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sql 通過某段時間求得改段時間內的工作時長,排除工作日

好的 char rup and while 思路 then task 工作日

CREATE FUNCTION Fun_GetTotalHourBySomeTime(@TaskId NVARCHAR(30),@Bu_trupstartDate NVARCHAR(50),@Bu_trupEndDate NVARCHAR(50))

RETURNS FLOAT
AS
BEGIN
--計算工作日的天數
DECLARE @tStartDate DATE
DECLARE @tEndDate DATE
SET @tStartDate=CONVERT(NVARCHAR(50),@Bu_trupstartDate,112)
SET @tEndDate=CONVERT(NVARCHAR(50),@Bu_trupEndDate,112)

DECLARE @DAYNum DATE ,
@COUNT INT;
SET @DAYNum = @tStartDate;
SET @COUNT = 0;
WHILE @DAYNum <= @tEndDate
BEGIN
SET @COUNT = @COUNT + ( CASE DATEPART(WEEKDAY, @DAYNum)
WHEN 1 THEN 0
WHEN 7 THEN 0
ELSE 1
END );
SET @DAYNum = DATEADD(DAY, 1, @DAYNum);
END
--工作日的天數先少計算一天;
SET @[email protected];

DECLARE @sResultHour FLOAT
DECLARE @Day INT
DECLARE @startTimeH INT
DECLARE @endTimeH INT
DECLARE @startTimeM INT
DECLARE @endTimeM INT
--SELECT @sResultHour=8
SELECT @Day=DATEDIFF(DAY,Bu_trupstartDate,Bu_trupEndDate),@startTimeH=startTimeH,@startTimeM=startTimeM,@endTimeH=endTimeH,@endTimeM=endTimeM
FROM dbo.VIEW_Apply_Bu_tripReport WHERE [email protected]
IF(@Day=0)
BEGIN
IF(@startTimeH<12 AND @endTimeH<=12)
BEGIN
SELECT @sResultHour=CAST(DATEDIFF(MINUTE,Bu_trupstartDate,Bu_trupEndDate)*1./60 AS DECIMAL(18,1)) FROM VIEW_Apply_Bu_tripReport WHERE [email protected]
END
IF(@startTimeH<12 AND @endTimeH>12)
BEGIN
SELECT @sResultHour=(DATEDIFF(HOUR,Bu_trupstartDate,Bu_trupEndDate))-1 FROM VIEW_Apply_Bu_tripReport WHERE [email protected]
END
IF(@startTimeH>=12)
BEGIN
SELECT @sResultHour=CAST(DATEDIFF(MINUTE,Bu_trupstartDate,Bu_trupEndDate)*1./60 AS DECIMAL(18,1)) FROM VIEW_Apply_Bu_tripReport WHERE [email protected]
END
END
IF(@Day>=1)
BEGIN
IF(@startTimeH<12 AND @endTimeH<=12)
BEGIN
SELECT @sResultHour=CAST((@COUNT*7.5*60+(@[email protected])*60+(@[email protected]))*1./60 AS DECIMAL(18,1))
FROM VIEW_Apply_Bu_tripReport WHERE [email protected]
END
IF(@startTimeH<12 AND @endTimeH>12)
BEGIN
SELECT @sResultHour=CAST((@COUNT*7.5*60+(@[email protected])*60+(@[email protected]))*1./60 AS DECIMAL(18,1))
FROM VIEW_Apply_Bu_tripReport WHERE [email protected]
END
IF(@startTimeH>=12)
BEGIN
SELECT @sResultHour=CAST((@COUNT*7.5*60+(@[email protected])*60+(@[email protected]))*1./60 AS DECIMAL(18,1))
FROM VIEW_Apply_Bu_tripReport WHERE [email protected]
END
END
RETURN @sResultHour
END

如有更好的解決思路,還請分享學習,謝謝!!!

sql 通過某段時間求得改段時間內的工作時長,排除工作日