问题描述
我的桌子是这样的:
id | 时间 |
---|---|
1 | 2021-07-17 17:44:26 |
2 | 2021-07-17 17:44:26 |
3 | 2021-07-17 17:44:26 |
4 | 2021-07-17 17:44:31 |
5 | 2021-07-17 17:44:31 |
6 | 2021-07-17 17:44:31 |
7 | 2021-07-17 17:44:36 |
8 | 2021-07-17 17:44:36 |
9 | 2021-07-17 17:44:36 |
10 | 2021-07-17 17:44:41 |
11 | 2021-07-17 17:44:41 |
12 | 2021-07-17 17:44:41 |
13 | 2021-07-17 17:44:51 |
14 | 2021-07-17 17:44:51 |
15 | 2021-07-17 17:44:51 |
16 | 2021-07-17 17:44:56 |
17 | 2021-07-17 17:44:56 |
18 | 2021-07-17 17:44:56 |
19 | 2021-07-17 17:45:02 |
20 | 2021-07-17 17:45:02 |
21 | 2021-07-17 17:45:02 |
总是接下来的 3 行有相同的时间,然后 beetwen 通常是 5 秒的时间间隔,如何找到所有超过 8 秒的间隔并计算间隔时间以获得类似的东西:
gap_id | gap_time_start | 间隙长度 |
---|---|---|
1 | 2021-07-17 17:44:41 | 10 |
解决方法
如果您使用的是 MySQL 8+,您可以像这样使用 LEAD()
窗口函数:
select * from (
SELECT id as gap_id,`time` as gap_start_time,timediff( lead(`time`) over W,`time`) as gap_length
from `myTable` window w as (order by `time` asc)
) T where T.gap_length > 5;
输出:
12,2021-07-17 17:44:41,00:00:10
18,2021-07-17 17:44:56,00:00:06
参考:LEAD()