1

扩展这个问题：这里

我想计算 t-sql 中两个日期之间的工作日，只是周末，而不是整个工作日（24 小时），假设每个工作日从上午 9 点到下午 5:30。

我如何使用例如 CMS 的方法来做到这一点：

 DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = '2008/10/01'
SET @EndDate = '2008/10/31'


SELECT
   (DATEDIFF(dd, @StartDate, @EndDate) + 1)
  -(DATEDIFF(wk, @StartDate, @EndDate) * 2)
  -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN 1 ELSE 0 END)
  -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END)
4

5 回答 5

1

我认为你很好，只需将 THEN 1 更改为 THEN -1

但是，我不完全确定您想对问题的时间部分做什么。

DECLARE @StartDate DATETIME
DECLARE @EndDate DATETIME
SET @StartDate = '2013/03/01'
SET @EndDate = '2013/03/31'


SELECT
   (DATEDIFF(dd, @StartDate, @EndDate) + 1)
  -(DATEDIFF(wk, @StartDate, @EndDate) * 2)
  -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN -1 ELSE 0 END)
  -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN -1 ELSE 0 END)
于 2013-03-18T22:08:02.790 回答
1

这就是我一直在寻找的，这是其他人需要的代码：

参数是：

@D1 (StartDate)
@D2 (EndDate)
@T1 (Business Hours - Start)
@T2 (Business Hours - End)

该函数返回一个带有小时数的浮点数。

CREATE FUNCTION [dbo].[WorkHoursDiff] (@D1 as datetime,@D2 as datetime,@T1 as int,@T2 as int)  
RETURNS float AS  
BEGIN 
    if @D1>@D2 return 0

    declare @whd as float

    declare @H1 as float
    declare @H2 as float


    if not (datepart(yyyy,@D1   )=datepart(yyyy,@D2) and datepart(mm,@D1)=datepart(mm,@D2) and datepart(dd,@D1)=datepart(dd,@D2))
        begin
            declare @DD as datetime
            SET @whd=0
            if datepart(dw,@D1)<>1 and datepart(dw,@D1)<>7 
                                        if dbo.GetDateHour(@D1)<@T1 SET @whd = @T2-@T1
            else if dbo.GetDateHour(@D1)>@T2 SET @whd = 0
            else if dbo.GetDateHour(@D1)>=@T1 and dbo.GetDateHour(@D1)<=@T2 SET @whd = @T2-dbo.GetDateHour(@D1)

            SET @DD=@D1 + 1
            while not (datepart(yyyy,@DD)=datepart(yyyy,@D2) and datepart(mm,@DD)=datepart(mm,@D2) and datepart(dd,@DD)=datepart(dd,@D2))
                begin
                    if not(datepart(dw,@DD)=1 or datepart(dw,@DD)=7)  
                         SET @whd=@whd + (@T2-@T1)

                    SET @DD=@DD + 1
                end
                if datepart(dw,@D2)<>1 and datepart(dw,@D2)<>7
                if dbo.GetDateHour(@D2)<@T1 SET @whd = @whd 
                else if dbo.GetDateHour(@D2)>@T2 SET @whd = @whd + @T2 - @T1
                else if dbo.GetDateHour(@D2)>=@T1  and dbo.GetDateHour(@D2)<=@T2 SET @whd = @whd +dbo.GetDateHour(@D2)-@T1

        end
    else
        begin
            if dbo.GetDateHour(@D1)<@T1 SET @H1=@T1
            else if dbo.GetDateHour(@D1)>@T2 SET @H1=@T2
            else SET @H1=dbo.GetDateHour(@D1)

            if dbo.GetDateHour(@D2)<@T1 SET @H2=@T1
            else if dbo.GetDateHour(@D2)>@T2 SET @H2=@T2
            else SET @H2=dbo.GetDateHour(@D2)

            if datepart(dw,@D1)=1 or datepart(dw,@D1)=7  SET @whd=0
            else SET @whd=@H2-@H1
        end
    return @whd

END
于 2013-03-19T13:14:58.373 回答
0

发现了这个。。

 CREATE FUNCTION dbo.GetWorkingDays
  ( @StartDate datetime,
    @EndDate datetime )
RETURNS INT
AS
BEGIN
  DECLARE @WorkDays int, @FirstPart int
  DECLARE @FirstNum int, @TotalDays int
  DECLARE @LastNum int, @LastPart int
  IF (DATEDIFF(day, @StartDate, @EndDate) < 2)
    BEGIN
      RETURN ( 0 )
    END
  SELECT
   @TotalDays = DATEDIFF(day, @StartDate, @EndDate) - 1,
   @FirstPart = CASE DATENAME(weekday, @StartDate)
                 WHEN 'Sunday' THEN 6
                 WHEN 'Monday' THEN 5
                 WHEN 'Tuesday' THEN 4
                 WHEN 'Wednesday' THEN 3
                 WHEN 'Thursday' THEN 2
                 WHEN 'Friday' THEN 1
                 WHEN 'Saturday' THEN 0
               END,
   @FirstNum = CASE DATENAME(weekday, @StartDate)
                 WHEN 'Sunday' THEN 5
                 WHEN 'Monday' THEN 4
                 WHEN 'Tuesday' THEN 3
                 WHEN 'Wednesday' THEN 2
                 WHEN 'Thursday' THEN 1
                 WHEN 'Friday' THEN 0
                 WHEN 'Saturday' THEN 0
               END
  IF (@TotalDays < @FirstPart)
     BEGIN
       SELECT @WorkDays = @TotalDays
     END
  ELSE
     BEGIN
       SELECT @WorkDays = (@TotalDays - @FirstPart) / 7
       SELECT @LastPart = (@TotalDays - @FirstPart) % 7
       SELECT @LastNum = CASE
         WHEN (@LastPart < 7) AND (@LastPart > 0) THEN @LastPart - 1
         ELSE 0
       END
       SELECT @WorkDays = @WorkDays * 5 + @FirstNum + @LastNum
     END
  RETURN ( @WorkDays )
END
GO
于 2013-03-19T08:11:27.887 回答
0

尝试这个

Declare @startdate datetime,
@enddate datetime,
@countSunday int,
@countTuesday int,
@countWednesday int,@countThursday int,@countFriday int,@countSaturday int

set @countSunday=0
set @countMonday=0
set @countTuesday =0
set @countWednesday =0
set @countThursday =0
set @countFriday =0
set @countSaturday=0
set @startdate='2018-10-01'
set @enddate='2018-10-20'

while @startdate<=@enddate
Begin
    IF DatePart(WEEKDAY,@startdate) = 1
    SET @countSunday=@countSunday+1
    else if  DatePart(WEEKDAY,@startdate) = 2
      SET @countMonday=@countMonday+1  
      else if  DatePart(WEEKDAY,@startdate) = 3
      SET @countTuesday=@countTuesday+1  
      else if  DatePart(WEEKDAY,@startdate) = 4
      SET @countWednesday=@countWednesday+1  
      else if  DatePart(WEEKDAY,@startdate) = 5
      SET @countThursday=@countThursday+1  
      else if  DatePart(WEEKDAY,@startdate) = 6
      SET @countFriday=@countFriday+1  
      else if  DatePart(WEEKDAY,@startdate) = 7
      SET @countSaturday=@countSaturday+1  
    SET @startdate=DateAdd(d,1,@startdate)
END


select @countSunday as Sunday ,@countMonday as Monday,@countTuesday as Tuesday,@countWednesday as Wednesday,@countThursday as Thursday,@countFriday as Friday,@countSaturday as Saturday
于 2018-10-11T06:10:31.103 回答
0

尝试这个

DECLARE @DateFrom DateTime ='20181001' , @DateTo DateTime = '20181031'  ;
WITH CTE(dt)
AS
(
SELECT @DateFrom
UNION ALL
SELECT DATEADD(d, 1, dt) FROM CTE
  WHERE dt < @DateTo
)
SELECT datename("dw", dt) as Days,count(datename("dw", dt)) as numberofDays FROM CTE 
group by datename ("dw", dt)
于 2018-10-11T07:53:41.947 回答