there is a solution ..
there is function written by Jeff Moden ( special thanks to Jeff author of so many useful articles )
it takes care of a lot of things .. please see the comments in his function
please click arrow to the left for Jeff's Function
CREATE FUNCTION dbo.fn_WorkDays
/***************************************************************************************
Purpose:
1. Given any valid start date and end date, this function will calculate and return
the number of workdays (Mon - Fri).
2. Given only a valid start date (end date has DEFAULT in it), this function will
return a 1 if the start date is a weekday and a 0 if not a weekday.
Usage:
1. MASTER.dbo.fn_WorkDays(@StartDate,@EndDate)
2. MASTER.dbo.fn_WorkDays(@StartDate,DEFAULT) --Always returns 1 or 0
3. MASTER.dbo.fn_WorkDays(@EndDate,@StartDate)
4. MASTER.dbo.fn_WorkDays(@StartDate,@StartDate) --Always returns 1 or 0
5. MASTER.dbo.fn_WorkDays(@EndDate,@EndDate) --Always returns 1 or 0
Notes:
1. Holidays are NOT considered.
2. Because of the way SQL Server calculates weeks and named days of the week, no
special consideration for the value of DATEFIRST is given. In other words, it
doesn't matter what DATEFIRST is set to for this function.
3. If the input dates are in the incorrect order, they will be reversed prior to any
calculations.
4. Only whole days are considered. Times are NOT used.
5. The number of workdays INCLUDES both dates
6. Inputs may be literal representations of dates, datetime datatypes, numbers that
represent the number of days since 1/1/1900 00:00:00.000, or anything else that can
be implicitly converted to or already is a datetime datatype.
7. Undocumented: The DATEPART(dw,date) does not actually count weeks... It counts the
transition to a Sunday regardless of the DATEFIRST setting. In essence, it counts
only whole weekends in any given date range.
8. This UDF does NOT create a tally table or sequence table to operate. Not only is
it set based, it is truly "tableless".
Error Indications:
1. If either the @StartDate or the @EndDate parameter is an invalid date, the
following error is returned...
"Server: Msg 242, Level 16, State 3, Line 3
The conversion of a char data type to a datetime data type resulted in an
out-of-range datetime value."
2. If either the @StartDate or the @EndDate parameter is a string not resembling a
date, the following error is returned...
"Server: Msg 241, Level 16, State 1, Line 3
Syntax error converting datetime from character string."
3. If only one parameter is passed, the following error is returned...
"Server: Msg 313, Level 16, State 2, Line 3
An insufficient number of arguments were supplied for the procedure or
function MASTER.dbo.fn_WorkDays."
Revisions:
Rev 00 - 12/12/2004 - Jeff Moden - Initial creation and test.
Rev 01 - 12/12/2004 - Jeff Moden - Load test, cleanup, document, release.
Rev 02 - 12/26/2004 - Jeff Moden - Return NULL if @StartDate is NULL or DEFAULT and
modify to be insensitive to DATEFIRST settings.
***************************************************************************************/
--======================================================================================
-- Presets
--======================================================================================
--===== Define the input parameters (ok if reversed by mistake)
(
@StartDate DATETIME,
@EndDate DATETIME = NULL --@EndDate replaced by @StartDate when DEFAULTed
)
--===== Define the output data type
RETURNS INT
AS
--======================================================================================
-- Calculate the RETURN of the function
--======================================================================================
BEGIN
--===== Declare local variables
--Temporarily holds @EndDate during date reversal
DECLARE @Swap DATETIME
--===== If the Start Date is null, return a NULL and exit
IF @StartDate IS NULL
RETURN NULL
--===== If the End Date is null, populate with Start Date value
-- so will have two dates (required by DATEDIFF below)
IF @EndDate IS NULL
SELECT @EndDate = @StartDate
--===== Strip the time element from both dates (just to be safe) by converting
-- to whole days and back to a date. Usually faster than CONVERT.
-- 0 is a date (01/01/1900 00:00:00.000)
SELECT @StartDate = DATEADD(dd,DATEDIFF(dd,0,@StartDate),0),
@EndDate = DATEADD(dd,DATEDIFF(dd,0,@EndDate) ,0)
--===== If the inputs are in the wrong order, reverse them
IF @StartDate > @EndDate
SELECT @Swap = @EndDate,
@EndDate = @StartDate,
@StartDate = @Swap
--===== Calculate and return the number of workdays using the
-- input parameters. This is the meat of the function.
-- This is really just one formula with a couple of parts
-- that are listed on separate lines for documentation
-- purposes.
RETURN (
SELECT
--Start with total number of days including weekends
(DATEDIFF(dd,@StartDate,@EndDate)+1)
--Subtact 2 days for each full weekend
-(DATEDIFF(wk,@StartDate,@EndDate)*2)
--If StartDate is a Sunday, Subtract 1
-(CASE WHEN DATENAME(dw,@StartDate) = 'Sunday'
THEN 1
ELSE 0
END)
--If EndDate is a Saturday, Subtract 1
-(CASE WHEN DATENAME(dw,@EndDate) = 'Saturday'
THEN 1
ELSE 0
END)
)
END
GO
please use it like this
i have not TESTED
select customerid, Date_of_Last_Attempt2, getdate()
from gbr_creditsolve
Where campaign = 'GBR AA' and dbo.fn_WorkDays(Date_of_Last_Attempt2,getdate()) = 12 and datediff(day,importdate,getdate()) = 0
Here is my testing from my own test data
select * from data a cross apply (select dbo.fn_WorkDays(a.date1,a.date2) as weekdays) as x
