MySQL - 一組時差的總和

Question

我想總結所有時差以顯示志願者的總工作時間。獲得一組時間差異是很容易的：

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

它給出了幾小時的時間列表，然後我想總結一下它的總和。

所以，如果結果集爲：

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

它只是總40小時。

這有

辦法做這樣的事情：

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' 

如果沒有，那麼也許：

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

Answer 2

嘗試類似的東西

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

Answer 3

你幾乎從馬修得到答案，你需要做的是增加投：

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