MySQL - 一组时差的SUM

Question

我想总结所有时间差异，以显示志愿者的总工作时数。获得时间差的结果集非常简单：

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

给出了按小时计算的时间列表，但是我希望将其总计为总计。

所以如果结果集是：

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

这将只有40个小时。

这可以做一些事情：

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

Answer 1

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

如果不是那么可能：

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

Answer 2

你几乎可以从 Matthew 得到答案，你需要做的就是添加演员：

Select CAST(SUM(timediff(timeOut, timeIn)) as time) as totalhours
FROM volHours 
WHERE username = 'skolcz'    

Answer 3

尝试类似的东西

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