Mysql 时间差和

我跟不上时代的总和差异

在数据库中，它们是重复记录。例如：

_Status________Information__________Time______
Start(1)  |  heating on 20°C        |  10:00
Start(1)  |  heating on 21°C        |  10:20
Stop(0)   |  Heating                |  11:00
Stop(0)   |  Wait for instructions  |  12:00

记录A（开始）和C（停止）后的差值为1小时

当我不使用SUM时，我会得到具有正确时差的行

查询

SET @end_time=0;

SELECT (TIME_TO_SEC(t2.date) - TIME_TO_SEC(t1.date)) AS difference, t1.date AS start, t2.date AS stop, @end_time:=t2.date

    FROM `heating_history` AS t1

    LEFT JOIN `heating_history`  AS t2 ON t2.date > t1.date and t2.status!=1

    WHERE t1.status=1 and TIME_TO_SEC(t1.date) > TIME_TO_SEC(@end_time)

SET @end_time=0;

SELECT SUM(TIME_TO_SEC(t2.date) - TIME_TO_SEC(t1.date)) AS difference, t1.date AS start, t2.date AS stop, @end_time:=t2.date

FROM `heating_history` AS t1

LEFT JOIN `heating_history`  AS t2 ON t2.date > t1.date and t2.status!=1

WHERE t1.status=1 and TIME_TO_SEC(t1.date) > TIME_TO_SEC(@end_time)

结果

difference  start                   stop                    @end_time:=t2.date
12703   2014-01-29 07:18:32     2014-01-29 10:50:15     2014-01-29 10:50:15
4079    2014-01-29 13:27:12     2014-01-29 14:35:11     2014-01-29 14:35:11
9839    2014-01-29 16:46:12     2014-01-29 19:30:11     2014-01-29 19:30:11
4810    2014-01-29 21:18:11     2014-01-29 22:38:21     2014-01-29 22:38:21

difference  start                   stop                    @end_time:=t2.date
258536  2014-01-29 07:18:32     2014-01-29 10:50:15     2014-01-29 10:50:15

---

但当我想要求和的时候，我得到了一个更好的结果

查询

SET @end_time=0;

SELECT (TIME_TO_SEC(t2.date) - TIME_TO_SEC(t1.date)) AS difference, t1.date AS start, t2.date AS stop, @end_time:=t2.date

    FROM `heating_history` AS t1

    LEFT JOIN `heating_history`  AS t2 ON t2.date > t1.date and t2.status!=1

    WHERE t1.status=1 and TIME_TO_SEC(t1.date) > TIME_TO_SEC(@end_time)

SET @end_time=0;

SELECT SUM(TIME_TO_SEC(t2.date) - TIME_TO_SEC(t1.date)) AS difference, t1.date AS start, t2.date AS stop, @end_time:=t2.date

FROM `heating_history` AS t1

LEFT JOIN `heating_history`  AS t2 ON t2.date > t1.date and t2.status!=1

WHERE t1.status=1 and TIME_TO_SEC(t1.date) > TIME_TO_SEC(@end_time)

结果

difference  start                   stop                    @end_time:=t2.date
12703   2014-01-29 07:18:32     2014-01-29 10:50:15     2014-01-29 10:50:15
4079    2014-01-29 13:27:12     2014-01-29 14:35:11     2014-01-29 14:35:11
9839    2014-01-29 16:46:12     2014-01-29 19:30:11     2014-01-29 19:30:11
4810    2014-01-29 21:18:11     2014-01-29 22:38:21     2014-01-29 22:38:21

difference  start                   stop                    @end_time:=t2.date
258536  2014-01-29 07:18:32     2014-01-29 10:50:15     2014-01-29 10:50:15

---

谢谢您的帮助。

您的问题来自于没有指定要求和的内容，因为没有group by子句，它只是求和您所有的内容。要确保只求和所需的差异，请以这种方式使用子选择查询

SELECT SUM(DIFFERENCE),start,stop
FROM
(
SELECT (TIME_TO_SEC(t2.date) - TIME_TO_SEC(t1.date)) AS difference, t1.date AS start, t2.date AS stop, @end_time:=t2.date
FROM `heating_history` AS t1
LEFT JOIN `heating_history`  AS t2 ON t2.date > t1.date and t2.status!=1
WHERE t1.status=1 and TIME_TO_SEC(t1.date) > TIME_TO_SEC(@end_time) and DATE(t1.date)=DATE_SUB(CURDATE(), INTERVAL 1 day)) as MyQuery

---

您不能在查询中抛出

SUM

，期望得到正确的结果。您需要指定

GROUP BY

子句或使用Sub Select对正确的结果进行

求和。
这是SQL Server还是MySQL或其他什么？还可以添加返回正确行的查询和返回错误SUMI use MySQL的查询。添加的查询