Isoweek in SQL Server 2005

Question

In SQL Server 2008 the isoweek can be found with this:

SELECT datepart(iso_week, getdate())

Before SQL Server 2008 there were no built-in f

Answer 1

There is a link here for other earlier attempts http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=60510

This is the OLD code for the function

CREATE function f_isoweek(@date datetime)
RETURNS INT
as
BEGIN
DECLARE @rv int

SELECT @rv = datediff(ww, dateadd(ww, datediff(d, 0, dateadd(yy, datediff(yy, 0, day4),3))/7,-4),day4)
FROM (SELECT dateadd(ww, datediff(day, 0, @date)/7, 3) day4) a

RETURN @rv
END

After combining @AndriyM 's brilliant answer with my own, we are down to 1 line. This is the NEW code.

CREATE function f_isoweek(@date datetime)
RETURNS INT
as
BEGIN

RETURN (datepart(DY, datediff(d, 0, @date) / 7 * 7 + 3)+6) / 7
-- replaced code for yet another improvement.
--RETURN (datepart(DY, dateadd(ww, datediff(d, 0, @date) / 7, 3))+6) / 7

END

Explanation for the old code (not going to explain the new code. It is fragments from my code and AndriyM's code):

Finding weekday 4 of the chosen date

dateadd(week, datediff(day, 0, @date)/7, 3) 

Finding isoyear - year of weekday 4 of a week is always the same year as the isoyear of that week

datediff(yy, 0, day4)

When adding 3 days to the first day of the isoyear a random day of the first isoweek of the isoyear is found

dateadd(yy, datediff(yy, 0, day4),3)

finding relative week of the first isoweek of the isoyear

datediff(d, 0, dateadd(yy, datediff(yy, 0, day4),3))/7

Finding the monday minus 4 days of the first isoweek results in thursday of the week BEFORE the first day of the first isoweek of the isoyear

dateadd(ww, datediff(d, 0, dateadd(yy, datediff(yy, 0, day4),3))/7,-4)

Knowing first thursday of the week before the first isoweek and first thursday of the chosen week, makes it is quite easy to calculate the week, it doesn't matter which setting datefirst has since the weekdays of both dates are thursdays.

datediff(ww, dateadd(ww, datediff(d, 0, dateadd(yy, datediff(yy, 0, day4),3))/7,-4),day4)