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间.获得时间差的结果集很容易:

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

它给出了按小时计算的时间列表,但是我希望将它总计为总计.

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.

这有一种方法可以做:

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