Question

我有一个MySQL DB表，用于记录用户登录和注销活动。我需要选择注销后20分钟内未重新登录的所有用户记录。

例如：表格：日志

ID | User | Event  | Time
-----------------------------
1  | 1    | LOGIN  | 10:00:00
2  | 2    | LOGIN  | 10:05:00
3  | 3    | LOGIN  | 10:15:00
4  | 1    | LOGOUT | 11:00:00
5  | 3    | LOGOUT | 11:01:00
6  | 2    | LOGIN  | 12:20:00
7  | 2    | LOGOUT | 12:30:00
8  | 1    | LOGIN  | 12:31:00
9  | 2    | LOGIN  | 12:55:00

根据此样本表，必须显示在注销和登录之间超出间隔20分钟的用户。用户1超过记录4和8之间的20分钟间隔用户2超过记录7和9之间的20分钟间隔

这应该显示

ID | User
----------
4  | 1
8  | 1
7  | 2
9  | 2

如何编写查询以完成此操作？

Answer 1

这回答了这个问题：

我需要选择注销后20分钟内未重新登录的所有用户记录。

select lo.*
from (select l.*,
             (select min(l2.time)
              from logs l2
              where l2.user = l.user and l2.time > l.time and
                    l2.event = 'LOGIN'
             ) as next_login_time
      from logs l
      where l.event  = 'LOGOUT'
     ) lo
where next_login_time > time + interval 20 minute;

您的样本结果包括登录结果。目前还不清楚如何根据您提出的问题来定义它们。

Answer 2

如果两行相距20分钟，您似乎有兴趣获取两行。以下尝试模拟LEAD和LAG：

allow

或者，尝试以下似乎可行的方法，除了它将两行放在一起：

SELECT * FROM (
    SELECT *, (
        SELECT CASE WHEN EVENT = 'LOGOUT' AND main.Event = 'LOGIN' THEN TimeDIFF(main.Time, Time) END
        FROM t AS prev
        WHERE User = main.User AND Time < main.Time
        ORDER BY Time DESC
        LIMIT 1
    ) AS diff_lag, (
        SELECT CASE WHEN EVENT = 'LOGIN' AND main.Event = 'LOGOUT' THEN TimeDIFF(Time, main.Time) END
        FROM t AS next
        WHERE User = main.User AND Time > main.Time
        ORDER BY Time ASC
        LIMIT 1
    ) AS diff_lead
    FROM t AS main
) x
WHERE diff_lag > '00:20:00' OR diff_lead > '00:20:00'

MySQL获取与其他记录有时差的记录

2 个答案: