Question

我只是在SQL Server中编写此函数。它计算工人的工作分钟数，不包括休息时间。

ALTER FUNCTION getMinutesWork
    ( @startT dateTime,
      @endT dateTime)
AS
BEGIN
    DECLARE @result int = DATEDIFF(mi, @startT, @endT)  --time difference including break times
    SET @startT = CAST(@startT AS TIME)
    SET @endT = CAST(@endT AS TIME)

    -- work schedule  
    DECLARE @t1 datetime = CAST('08:00:00' AS TIME),  --work
            @t2 datetime = CAST('10:00:00' AS TIME),  --break                   
            @t3 datetime = CAST('10:15:00' AS TIME),  --w
            @t4 datetime = CAST('12:40:00' AS TIME),  --b
            @t5 datetime = CAST('13:25:00' AS TIME),  --w
            @t6 datetime = CAST('16:30:00' AS TIME),  --b 
            @t7 datetime = CAST('16:45:00' AS TIME),  --w
            @t8 datetime = CAST('18:15:00' AS TIME);  --b

    --excluding break times
    IF ((@startT <= @t2 AND @endT >@t2) AND (@startT < @t3 AND @endT >= @t3))
       SET @result = @result - DATEDIFF(mi, @t2, @t3) 

    IF ((@startT <= @t4 AND @endT >@t4) AND (@startT < @t5 AND @endT >= @t5))
       SET @result = @result - DATEDIFF(mi, @t4, @t5)

    IF ((@startT <= @t6 AND @endT >@t6) AND (@startT < @t7 AND @endT >= @t7))
       SET @result = @result - DATEDIFF(mi, @t6, @t7)

    IF ((@startT > @t8) AND (@endT > @t1))
       SET @result = @result - DATEDIFF(mi, @t1, @startT)

    --excluding break times if worker starts on break time
    SET @result = (CASE 
                     WHEN (@startT < @t1) 
                       THEN @result - DATEDIFF(mi, @startT, @t1)
                     WHEN (@startT > @t2 AND @startT < @t3) 
                       THEN @result - DATEDIFF(mi, @startT, @t3)
                     WHEN (@startT > @t4 AND @startT < @t5) 
                       THEN @result - DATEDIFF(mi, @startT, @t5)
                     WHEN (@startT > @t6 AND @startT < @t7) 
                       THEN @result - DATEDIFF(mi, @startT, @t7)
                     WHEN (@startT > @t8) AND (@endT < @t1)
                       THEN 0
                     ELSE @result
                 END)

----------'2004-10-19 20:00:00','2004-10-18 10:00:00'------------
    --excluding break times if worker finish his work on break time
    SET @result = (
CASE 
    WHEN (@endT < @t1) 
        THEN 0
    WHEN (@endT >= @t2 AND @endT < @t3) 
        THEN @result - DATEDIFF(mi,@t2,@endT)
    WHEN (@endT >= @t4 AND @endT < @t5) 
        THEN @result - DATEDIFF(mi,@t4,@endT)
    WHEN (@endT >= @t6 AND @endT < @t7) 
        THEN @result - DATEDIFF(mi,@t6,@endT)
    WHEN (@endT > @t8)
        THEN @result - DATEDIFF(mi,@t8,@endT)
    ELSE @result
END )

-----------------------------------------------------------------------------------

IF @result <0
    SET @result = 0
-----------------------------------------------------------------------------------

RETURN @result
End

当工人在同一天来来往往时工作正常。但是当工人来到晚上11点工作到明天上午11点时，它并没有排除休息时间。有什么好主意吗？很抱歉很长一点。

编辑：我发现了问题，但无法解决问题。如果有@startT = '20:00:00'和@endT = '10:20:00'等过时工作，则此部分无法正常工作：

IF ((@startT <= @t2 AND @endT >@t2) AND (@startT < @t3 AND @endT >= @t3))
    SET @result = @result - DATEDIFF(mi,@t2,@t3) 
IF ((@startT <= @t4 AND @endT >@t4) AND (@startT < @t5 AND @endT >= @t5))
    SET @result = @result - DATEDIFF(mi,@t4,@t5)
IF ((@startT <= @t6 AND @endT >@t6) AND (@startT < @t7 AND @endT >= @t7))
    SET @result = @result - DATEDIFF(mi,@t6,@t7)

我该如何解决这个问题？

Answer 1

请检查一下这是否有效？我认为这解决了逻辑问题。

＆＃13;

[[2,0,0,0],
[1,1,0,0],
[1,0,1,0],
…
[0,0,1,1],
[0,0,0,2]]

＆＃13;

Answer 2

我编辑了if语句部分并且它有效：

--total work time if worker stay overnight including break times
IF ((@startT <= @t8) AND (@endT < @startT))
BEGIN
    SET @result = DATEDIFF(mi,@startT,'23:59:59') + DATEDIFF(mi,'00:00:01',@endT) +1
    SET @result = @result - DATEDIFF(mi,@t8,'23:59:59') - DATEDIFF(mi,'00:00:01',@t1)-1     
END

-----------------------------------------------------------------------------------
    --excluding break times

IF (((@startT > @t8) AND (@endT > @t1)))        
    SET @result = @result - DATEDIFF(mi,@startT,'23:59:59') - DATEDIFF(mi,'00:00:01',@t1)-1     
IF ((@startT <= @t2 AND @endT >= @t3)) OR ((@startT > @t8) AND (@endT >= @t3))
    SET @result = @result - DATEDIFF(mi,@t2,@t3) 
IF ((@startT <= @t4 AND @endT >= @t5)) OR ((@startT > @t8) AND (@endT >= @t5))
    SET @result = @result - DATEDIFF(mi,@t4,@t5)
IF ((@startT <= @t6 AND @endT >= @t7)) OR ((@startT > @t8) AND (@endT >= @t7))
    SET @result = @result - DATEDIFF(mi,@t6,@t7)

SQL Server：工作人员不包括休息时间

2 个答案: