Question

我已经有一个现有的问题，但是即使答案有帮助，我仍然无法解决我的问题。

Exclude weekends and public holiday for SQL datediff

不是使用2 while循环，而是使用1 while循环和更多的IF-ELSE语句，因为我对此更加熟悉。但是，我意识到我应该数数的方式也不同

问题

我认为我的代码将给我7天+时间，但正确的答案应该是6天+时间。我尝试减掉一天的代码，它在某些情况下有效，但在其他情况下失败。
当StartDate大约是14:00 PM而ENDDate是8:00 AM时，获取变量“时间”的奇怪答案。（开始日期超过ENDDate）

条件：排除（周末和公众假期）（数据库中已存在PH表，代码显示为HolidayTable，此问题已省略）

    Declare @numdays int=0
    DECLARE @numDaysExclude int = 0


    /*1. Calculate only time difference*/
    DECLARE @time AS DECIMAL(18,2) = CAST(DATEDIFF(second, CONVERT(time, @StartDate), CONVERT(time, @ENDDate)) as Decimal(18,2))/86400.00

    /*2. Calculate only datediff*/
    SET @numDays = DATEDIFF(day,@StartDate,@ENDDate)



    /*For cases such as (20-12-2017, 1-1-2018)*/
    IF DATEDIFF(day,@StartDate,@ENDDate)>0 
    BEGIN
        WHILE DATEDIFF(day,@StartDate,@ENDDate)>0
        BEGIN
        IF exists (SELECT Holiday_Date FROM Holiday where Holiday_Date=CONVERT(date, @StartDate)) or DATENAME(DW,@StartDate)='saturday' or DATENAME(DW,@StartDate)='sunday'
            BEGIN
                SET @numDaysExclude = @numDaysExclude +1
            END
        SET @StartDate = DATEADD(d,1,@StartDate)
        END

        /*
        IF DATENAME(DW,@ENDDate)='sunday' 
        SET @numDays=@numDays-1 */ 

        DECLARE @result AS DECIMAL(18,2) =  (@numDays-@numDaysExclude) + @time
    END




    /*For cases such as (1-1-2018, 10-12-2017)*/
    ELSE IF DATEDIFF(day,@StartDate,@ENDDate)<0
    BEGIN
    /*Set @time = 1-@time
    IF @time > 1 
        SET @time = CAST(DATEDIFF(second, CONVERT(time, @ENDDate), CONVERT(time, @StartDate)) as Decimal(18,2))/86400.00

    IF DATENAME(DW,@ENDDate)='sunday' 
    SET @numDays=@numDays-1 */

        WHILE DATEDIFF(day,@StartDate,@ENDDate)<0
        BEGIN
        IF exists (SELECT Holiday_Date FROM Holiday where Holiday_Date=CONVERT(date, @ENDDate)) or DATENAME(DW,@ENDDate)='saturday' or DATENAME(DW,@ENDDate)='sunday'
            BEGIN
                SET @numDaysExclude = @numDaysExclude +1
            END
        SET @ENDDate = DATEADD(d,1,@ENDDate)
        END

        SET @result = @numDays + @numDaysExclude + @time

    END

    RETURN @result
    END

有人可以帮忙看看吗？我在这里呆了很长时间，尝试了许多我知道的方法...这对我来说是最容易理解的，因为我不太擅长编程。谢谢！

  [1]: https://i.stack.imgur.com/SOije.jpg

使用SQL获取两个日期之间的差异

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