MySQL 中 2 行之间的时间差

如何解决MySQL 中 2 行之间的时间差

交易ID	用户名	交易日期时间
1	1	'2021-04-22 11:00:00'
2	2	'2021-04-22 11:00:11'
3	1	'2021-04-22 11:00:22'
4	3	'2021-04-22 11:00:33'
5	3	'2021-04-22 11:00:44'
6	1	'2021-04-22 11:00:55'

我想查看每个 UserID 的事务之间的时间差。像这样：

交易ID	用户名	交易日期时间	时差
1	1	'2021-04-22 11:00:00'	NULL
2	2	'2021-04-22 11:00:11'	NULL
3	1	'2021-04-22 11:00:22'	00:22
4	3	'2021-04-22 11:00:33'	NULL
5	3	'2021-04-22 11:00:44'	00:11
6	1	'2021-04-22 11:00:55'	00:33

有没有办法做到这一点？

解决方法

SELECT *,SEC_TO_TIME(TIMESTAMPDIFF(SECOND,TransactionDateTime,LAG(TransactionDateTime) OVER (PARTITION BY UserID
                                                                ORDER BY TransactionDateTime))) TimeDifference
FROM table
ORDER BY TransactionDateTime

但我使用的是 MySQL 5.5 版本，不支持 PARTITION BY 函数。也许还有其他方法？ – 瓦鲁赞·斯捷潘扬

SELECT t1.*,t2.TransactionDateTime,t1.TransactionDateTime)) TimeDifference
FROM table t1
LEFT JOIN table t2 ON t1.UserID = t2.UserID 
                 AND t1.TransactionDateTime > t2.TransactionDateTime
WHERE NOT EXISTS ( SELECT NULL
                   FROM table t3
                   WHERE t1.UserID = t3.UserID 
                     AND t1.TransactionDateTime > t3.TransactionDateTime
                     AND t3.TransactionDateTime > t2.TransactionDateTime )
ORDER BY t1.TransactionID;

https://dbfiddle.uk/?rdbms=mysql_5.5&fiddle=b7d43521afc8fe6623f152343bb88d4b

我建议使用相关子查询来获取旧版本 MySQL 中的上一个日期/时间：

select t.*,timediff(TransactionDateTime,prev_TransactionDateTime) as timedifference
from (select t.*,(select max(t2.TransactionDateTime)
              from t t2
              where t2.UserId = t.UserId and
                    t2.TransactionDateTime < t.TransactionDateTime
             ) as prev_TransactionDateTime
      from t
     ) t;