (select T.user_id, min(T.activity_date) as first_date
from Traffic as T
where T.activity = 'login'
group by T.user_id) as tmp
where datediff(tmp.first_date, '2019-06-30') <= 90
group by tmp.first_date
order by tmp.first_date;