執行日期/時間的減法運算

Question

我有一個表四個方面：

date entered, time entered, date completed, time completed 

我想知道最新的區別/時間輸入的日期/時間內完成

例如

date entered = 1/1/2001 
time entered = 10:00 
time completed = 1/2/2001 
time completed = 11:00 

區別是25小時。

我該如何使用select語句執行此計算？

我只是嘗試這樣做：

DATEDIFF(hh,dateadd(hh,[Time Entered],[Date Entered]),dateadd(hh,[Time Completed],[Date Completed])) AS [Hours] 

，並得到了以下錯誤：

Msg 8116, Level 16, State 1, Line 2 
Argument data type time is invalid for argument 2 of dateadd function. 

Answer 1

Declare @dateentered date = '20010101' 
Declare @timeentered time = '10:00' 
Declare @datecompleted date = '20010102' 
Declare @timecompleted time = '11:00' 

select datediff(hh, @dateentered + cast(@timeentered as datetime), 
        @datecompleted + cast(@timecompleted as datetime)) 

所以，在你的表列方面：

select datediff(hh, [date entered] + cast([time entered] as datetime), 
        [date completed] + cast([time completed] as datetime)) as [Hours] 

Answer 2

使用DateDiff

Answer 3

select datediff(hour,'1/1/2001 10:00','1/2/2001 11:00') 

Answer 4

嘗試

select DateDiff(ss, [Date Entered] + convert(datetime, [Time Entered]), 
        [Date Completed] + convert(datetime, [Time Completed])) 
from myTable 

以在幾秒鐘內獲得結果。

這裏有一個獨立的例子：

declare @dateentered date = '1/1/2001' 
declare @timeentered time = '10:00' 
declare @datecompleted date = '1/2/2001' 
declare @timecompleted time = '11:00' 

select DateDiff(ss, 
    @dateentered + convert(datetime, @timeentered), 
    @datecompleted + convert(datetime, @timecompleted)) 

當然，你可以指定不同的日期部分爲DATEDIFF指定。

Answer 5

DECLARE @StartDate DATETIME 
Declare @EndDate DATETIME 
declare @startime datetime 
declare @endime datetime 

SET @StartDate = '2001-01-01' 
set @startime = '10:00' 
SET @EndDate = '2001-01-02' 
set @endime = '11:00' 

set @StartDate = @StartDate + @startime 
set @EndDate = @EndDate + @endime 

--To get only Hours 
SELECT DATEDIFF(hh, @StartDate,@EndDate) AS [Hours];