Comments on: SQL SERVER – UDF – User Defined Function to Find Weekdays Between Two Dates

By: Vams

Vams — Mon, 29 Aug 2011 14:42:24 +0000

@BT-Microlink

The latest function still has missing logic.Can you please repost the complete function again.
Thanks!

By: BT-MicroLink

BT-MicroLink — Wed, 27 Oct 2010 17:57:58 +0000

Test for revised function:

DECLARE @Date AS datetime
SET @Date = ’10-27-2010′

SELECT dbo.WeekDayDiff(DATEADD(d,-10,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,-9,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,-8,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,-7,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,-6,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,-5,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,-4,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,-3,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,-2,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,-1,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,0,@Date), @Date);

SELECT dbo.WeekDayDiff(DATEADD(d,10,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,9,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,8,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,7,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,6,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,5,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,4,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,3,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,2,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,1,@Date), @Date);
SELECT dbo.WeekDayDiff(DATEADD(d,0,@Date), @Date);

By: BT-MicroLink

BT-MicroLink — Wed, 27 Oct 2010 17:57:32 +0000

Duly noted. Some of my own testing revealed issues when the @StartDate was after the @EndDate. Revised function:

CREATE FUNCTION [dbo].[WeekDayDiff]

(
@StartDate datetime,
@EndDate datetime
)
RETURNS int

BEGIN

DECLARE @TotalDays INT
DECLARE @WorkDays INT
DECLARE @WeekendStartDate INT
DECLARE @WeekendEndDate INT
DECLARE @Weeks INT
DECLARE @FinalWeekDays INT

SET @TotalDays = DATEDIFF(day, @StartDate, @EndDate)
SELECT @WeekendStartDate = CASE DATENAME(weekday, @StartDate)
WHEN ‘Saturday’ THEN 1
WHEN ‘Sunday’ THEN 1
ELSE 0
END
SELECT @WeekendEndDate = CASE DATENAME(weekday, @EndDate)
WHEN ‘Saturday’ THEN 1
WHEN ‘Sunday’ THEN 1
ELSE 0
END
SET @TotalDays = CASE WHEN DATEPART(dw,@StartDate) > DATEPART(dw,@EndDate) AND @StartDate < @EndDate THEN @TotalDays – 2
WHEN DATEPART(dw,@StartDate) <= DATEPART(dw,@EndDate) AND @StartDate <= @EndDate THEN @TotalDays – (@WeekendStartDate + @WeekendEndDate)
WHEN DATEPART(dw,@StartDate) @EndDate THEN @TotalDays + 2
WHEN DATEPART(dw,@StartDate) >= DATEPART(dw,@EndDate) AND @StartDate >= @EndDate THEN @TotalDays + (@WeekendStartDate + @WeekendEndDate)
END
SET @Weeks = @TotalDays / 7;
SET @FinalWeekDays = @TotalDays % 7;
SET @WorkDays = CASE WHEN @Weeks = 0 AND (@StartDate = @EndDate OR @WeekendStartDate + @WeekendEndDate = 2) THEN 0
WHEN @Weeks = 0 AND @StartDate @EndDate THEN @TotalDays
ELSE (@Weeks * 5) + @FinalWeekDays
END

RETURN @WorkDays

END

By: pinaldave

pinaldave — Thu, 14 Oct 2010 06:00:06 +0000

Hi,

If you try your query you will get a bit different result.

SELECT dbo.WeekDayDiff(GETDATE()-10, GETDATE());
SELECT dbo.WeekDayDiff(GETDATE()-9, GETDATE());
SELECT dbo.WeekDayDiff(GETDATE()-8, GETDATE());
SELECT dbo.WeekDayDiff(GETDATE()-7, GETDATE());
SELECT dbo.WeekDayDiff(GETDATE()-6, GETDATE());
SELECT dbo.WeekDayDiff(GETDATE()-5, GETDATE());
SELECT dbo.WeekDayDiff(GETDATE()-4, GETDATE());
SELECT dbo.WeekDayDiff(GETDATE()-3, GETDATE());
SELECT dbo.WeekDayDiff(GETDATE()-2, GETDATE());
SELECT dbo.WeekDayDiff(GETDATE()-1, GETDATE());

Result:
———–
8
———–
7
———–
6
———–
5
———–
4
———–
3
———–
4
———–
3
———–
2
———–
1

Would you please check?

Kind Regards,

By: BT-MicroLink

BT-MicroLink — Wed, 13 Oct 2010 20:41:45 +0000

–Extending on the contribution by Muthuvel to accommodate date variances less than 1 week or when dates are less than 1 week and are same day or Sat. to Sun.

FUNCTION [dbo].[WeekDayDiff]

(
@StartDate datetime,
@EndDate datetime
)
RETURNS int

BEGIN

DECLARE @TotalDays INT
DECLARE @WorkDays INT
DECLARE @WeekendEndDate INT
DECLARE @Weeks INT
DECLARE @FinalWeekDays INT

SET @TotalDays = DATEDIFF(day, @StartDate, @EndDate)
SELECT @WeekendEndDate = CASE DATENAME(weekday, @EndDate)
WHEN ‘Saturday’ THEN 1
WHEN ‘Sunday’ THEN 2
ELSE 0
END
SET @TotalDays = CASE WHEN DATEPART(dw,@StartDate) > DATEPART(dw,@EndDate) THEN @TotalDays – @WeekendEndDate – 2
ELSE @TotalDays – @WeekendEndDate
END
SET @Weeks = @TotalDays / 7;
SET @FinalWeekDays = @TotalDays % 7;
SET @WorkDays = CASE WHEN @Weeks = 0 AND (@StartDate = @EndDate OR (DATENAME(weekday, @StartDate) = ‘Saturday’ AND DATENAME(weekday, @EndDate) = ‘Sunday’)) THEN 0
WHEN @Weeks = 0 AND @StartDate @EndDate THEN @TotalDays
ELSE (@Weeks * 5) + @FinalWeekDays
END

RETURN @WorkDays

END

By: Muthuvel

Muthuvel — Wed, 10 Mar 2010 11:07:20 +0000

DECLARE @TotalDays INT,@WorkDays INT
DECLARE @ReducedDayswithEndDate INT
DECLARE @WeekPart INT
DECLARE @DatePart INT

SET @TotalDays= DATEDIFF(day, @StartDate, @EndDate) +1
SELECT @ReducedDayswithEndDate = CASE DATENAME(weekday, @EndDate)
WHEN ‘Saturday’ THEN 1
WHEN ‘Sunday’ THEN 2
ELSE 0 END
SET @TotalDays=@TotalDays-@ReducedDayswithEndDate
SET @WeekPart=@TotalDays/7;
SET @DatePart=@TotalDays%7;
SET @WorkDays=(@WeekPart*5)+@DatePart

RETURN @WorkDays

By: HM

HM — Mon, 17 Nov 2008 21:05:45 +0000

I would like to search for DataLength(AcctNo) = ’2′ and if first position is = ’0′ then make AcctNo = 10x where x is second position character of acctno.

By: red

red — Fri, 06 Jun 2008 17:51:09 +0000

if you want to know weekdays fractional to the second …

create function dbo.WeekDays(@StartDate datetime, @EndDate datetime)
returns float
as
begin
return (
(cast(datediff(s, @StartDate, @EndDate) as float) / 86400)
-(2*DateDiff(ww,@StartDate,@EndDate))+
case when datepart(dw, @StartDate)=7 then 1 else 0 end
)
end

By: Wieland

Wieland — Mon, 14 Apr 2008 15:51:42 +0000

I think the missing sign should be a ‘>’, not an ‘=’.

i.e.
WHEN (@LastPart > 0) THEN @LastPart – 1

By: shms

shms — Wed, 12 Mar 2008 03:42:18 +0000

how do i taking holidays when calculating number of working days. because i already create table holiday for store all weekend holidays and public holiday.

By: Ben

Ben — Thu, 15 Nov 2007 12:25:21 +0000

I did one like this:

CREATE FUNCTION [dbo].[WeekdaysCount]
( @StartDate datetime,
@EndDate datetime )
RETURNS INT
AS
BEGIN

DECLARE @TotalDays int, @WorkDays int, @Sats int, @Suns int

IF (DATEDIFF(day, @StartDate, @EndDate) = 0)
BEGIN
RETURN ( 0 )
END

SET @TotalDays = DATEDIFF(day, @StartDate, @EndDate)
SELECT @Sats = DATEDIFF(ww,@StartDate,@EndDate) + CASE WHEN DATENAME(dw,@StartDate) = ‘sunday’ THEN 1 ELSE 0 END,
@Suns = DATEDIFF(ww,@StartDate,@EndDate) + CASE WHEN DATENAME(dw,@EndDate) = ‘saturday’ THEN 1 ELSE 0 END

SET @WorkDays = @TotalDays – @Sats – @Suns

RETURN ( @WorkDays )
END
GO

By: Clinton

Clinton — Wed, 15 Aug 2007 07:26:30 +0000

Hi. I get the following message when attempting to add your function. Could you please confirm this works for you

Msg 102, Level 15, State 1, Procedure GetWorkingDays, Line 18
Incorrect syntax near ‘@FirstPart’.
Msg 102, Level 15, State 1, Procedure GetWorkingDays, Line 51
Incorrect syntax near ‘END’.

By: Chris

Chris — Mon, 13 Aug 2007 19:34:56 +0000

Missing “=” from line
WHEN (@LastPart 0) THEN @LastPart – 1
Should be
WHEN (@LastPart = 0) THEN @LastPart – 1

By: mike

mike — Thu, 26 Jul 2007 02:32:20 +0000

also if you want this to be inclusive, add this right below the begin statement and before the first declare

select @StartDate = @StartDate -1
select @EndDate = @EndDate + 1

By: mike

mike — Thu, 26 Jul 2007 02:20:47 +0000

here is the complete text…

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 0) THEN @LastPart – 1
ELSE 0
END
SELECT @WorkDays = @WorkDays * 5 + @FirstNum + @LastNum
END
RETURN ( @WorkDays )
END
GO

By: Dal

Dal — Sat, 05 May 2007 18:48:27 +0000

Alexander is good author. Thanks for bringing this to here.