1141. 查询近30天活跃用户数

发表于 更新于

难度系数： 简单
活动记录表：Activity

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+---------------+---------+
| 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。

示例 1：

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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。

SQL结构

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Create table If Not Exists Activity (user_id int, session_id int, activity_date date, activity_type ENUM('open_session', 'end_session', 'scroll_down', 'send_message'));
Truncate table Activity;
insert into Activity (user_id, session_id, activity_date, activity_type) values ('1', '1', '2019-07-20', 'open_session');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('1', '1', '2019-07-20', 'scroll_down');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('1', '1', '2019-07-20', 'end_session');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('2', '4', '2019-07-20', 'open_session');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('2', '4', '2019-07-21', 'send_message');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('2', '4', '2019-07-21', 'end_session');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('3', '2', '2019-07-21', 'open_session');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('3', '2', '2019-07-21', 'send_message');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('3', '2', '2019-07-21', 'end_session');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('4', '3', '2019-06-25', 'open_session');
insert into Activity (user_id, session_id, activity_date, activity_type) values ('4', '3', '2019-06-25', 'end_session');

解法：

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SELECT activity_date AS `day`, COUNT(DISTINCT user_id) AS active_users
FROM Activity
WHERE activity_date BETWEEN '2019-06-28' AND '2019-07-27'
GROUP BY activity_date;

原题链接：https://leetcode.cn/problems/user-activity-for-the-past-30-days-i/