SQL面试题练习 —— 找出所有连续未登录5天及以上的用户并提取出这些用户最近一次登录的日期

1 题目

找出所有连续未登录5天及以上的用户并提取出这些用户最近一次登录的日期

样例数据

+---------------------+------------------------+--+
| user_login.user_id  | user_login.login_date  |
+---------------------+------------------------+--+
| 1                   | 2022-01-01             |
| 1                   | 2022-01-02             |
| 1                   | 2022-01-03             |
| 1                   | 2022-01-05             |
| 1                   | 2022-01-06             |
| 1                   | 2022-01-09             |
| 1                   | 2023-01-01             |
| 2                   | 2022-01-01             |
| 2                   | 2022-01-03             |
| 2                   | 2022-01-04             |
| 2                   | 2022-01-06             |
| 2                   | 2022-01-07             |
| 2                   | 2022-01-08             |
| 3                   | 2022-01-01             |
| 3                   | 2022-01-02             |
| 3                   | 2022-01-04             |
| 3                   | 2022-01-05             |
| 3                   | 2022-01-07             |
| 3                   | 2022-01-08             |
+---------------------+------------------------+--+

2 建表语句

-- 创建用户登录数据表
CREATE TABLE user_login(user_id INT,login_date DATE
);-- 插入模拟数据
INSERT INTO user_login VALUES
(1, '2022-01-01'),
(1, '2022-01-02'),
(1, '2022-01-03'),
(1, '2022-01-05'),
(1, '2022-01-06'),
(1, '2022-01-09'),
(1, '2023-01-01'),
(2, '2022-01-01'),
(2, '2022-01-03'),
(2, '2022-01-04'),
(2, '2022-01-06'),
(2, '2022-01-07'),
(2, '2022-01-08'),
(3, '2022-01-01'),
(3, '2022-01-02'),
(3, '2022-01-04'),
(3, '2022-01-05'),
(3, '2022-01-07'),
(3, '2022-01-08');

3 题解

1. 计算本次登录日期与上一次登录日期差值

select user_id,login_date,datediff(login_date, lag(login_date) over (partition by user_id order by login_date )) as dt
from user_login

执行结果

+----------+-------------+-------+--+
| user_id  | login_date  |  dt   |
+----------+-------------+-------+--+
| 1        | 2022-01-01  | NULL  |
| 1        | 2022-01-02  | 1     |
| 1        | 2022-01-03  | 1     |
| 1        | 2022-01-05  | 2     |
| 1        | 2022-01-06  | 1     |
| 1        | 2022-01-09  | 3     |
| 1        | 2023-01-01  | 357   |
| 2        | 2022-01-01  | NULL  |
| 2        | 2022-01-03  | 2     |
| 2        | 2022-01-04  | 1     |
| 2        | 2022-01-06  | 2     |
| 2        | 2022-01-07  | 1     |
| 2        | 2022-01-08  | 1     |
| 3        | 2022-01-01  | NULL  |
| 3        | 2022-01-02  | 1     |
| 3        | 2022-01-04  | 2     |
| 3        | 2022-01-05  | 1     |
| 3        | 2022-01-07  | 2     |
| 3        | 2022-01-08  | 1     |
+----------+-------------+-------+--+

2. 计算每个用户最近一次登录日期

select  user_id,max(login_date) recent_login_date
from user_login
group by user_id

执行结果

+----------+--------------------+--+
| user_id  | recent_login_date  |
+----------+--------------------+--+
| 1        | 2023-01-01         |
| 2        | 2022-01-08         |
| 3        | 2022-01-08         |
+----------+--------------------+--+

3. 合并上述两张表

select t1.user_id,t1.login_date,t1.dt,t2.user_id,t2.recent_login_date
from (select user_id,login_date,datediff(login_date, lag(login_date) over (partition by user_id order by login_date )) as dtfrom user_login) t1left join(select  user_id,max(login_date) recent_login_datefrom user_logingroup by user_id) t2on t1.user_id = t2.user_id

执行结果

+-------------+----------------+--------+-------------+-----------------------+--+
| t1.user_id  | t1.login_date  | t1.dt  | t2.user_id  | t2.recent_login_date  |
+-------------+----------------+--------+-------------+-----------------------+--+
| 1           | 2022-01-01     | NULL   | 1           | 2023-01-01            |
| 1           | 2022-01-02     | 1      | 1           | 2023-01-01            |
| 1           | 2022-01-03     | 1      | 1           | 2023-01-01            |
| 1           | 2022-01-05     | 2      | 1           | 2023-01-01            |
| 1           | 2022-01-06     | 1      | 1           | 2023-01-01            |
| 1           | 2022-01-09     | 3      | 1           | 2023-01-01            |
| 1           | 2023-01-01     | 357    | 1           | 2023-01-01            |
| 2           | 2022-01-01     | NULL   | 2           | 2022-01-08            |
| 2           | 2022-01-03     | 2      | 2           | 2022-01-08            |
| 2           | 2022-01-04     | 1      | 2           | 2022-01-08            |
| 2           | 2022-01-06     | 2      | 2           | 2022-01-08            |
| 2           | 2022-01-07     | 1      | 2           | 2022-01-08            |
| 2           | 2022-01-08     | 1      | 2           | 2022-01-08            |
| 3           | 2022-01-01     | NULL   | 3           | 2022-01-08            |
| 3           | 2022-01-02     | 1      | 3           | 2022-01-08            |
| 3           | 2022-01-04     | 2      | 3           | 2022-01-08            |
| 3           | 2022-01-05     | 1      | 3           | 2022-01-08            |
| 3           | 2022-01-07     | 2      | 3           | 2022-01-08            |
| 3           | 2022-01-08     | 1      | 3           | 2022-01-08            |
+-------------+----------------+--------+-------------+-----------------------+--+

4. 找出所有连续未登录5天及以上的用户

select t1.user_id,t2.recent_login_date
from (select user_id,login_date,datediff(login_date, lag(login_date) over (partition by user_id order by login_date )) as dtfrom user_login) t1left join(select  user_id,max(login_date) recent_login_datefrom user_logingroup by user_id) t2on t1.user_id = t2.user_id
where t1.dt >= 5;

执行结果

+-------------+-----------------------+--+
| t1.user_id  | t2.recent_login_date  |
+-------------+-----------------------+--+
| 1           | 2023-01-01            |
+-------------+-----------------------+--+