On the Trail of the ISO Week
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Comments posted to this topic are about the content posted at http://www.sqlservercentral.com/columnists/chedgate/onthetrailoftheisoweek.asp
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I have some problems when I try to implement this solution. I get results in the fourties and up. When I eg. input 2005-01-27 a normal datepart with wk gives the result 5, but this function gives the result 44.
Declare @startdate as smalldatetime
set @startdate ='2005-27-01'
SELECT @startdate,DATEPART(wk, @startdate) as Sql_weeknr,
CASE
-- Exception where @startdate is part of week 52 (or 53) of the previous year
WHEN @startdate <
CASE (DATEPART(dw, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04') + @@DATEFIRST - 1) % 7
WHEN 1 THEN CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04'
WHEN 2 THEN DATEADD(d, -1, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
WHEN 3 THEN DATEADD(d, -2, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
WHEN 4 THEN DATEADD(d, -3, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
WHEN 5 THEN DATEADD(d, -4, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
WHEN 6 THEN DATEADD(d, -5, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
ELSE DATEADD(d, -6, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
END
THEN
(DATEDIFF(d,
CASE (DATEPART(dw, CAST(YEAR(@startdate) - 1 AS CHAR(4)) + '-01-04') + @@DATEFIRST - 1) % 7
WHEN 1 THEN CAST(YEAR(@startdate) - 1 AS CHAR(4)) + '-01-04'
WHEN 2 THEN DATEADD(d, -1, CAST(YEAR(@startdate) - 1 AS CHAR(4)) + '-01-04')
WHEN 3 THEN DATEADD(d, -2, CAST(YEAR(@startdate) - 1 AS CHAR(4)) + '-01-04')
WHEN 4 THEN DATEADD(d, -3, CAST(YEAR(@startdate) - 1 AS CHAR(4)) + '-01-04')
WHEN 5 THEN DATEADD(d, -4, CAST(YEAR(@startdate) - 1 AS CHAR(4)) + '-01-04')
WHEN 6 THEN DATEADD(d, -5, CAST(YEAR(@startdate) - 1 AS CHAR(4)) + '-01-04')
ELSE DATEADD(d, -6, CAST(YEAR(@startdate) - 1 AS CHAR(4)) + '-01-04')
END,
@startdate
) / 7) + 1
-- Exception where @startdate is part of week 1 of the following year
WHEN @startdate >= CASE (DATEPART(dw, CAST(YEAR(@startdate) + 1 AS CHAR(4)) + '-01-04') + @@DATEFIRST - 1) % 7
WHEN 1 THEN CAST(YEAR(@startdate) + 1 AS CHAR(4)) + '-01-04'
WHEN 2 THEN DATEADD(d, -1, CAST(YEAR(@startdate) + 1 AS CHAR(4)) + '-01-04')
WHEN 3 THEN DATEADD(d, -2, CAST(YEAR(@startdate) + 1 AS CHAR(4)) + '-01-04')
WHEN 4 THEN DATEADD(d, -3, CAST(YEAR(@startdate) + 1 AS CHAR(4)) + '-01-04')
WHEN 5 THEN DATEADD(d, -4, CAST(YEAR(@startdate) + 1 AS CHAR(4)) + '-01-04')
WHEN 6 THEN DATEADD(d, -5, CAST(YEAR(@startdate) + 1 AS CHAR(4)) + '-01-04')
ELSE DATEADD(d, -6, CAST(YEAR(@startdate) + 1 AS CHAR(4)) + '-01-04')
END
THEN 1
ELSE
-- Calculate the ISO week number for all dates that are not part of the exceptions above
(DATEDIFF(d,
CASE (DATEPART(dw, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04') + @@DATEFIRST - 1) % 7
WHEN 1 THEN CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04'
WHEN 2 THEN DATEADD(d, -1, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
WHEN 3 THEN DATEADD(d, -2, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
WHEN 4 THEN DATEADD(d, -3, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
WHEN 5 THEN DATEADD(d, -4, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
WHEN 6 THEN DATEADD(d, -5, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
ELSE DATEADD(d, -6, CAST(YEAR(@startdate) AS CHAR(4)) + '-01-04')
END,
@startdate
) / 7) + 1
END AS IsoWeek
A bit puzled about this
Best regards
John Valore
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I tried John's code - I had to change the @StartDate to '2005-01-27'.
I had no problems - I received an ISO Week of "4"
Regards,
Nic Washington
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Thanks for the reply Nic,
I took some time before someone responded, so I had actually abandoned the approach and settled with the standard datepart function.
Thanks to your response, I looked at the problem again and have now solved the problem. I still have to input the date in the format yyyy-mm-dd :
set @startdate =CAST('2005-01-27' as smalldatetime)
but if I include a statement :
set dateformat mdy
(default american date input/output), the code from Chris Hedgate now works.
The devil is apparently still in the details and the several datetime formats.
Thank you very much.
John Valore
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Hi,
Just an update.
I was going to use the functionality in an udf environment, but "forgot" that it is not possible to issue SET commands in an udf.
So I had to look at Chris code again when it suddenly strooke me, that I should replace all the '-01-04' with '-04-01'.
Bingo.
John Valore
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Cris:
If you use the ISO format for dates (yyyymmdd), not the yyyy-mm-dd format, SQL Server is not vulnerable for myd or dmy dates.
See CONVERT ISO 112 (or 12 for yymmdd) in BOL.
SQL server accepts '20050323' (today, y=2005, m=03, d=23) either in Europe or USA.
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Here is a version which uses CONVERT to allow using the yyyy-mm-dd format. This makes the code independent from SQL Server installation flavors.
Ralf
-----
DECLARE @date datetime
SET @date = CONVERT(datetime, '2003-12-31', 120) -- 1
--SET @date = CONVERT(datetime, '2005-01-01', 120) -- 53
--SET @date = CONVERT(datetime, '2005-01-05', 120) -- 1
--SET @date = CONVERT(datetime, '2006-01-01', 120) -- 52
SELECT @date,
CASE
-- Exception where @date is part of week 52 (or 53) of the previous year
WHEN @date =
CASE (DATEPART(dw, CONVERT(datetime, CAST(YEAR(@date) + 1 AS CHAR(4)) + '-01-04', 120)) + @@DATEFIRST - 1) % 7
WHEN 1 THEN CONVERT(datetime, CAST(YEAR(@date) + 1 AS CHAR(4)) + '-01-04', 120)
WHEN 2 THEN DATEADD(d, -1, CONVERT(datetime, CAST(YEAR(@date) + 1 AS CHAR(4)) + '-01-04', 120))
WHEN 3 THEN DATEADD(d, -2, CONVERT(datetime, CAST(YEAR(@date) + 1 AS CHAR(4)) + '-01-04', 120))
WHEN 4 THEN DATEADD(d, -3, CONVERT(datetime, CAST(YEAR(@date) + 1 AS CHAR(4)) + '-01-04', 120))
WHEN 5 THEN DATEADD(d, -4, CONVERT(datetime, CAST(YEAR(@date) + 1 AS CHAR(4)) + '-01-04', 120))
WHEN 6 THEN DATEADD(d, -5, CONVERT(datetime, CAST(YEAR(@date) + 1 AS CHAR(4)) + '-01-04', 120))
ELSE DATEADD(d, -6, CONVERT(datetime, CAST(YEAR(@date) + 1 AS CHAR(4)) + '-01-04', 120))
END
THEN 1
-- Calculate the ISO week number for all dates that are not part of the exceptions above
ELSE (DATEDIFF(d,
CASE (DATEPART(dw, CONVERT(datetime, CAST(YEAR(@date) AS CHAR(4)) + '-01-04', 120)) + @@DATEFIRST - 1) % 7
WHEN 1 THEN CONVERT(datetime, CAST(YEAR(@date) AS CHAR(4)) + '-01-04', 120)
WHEN 2 THEN DATEADD(d, -1, CONVERT(datetime, CAST(YEAR(@date) AS CHAR(4)) + '-01-04', 120))
WHEN 3 THEN DATEADD(d, -2, CONVERT(datetime, CAST(YEAR(@date) AS CHAR(4)) + '-01-04', 120))
WHEN 4 THEN DATEADD(d, -3, CONVERT(datetime, CAST(YEAR(@date) AS CHAR(4)) + '-01-04', 120))
WHEN 5 THEN DATEADD(d, -4, CONVERT(datetime, CAST(YEAR(@date) AS CHAR(4)) + '-01-04', 120))
WHEN 6 THEN DATEADD(d, -5, CONVERT(datetime, CAST(YEAR(@date) AS CHAR(4)) + '-01-04', 120))
ELSE DATEADD(d, -6, CONVERT(datetime, CAST(YEAR(@date) AS CHAR(4)) + '-01-04', 120))
END,
@date) / 7) + 1
END AS IsoWeek
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Hmmm that is my function similiar to above but IMHO easier to understand
CREATE FUNCTION ISOweek(@date datetime)
RETURNS int
AS
BEGIN
declare @dayOfWeek int
declare @1ThISOWeek1ThDay datetime
declare @4ThJan datetime
declare @1ThJan datetime
declare @31ThDec datetime
declare @numberOfISOweeks int
declare @Tmp int
declare @returnValue as int
set @4ThJan = cast(CAST(YEAR(@date) AS CHAR(4)) + '-01-04' as datetime)
set @1ThJan = cast(CAST(YEAR(@date) AS CHAR(4)) + '-01-01' as datetime)
set @31ThDec = cast(CAST(YEAR(@date) AS CHAR(4)) + '-12-31' as datetime)
--"rule of thursday"
if (DATEPART(dw, @1ThJan) = 5 or DATEPART(dw, @31ThDec) = 5 )
set @numberOfISOweeks = 53
else
set @numberOfISOweeks = 52
set @dayOfWeek = (select DATEPART(dw, @4ThJan) - 1)
if @dayOfWeek = 0
set @dayOfWeek = 7
set @1ThISOWeek1ThDay = (select DATEADD(day, -(@dayOfWeek -1) ,@4ThJan))
if @date @numberOfISOweeks
set @returnValue = 1 -- or -1 as an error
return @returnValue
END
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Sombody now how to calculate first monday day from given week
For example I need get the day from a given week:
ISO date ‘20050101’ is the week 53 in ISO 8601
If a have the week number 53 how I can calculate the ISO date?
Thank for all in advance.
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Better late than never, I guess...
DECLARE @Year CHAR(4)
DECLARE @NextYear CHAR(4)
DECLARE @ISOWeek VARCHAR(2)
SET @Year = '2001'
SET @ISOWeek = '53'
SET @NextYear = @Year+1
SELECT
CASE
WHEN @ISOWeek > 0
AND DATEADD(wk,DATEDIFF(wk,0,'01/04/'+@Year),0)+((@ISOWeek-1)*7)
< DATEADD(wk,DATEDIFF(wk,0,'01/04/'+@NextYear),0)
THEN DATEADD(wk,DATEDIFF(wk,0,'01/04/'+@Year),0)+((@ISOWeek-1)*7)
ELSE 0
END AS StartDate,
DATEADD(wk,DATEDIFF(wk,0,'01/04/'+@Year),0)+((@ISOWeek)*7)-1 AS EndDate
--todo... needs some error checking for the week number...
It returns a "0" or 01/01/1900 for the start date if the week was not found for the year...
I'm thinking that there's no ISO week 53 in 2001 but I might be using the wrong ISO "standard".
--Jeff Moden
RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.Change is inevitable... Change for the better is not.
Helpful Links:
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally) -
Hi,
The ISO Week problem arised last year for us. I programmed, tested and benchmarked a set of various conversion routines. Speed was essential in my case.
Above routine: 1360ms for 10000 conversions
fnISOWeek: 513ms for 10000 conversions
In general I first group results by SQL Year/Week then convert the grouped results using fnISOWeekFromSQL (very fast). Finally I regroup, as SQL separates the end and beginning weeks over a year end. This is faster in my case (millions of rows) than converting every row, especially in crosstabs.
I thought I'd share them, though please do credit me if you use them!:
fnISOWeek (@datetime) returns ISO week as an integer from SQL datetime
fnISOYear (@datetime) returns ISO Year as an integer from SQL datetime
fnISOYearWeek (@datetime) Returns the ISO Year Week in YYYYWW format of the given SQL date (useful for ordering data)
fnISOWeekFromSQL (@week, @year) returns ISO week given the SQL DATEPART week/year
fnISOYearWeekFromSQL (@week, @year) Returns the ISO Year Week in YYYYWW format given the SQL DATEPART Year and Week
fnISOWeekStartDate (@week, @year) Returns the date of the first day of the given ISO Week/Year (Monday)
As we're talking date functions, I also have a very fast midnight calculation if anybody is interested (calculates "12/1/2006 00:00:00" from "12/1/2006 xx:xx:xx")
Adrian Peakman
PS (Added)
Oh, the use of DATEPART(dw) is sensitive to the value of DATEFIRST. (I think everyone's routines are affected by this... )
My routines are coded to use the ISO Standard of Monday as the first day of the week (SET DATEFIRST 1). US installation is DATEFIRST 7 (Sunday) by default I believe. They can easily be altered to conform to other requirements.
---------------------------------------
CREATE FUNCTION fnISOWeek
--Adrian Peakman 6/12/2004
--Returns the ISO Week of the given date
(@d datetime)
returns int
AS
BEGIN
DECLARE @ISOWeek int
SET @ISOWeek= DATEPART(wk,@d)
IF DATEPART(dw,CAST(DATEPART(Year,@d) as varchar)) >4
begin
SET @ISOWeek=@ISOWeek-1
IF @ISOWeek=0
begin
SET @d = DATEADD(day,-3,@d)
SET @ISOWeek= DATEPART(wk,@d)
IF DATEPART(dw,CAST(DATEPART(Year,@d) as varchar)) >4
SET @ISOWeek=@ISOWeek-1
end
end
IF @ISOWeek=53
IF (DATEPART(dd,@d)-DATEPART(dw,@d))>= 28 SET @ISOweek=1
Return (@ISOWeek)
END
--------------------------------
CREATE FUNCTION fnISOYear
--Adrian Peakman 6/12/2004
--Returns the ISO Year of the given date
(@d datetime)
returns int
AS
BEGIN
DECLARE @ISOWeek int
DECLARE @iDayOfYear int
DECLARE @iYear int
SET @iDayOfYear=DATEPART(dy,@d)
SET @iYear=DATEPART(Year,@d)
IF @iDayOfYear > 3 AND @iDayOfYear < 363 RETURN(@iYear)
--Okay now this date may be part of the current year, the last year or the next year!
SET @ISOWeek= dbo.fnISOWeek(@d)
IF @iDayOfYear4
begin
SET @ISOWeek=@ISOWeek-1
IF @ISOWeek=0
begin
--SET @ISOWeek=dbo.fnISOWeek(DATEADD(day,-3,@d))
SET @d = DATEADD(day,-3,@d)
SET @ISOWeek= DATEPART(wk,@d)
IF DATEPART(dw,CAST(DATEPART(Year,@d) as varchar)) >4
SET @ISOWeek=@ISOWeek-1
IF @ISOWeek >0 SET @ISOYear=@ISOYear-1
end
end
IF @ISOWeek=53
IF (DATEPART(dd,@d)-DATEPART(dw,@d))>= 28
BEGIN
SET @ISOweek=1
SET @ISOYear=@ISOYear+1
END
RETURN(@ISOYear*100+@ISOWeek)
END
-------------------------------
CREATE FUNCTION fnISOWeekFromSQL
--Adrian Peakman 6/12/2004
--Returns the ISO Week given the SQL Week
(
@IntWeekNumber int,
@IntYear int
)
RETURNS int
As
BEGIN
DECLARE @ISOWeek int
DECLARE @Jan1Day datetime
SET @Jan1Day=DATEPART(dw,CONVERT(varchar,@IntYear))
SET @ISOWeek=@IntWeekNumber
IF @Jan1Day>4
SET @ISOWeek=@ISOWeek-1
IF @ISOWeek>52
begin
IF DATEPART(dw,CONVERT(varchar,@IntYear+1))0
RETURN(@IsoWeek)
SET @ISOWeek=53
IF DATEPART(dw,CONVERT(varchar,@IntYear-1)) >4
SET @ISOWeek=@ISOWeek-1
RETURN(@ISOWeek)
END
---------------------------------
CREATE FUNCTION fnISOYearWeekFromSQL
--Adrian Peakman 6/12/2004
--Returns the ISO Year Week in YYYYWW format given the SQL DATEPART Year and Week
(
@intWeekNumber int,
@intYear int
)
RETURNS int
As
BEGIN
DECLARE @ISOWeek int
DECLARE @ISOYear int
DECLARE @Jan1Day datetime
SET @Jan1Day=DATEPART(dw,CONVERT(varchar,@intYear))
SET @ISOWeek=@intWeekNumber
SET @ISOYear=@intYear
IF @Jan1Day>4
SET @ISOWeek=@ISOWeek-1
IF @ISOWeek>52
begin
IF DATEPART(dw,CONVERT(varchar,@intYear+1))0
RETURN(@ISOYear*100+@IsoWeek)
SET @ISOWeek=53
SET @ISOYear=@ISOYear-1
IF DATEPART(dw,CONVERT(varchar,@IntYear-1)) >4
begin
SET @ISOWeek=@ISOWeek-1
end
RETURN(@ISOYear*100+@ISOWeek)
END
----------------------------------
CREATE FUNCTION fnISOWeekStartDate
--Adrian Peakman 6/12/2004
--Returns the date of the first day of the given ISO Week/Year (Monday)
(
@IntWeekNumber int,
@IntYear int
)
RETURNS DateTime
As
BEGIN
DECLARE @StartDate DateTime
DECLARE @Jan1Day int
SET @StartDate=CONVERT(varchar,@IntYear)
SET @Jan1Day=DATEPART(dw,@StartDate)
IF @Jan1Day>4
Set @StartDate=DATEADD(day,1-@Jan1Day+7*@IntWeekNumber,@StartDate)
ELSE
Set @StartDate=DATEADD(day,1-@Jan1Day+7*(@IntWeekNumber-1),@StartDate)
RETURN(@StartDate)
END
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Incidentally the routine posted by rkintrup fails for me every few years:
Can't test Tomasz's routine as if fails on my installation (British dates). I'd be interested to know if my routines fail on other installations than British, hopefully they shouldn't! (erk!)
Oh, the use of DATEPART(dw) is sensitive to the value of DATEFIRST. (I think everyone's routines are affected by this... )
My routines are coded to use the ISO Standard of Monday as the first day of the week (SET DATEFIRST 1). US installation is DATEFIRST 7 (Sunday) by default I believe. They can easily be altered to conform to other requirements.
Thanks,
Adrian
Date fnISOWeek Rkintrup
002/01/1970 1 53
09/01/1970 2 54
16/01/1970 3 55
23/01/1970 4 56
30/01/1970 5 57
06/02/1970 6 58
13/02/1970 7 59
20/02/1970 8 60
27/02/1970 9 61
06/03/1970 10 62
13/03/1970 11 63
20/03/1970 12 64
02/01/1976 1 53
09/01/1976 2 54
16/01/1976 3 55
23/01/1976 4 56
30/01/1976 5 57
06/02/1976 6 58
13/02/1976 7 59
20/02/1976 8 60
27/02/1976 9 61
05/03/1976 10 62
12/03/1976 11 63
19/03/1976 12 64
02/01/1981 1 53
09/01/1981 2 54
16/01/1981 3 55
23/01/1981 4 56
30/01/1981 5 57
06/02/1981 6 58
13/02/1981 7 59
20/02/1981 8 60
27/02/1981 9 61
06/03/1981 10 62
13/03/1981 11 63
20/03/1981 12 64
02/01/1987 1 53
09/01/1987 2 54
16/01/1987 3 55
23/01/1987 4 56
30/01/1987 5 57
06/02/1987 6 58
13/02/1987 7 59
20/02/1987 8 60
27/02/1987 9 61
06/03/1987 10 62
13/03/1987 11 63
20/03/1987 12 64
02/01/1998 1 53
09/01/1998 2 54
16/01/1998 3 55
23/01/1998 4 56
30/01/1998 5 57
06/02/1998 6 58
13/02/1998 7 59
20/02/1998 8 60
27/02/1998 9 61
06/03/1998 10 62
13/03/1998 11 63
20/03/1998 12 64
02/01/2004 1 53
09/01/2004 2 54
16/01/2004 3 55
23/01/2004 4 56
30/01/2004 5 57
06/02/2004 6 58
13/02/2004 7 59
20/02/2004 8 60
27/02/2004 9 61
05/03/2004 10 62
12/03/2004 11 63
19/03/2004 12 64
02/01/2009 1 53
09/01/2009 2 54
16/01/2009 3 55
23/01/2009 4 56
30/01/2009 5 57
06/02/2009 6 58
13/02/2009 7 59
20/02/2009 8 60
27/02/2009 9 61
06/03/2009 10 62
13/03/2009 11 63
20/03/2009 12 64
02/01/2015 1 53
09/01/2015 2 54
16/01/2015 3 55
23/01/2015 4 56
30/01/2015 5 57
06/02/2015 6 58
13/02/2015 7 59
20/02/2015 8 60
27/02/2015 9 61
06/03/2015 10 62
13/03/2015 11 63
20/03/2015 12 64
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FYI, the article sites ISO6801, I believe the author is referring to ISO8601, and I have certainly run into this same kind of issue in respect to ensuring applications developed in different environments are compatible with 8601 defined week numbers and weekday numbers.
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Where does it say ISO 6801 in the article? I can't find any references to it. I know I had an error a long time ago, but that was changed to 8601.
Re: the setting of DATEFIRST. Note this section from my article:
First you should note that whenever the day of the week is checked for a specific date, we always make sure that the ISO standard of using Monday as the starting day of the week is used, regardless of the current setting of @@datefirst. This is accomplished by adding @@datefirst - 1 to the value returned by DATEPART(dw, ) and then dividing this number by 7. The remainder of the division is the day of the week for the specified date using Monday as the starting day of the week. Note though that Sunday becomes 0 with this expression, so if you need it to be 7 (for some calculation or whatever) you must exchange it for that.
Finally, regarding performance, the fastest way I have found to get ISO 8601 week numbers is to use the built-in routines in Visual Basic.NET. Creating a user-defined function in Visual Basic.NET is very easy, so if you're running SQL Server 2005 that is an option.
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I like your method Tomasz, it leads to much cleaner looking queries.
However, your code doesn't quite work.
In particular you are subtracting 1 from the day of the week twice
This modified version works
CREATE FUNCTION dbo.fnISOweek(@date datetime)
RETURNS int
AS
BEGIN
declare @dayOfWeek int
declare @1ThISOWeek1ThDay datetime
declare @4ThJan datetime
declare @1StJan datetime
declare @31StDec datetime
declare @numberOfISOweeks int
declare @Tmp int
declare @returnValue as int
set @4ThJan = cast(CAST(YEAR(@date) AS CHAR(4)) + '0104' as datetime)
set @1StJan = cast(CAST(YEAR(@date) AS CHAR(4)) + '0101' as datetime)
set @31StDec = cast(CAST(YEAR(@date) AS CHAR(4)) + '1231' as datetime)
--"rule of thursday"
if (DATEPART(dw, @1StJan) = 5 or DATEPART(dw, @31StDec) = 5 )
set @numberOfISOweeks = 53
else
set @numberOfISOweeks = 52
set @dayOfWeek = DATEPART(dw, @4ThJan)
set @1ThISOWeek1ThDay = DATEADD(day, -(@dayOfWeek -1) ,@4ThJan)
if @date @numberOfISOweeks
set @returnValue = 1 -- or -1 as an error
return @returnValue
END
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