不同用户的多条操作时间差值平均值

例如: 一站表

id	uid	login_time
1	1	2018-04-10 18:19:14
2	1	2019-04-10 18:19:14
3	1	2020-04-10 18:19:14
4	2	2019-04-10 18:19:14

那么，最终结果类似于:

uid	average
1	(2020-04-10 18:19:14 - 2018-04-10 18:19:14) / 2

select uid, timestampdiff(day, min(login_time), max(login_time)) / (count(*) - 1) from table_name group by uid having count(*) > 1;

需要买本高级查询的书了-_-!

赏

谢谢你请我吃糖果

支付宝