仅限WeekDays的T-SQL日期差异

时间:2013-05-08 11:09:17

标签: sql-server tsql date

我试图在给定日期之前得到日期差异,不包括工作日。

这就是我所拥有的:

SELECT DATEADD (w, -4, GETDATE())

这将返回 2013-05-04 19:01:53.170 ,这意味着它也会计算周末。

相同
 SELECT DATEADD (dw, -4, GETDATE())

任何帮助将不胜感激。

提前致谢。

3 个答案:

答案 0 :(得分:3)

我正在使用这些函数返回两个日期之间的非周末秒:

CREATE FUNCTION [dbo].[DateDiff_NoWeekends](
    @date1 DATETIME,
    @date2 DATETIME
)

RETURNS INT AS BEGIN
    DECLARE @retValue INT

    SET @date1 = dbo.__CorrectDate(@date1, 1)
    SET @date2 = dbo.__CorrectDate(@date2, 0)

    IF (@date1 >= @date2)
        SET @retValue = 0
    ELSE BEGIN
        DECLARE @days INT, @weekday INT
        SET @days = DATEDIFF(d, @date1, @date2)
        SET @weekday = DATEPART(dw, @date1) - 1

        SET @retValue = DATEDIFF(s, @date1, @date2) - 2 * 24 * 3600 * ((@days + @weekday) / 7) 
    END

    RETURN @retValue
END

GO


CREATE FUNCTION [dbo].[__CorrectDate](
    @date DATETIME,
    @forward INT
)

RETURNS DATETIME AS BEGIN
    IF (DATEPART(dw, @date) > 5) BEGIN

        IF (@forward = 1) BEGIN
            SET @date = @date + (8 - DATEPART(dw, @date))
            SET @date = DateAdd(Hour, (8 - DatePart(Hour, @date)), @date)
        END ELSE BEGIN
            SET @date = @date - (DATEPART(dw, @date)- 5)
            SET @date = DateAdd(Hour, (18 - DatePart(Hour, @date)), @date)
        END
        SET @date = DateAdd(Minute, -DatePart(Minute, @date), @date)
        SET @date = DateAdd(Second, -DatePart(Second, @date), @date)
    END

    RETURN @date
END

这是四月(22)所有非周末日的sql-fiddle demo

SELECT [no weekend days in april] =
    (dbo.DateDiff_NoWeekends('2013-04-01','2013-05-01')
         / 3600 / 24)

答案 1 :(得分:1)

下面的查询仅给出了工作日的差异,即计算两天之间的无天数并减去周末天数,

    DECLARE @StartDate DATETIME,
    @EndDate DATETIME
    SELECT  @StartDate = '01-July-2008',
    @EndDate = '30-July-2008' 
    ;WITH DATE (Date1)
    AS (
    SELECT  DATEADD(DAY, DATEDIFF(DAY, '19000101', @StartDate), '19000101')
    UNION ALL
    SELECT  DATEADD(DAY, 1, Date1)
    FROM    DATE
    WHERE   Date1 < @EndDate
    )

    SELECT count(*) -
    (

    SELECT count(*)  
    --CONVERT(VARCHAR(15),d1.DATE1 ,110) as [Working Date],
    --DATENAME(weekday, d1.Date1) [Working Day] 
    from DATE d1 where (DATENAME(weekday, d1.Date1))     in ('Saturday','Sunday')

    )
    --CONVERT(VARCHAR(15),d1.DATE1 ,110) as [Working Date],
    --DATENAME(weekday, d1.Date1) [Working Day] 
    from DATE d1 where (DATENAME(weekday, d1.Date1))  not  in ('Saturday','Sunday')

请告知我任何澄清

答案 2 :(得分:0)

也许我仍然缺少一些完整的测试,但这也适用于我:在几天内采取差异,然后在每个周末减去2天

DateDiff(d, d1, d2) - 2*DateDiff(wk, d1, d2) 

也可以放入一个函数