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Jeff Moden (3/22/2010) Lookup CREATE FUNCTION in Books Online... Example "A" is a nice, short function that determines ISO week. It could probably be optimized to be an INLINE function for performance reasons or might be able to be optimized to simply be a formula instead of a function, but it's a whole lot shorter and easier to understand than some of the example code I've seen on this thread.
The function in BOL depends on the setting of DATEFIRST, so it isn't all that useful as a universal function.
This is a bit longer, but does not depend on the setting of DATEFIRST: ISO Week of Year Function http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=60510




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I needed the iso week number too in SQL 2005. Looked for it in all the usual places (bol, forums, google, etc) and found many different implementations. Most however have shortcomings. Usually a combination of one or more of:  using date string manipulations (dependent on locale) or  dependent on datefirst setting or  implemented procedural. It must be possible to do better than that. So this is what I wrote.
The following query calculates for a given (set of) date(s), the iso week number plus it's iso year. And it does not use any string manipulations/conversions, it is independent of datefirst and it can be put into a single select statement, a view or a cte. Even though SQL 2008 now supports datepart(isowk, ) which does the same thing, I hope someone will still find it useful for older versions:
select d.date ,case when d.date < x.thisfirstmonday then datepart(year, x.prevjan4) when d.date >= x.nextfirstmonday then datepart(year, x.nextjan4) else datepart(year, x.thisjan4) end as isoyear ,case when d.date < x.thisfirstmonday then 1 + datediff(day, x.prevfirstmonday, d.date) / 7 when d.date >= x.nextfirstmonday then 1 else 1 + datediff(day, x.thisfirstmonday, d.date) / 7 end isoweeknumber from (  To demonstrate show week numbers for 14 days around today's date,  Januari 1st and December 31st. select dateadd(day, 7  t.n, dt.date) as date from ( select dateadd(day, datediff(day, 0, getdate()), 0) as date union select dateadd(year, datediff(year, 0, getdate()), 0) union select dateadd(day, 1, dateadd(year, 1 + datediff(year, 0, getdate()), 0)) ) dt cross join (select 1 as n union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9 union all select 10 union all select 11 union all select 12 union all select 13 union all select 14 ) t ) d cross apply ( select max(case t.n when 1 then x1.jan4 else 0 end) as prevjan4 ,max(case t.n when 0 then x1.jan4 else 0 end) as thisjan4 ,max(case t.n when 1 then x1.jan4 else 0 end) as nextjan4 ,max(case t.n when 1 then x2.jan4weekday else 0 end) as prevjan4weekday ,max(case t.n when 0 then x2.jan4weekday else 0 end) as thisjan4weekday ,max(case t.n when 1 then x2.jan4weekday else 0 end) as nextjan4weekday ,max(case t.n when 1 then x3.firstmonday else 0 end) as prevfirstmonday ,max(case t.n when 0 then x3.firstmonday else 0 end) as thisfirstmonday ,max(case t.n when 1 then x3.firstmonday else 0 end) as nextfirstmonday from ( select 1 as n union all select 0 union all select 1 ) t cross apply ( select dateadd(year, t.n + datediff(year, 0, d.date), dateadd(day, 3, 0)) as jan4 ) x1 cross apply ( select (2 + datepart(dw, x1.jan4) + @@datefirst) % 7 + 1 as jan4weekday ) x2 cross apply ( select dateadd(day, 1  x2.jan4weekday, x1.jan4) as firstmonday ) x3 ) x order by 1;
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I've been struggling through the logic of ISO 8601 myself. My solution does not look like anything I've seen in this thread yet, so I am posting my solution for consideration. The approach I took boils down to the following steps:
1. Let TargetDate equal the date for which we want to know the ISO week value.
2. Let CurrentThursday equal the date of the Thursday associated with the same week as the TargetDate (taking into account that Sunday belongs with the prior Thursday).
3. Let FirstThursday equal the date of the Thursday of the week containing January 4th, basing that on the year value of the CurrentThursday (not the TargetDate) and still taking into account that Sunday belongs with the prior Thursday. (Why January 4th? January 4th is always in the first ISO week of the year, which is not true for January 1st.)
4. Let ISO_Week_Value equal the difference, in weeks, between the FirstThursday and the CurrentThursday, plus one.
Here is a query that uses variables to reflect each step of the algorithm and includes two AllInOne versions that rely on only the @TargetDate variable and no others. The @DateOffset variable is used to make sure that Monday  Wednesday are always in the same week as the following Thursday and that Friday  Sunday are always in the same week as the prior Thursday no matter what the @@datefirst value is. The @ThursdayOffset variable is used to find Thursday no matter what the @@datefirst value is.
declare @DateOffset as smallint, @ThursdayOffset as smallint, @TargetDate as datetime, @CurrentThursday as datetime, @FirstThursday as datetime, @ISO_WeekNumber as tinyint
set @DateOffset = case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end set @ThursdayOffset = case when @@datefirst < 5 then 5  @@datefirst else 12  @@datefirst end
set @TargetDate = getdate()
set @CurrentThursday = dateadd(day, @ThursdayOffset  datepart(dw, @TargetDate + @DateOffset), @TargetDate + @DateOffset)
set @FirstThursday = dateadd(day, @ThursdayOffset  datepart(dw, cast('1/4/' + datename(year, @CurrentThursday) as datetime) + @DateOffset), cast('1/4/' + datename(year, @CurrentThursday) as datetime) + @DateOffset)
set @ISO_WeekNumber = datediff(week, @FirstThursday, @CurrentThursday) + 1
select TestDate = @TargetDate, CurrentThursday = @CurrentThursday, FirstThursday = @FirstThursday, ISO_WeekNumber = @ISO_WeekNumber, ISO_WeekYear = year(@CurrentThursday),
ISO_WeekNumber_AllInOne = datediff(week, dateadd(day, case when @@datefirst < 5 then 5  @@datefirst else 12  @@datefirst end  datepart(dw, cast('1/4/' + datename(year, dateadd(day, case when @@datefirst < 5 then 5  @@datefirst else 12  @@datefirst end  datepart(dw, @TargetDate + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end), @TargetDate + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end)) as datetime) + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end), cast('1/4/' + datename(year, dateadd(day, case when @@datefirst < 5 then 5  @@datefirst else 12  @@datefirst end  datepart(dw, @TargetDate + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end), @TargetDate + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end)) as datetime) + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end), dateadd(day, case when @@datefirst < 5 then 5  @@datefirst else 12  @@datefirst end  datepart(dw, @TargetDate + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end), @TargetDate + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end)) + 1,
ISO_WeekYear_AllInOne = year(dateadd(day, case when @@datefirst < 5 then 5  @@datefirst else 12  @@datefirst end  datepart(dw, @TargetDate + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end), @TargetDate + case when @@datefirst > 4 then @@datefirst  8 when @@datefirst > 1 then @@datefirst  1 else 0 end))




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This one is fairly short and I did a lot of testing to verify it works OK: ISO Week of Year Function http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=60510
This has the week and day of week included: ISO Year Week Day of Week Function http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=60515Returns the ISO 8601 Year Week Day of Week in format YYYYW01D for the date passed.
There are a number of columns with ISO week support in this date table function: Date Table Function F_TABLE_DATE: http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=61519 ISO_YEAR_WEEK_NO  ISO 8601 year and week in format YYYYWW Example = 200403
ISO_WEEK_NO  ISO 8601 week of year in format WW Example = 52
ISO_DAY_OF_WEEK  ISO 8601 Day of week number, Mon=1, Tue=2, Wed=3, Thu=4, Fri=5, Sat=6, Sun=7
ISO_YEAR_WEEK_NAME  ISO 8601 year and week in format YYYYWNN Example = 2004W52
ISO_YEAR_WEEK_DAY_OF_WEEK_NAME  ISO 8601 year, week, and day of week in format YYYYWNND Example = 2004W522
Edit: Didn't see that this is an old thread and that I had already posted on it.



