Question

我有下表：

mysql> SELECT id,start1,stop1,start2,stop2 FROM times;
+----+---------------------+---------------------+---------------------+---------------------+
| id | start1              | stop1               | start2              | stop2               |
+----+---------------------+---------------------+---------------------+---------------------+
|  4 | 2010-04-23 08:05:00 | 2010-04-23 12:15:00 | 2010-04-23 12:45:00 | 2010-04-23 16:50:00 | 
|  2 | 2010-04-26 09:30:00 | 2010-04-26 12:10:00 | 2010-04-26 12:50:00 | 2010-04-26 16:50:00 | 
|  7 | 2010-04-28 08:45:00 | 2010-04-28 11:45:00 | 2010-04-28 13:10:00 | 2010-04-28 17:29:00 | 
|  6 | 2010-04-27 09:30:00 | 2010-04-27 12:15:00 | 2010-04-27 12:55:00 | 2010-04-27 18:44:00 | 
+----+---------------------+---------------------+---------------------+---------------------+

我想将总工作时间和差异归结为“所需工时”。它适用于下面的语句，但由于未知原因，它不适用于id 6. start * / stop *字段的格式为datetime。

SELECT *, TIME_FORMAT(TIMEDIFF(totaltime,'08:24'),'%H:%i') AS diff,
totaltime > '08:24' AS redorgreen FROM
(
SELECT 
    DATE_FORMAT(start1,'%a %e. %M %Y') AS date,
    TIME_FORMAT(SUM(TIMEDIFF(stop1,start1) + TIMEDIFF(stop2,start2)),'%H:%i') AS totaltime,
    TIME_FORMAT(start1,'%H:%i') AS start1,
    TIME_FORMAT(stop1,'%H:%i') AS stop1,
    TIME_FORMAT(start2,'%H:%i') AS start2,
    TIME_FORMAT(stop2,'%H:%i') AS stop2,
    id as id
FROM times GROUP BY id ASC
) AS somethingwedontneed;

结果如下：

select id,
       TIME_FORMAT(SUM(TIMEDIFF(stop1,start1) + TIMEDIFF(stop2,start2)),'%H:%i') 
       AS totaltime from times group by id;
+----+-----------+
| id | totaltime |
+----+-----------+
|  2 | 06:40     | 
|  4 | 08:15     | 
|  6 | NULL      | 
|  7 | 07:19     | 
+----+-----------+

提前感谢每一个提示。

Answer 1

SELECT id，TIMEDIFF（stop1，start1），TIMEDIFF（stop2，start2），ADDTIME（TIMEDIFF（stop1，start1），TIMEDIFF（stop2，start2）），TIME_FORMAT（ADDTIME（TIMEDIFF（stop1，start1），TIMEDIFF（ stop2，start2）），'％H：％i'）AS总时间从时而来 GROUP BY id

Answer 2

尝试将'08:24'包裹在TIME()中，即TIME('08:24')。

MySQL TIMEDIFF的时间计算

2 个答案: