MySQL-一组时差之和 [英] MySQL - SUM of a group of time differences

问题描述

我想总结所有时间差，以显示志愿者的总工作时间.获得时差结果集很容易:

I want to sum all the time differences to show the total hours worked by a volunteer. Getting a result set of time differences is easy:

Select timediff(timeOut, timeIn) 
FROM volHours 
WHERE username = 'skolcz'

它提供了按小时列出的时间列表，但是我想将其总计为一个总计.

which gives the list of times by hours but then I want to sum it up to a grand total.

因此，如果结果集为:

12:00:00
10:00:00
10:00:00
08:00:00

总共需要40个小时.

这有一种做类似的事情的方式:

This there a way to do something like:

SELECT SUM(Select timediff(timeOut,timeIn) 
FROM volHours 
WHERE username = 'skolcz') as totalHours

?

推荐答案

Select  SEC_TO_TIME(SUM(TIME_TO_SEC(timediff(timeOut, timeIn)))) AS totalhours
FROM volHours 
WHERE username = 'skolcz'

如果没有，那么也许:

Select  SEC_TO_TIME(SELECT SUM(TIME_TO_SEC(timediff(timeOut, timeIn))) 
FROM volHours 
WHERE username = 'skolcz') as totalhours

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