Determine 'this week' in T-SQL

Question

This is locale specific to the US wherein it considered that the start of a week is Sunday; I want to be able to ask SQL to give me the date of the next Sunday relative to today [getDate()]. If today is Jan 15 it should return Jan 18; if today were Sunday it should return the following Sunday which is the 25th. This would be trivial to write a UDF for but I was curious if anyone had other tricks/ideas?

Answer 1

DECLARE @d AS datetime
SET @d = '1/15/2009'
PRINT @d
PRINT DATEADD(day, 8 - DATEPART(weekday, @d), @d)
SET @d = '1/18/2009'
PRINT @d
PRINT DATEADD(day, 8 - DATEPART(weekday, @d), @d)

-- So it should be able to be used inline pretty efficiently:
DATEADD(day, 8 - DATEPART(weekday, datecolumn), datecolumn)

-- If you want to change the first day for a different convention, simply use SET DATEFIRST before performing the operation
-- e.g. for Monday: SET DATEFIRST 1
-- e.g. for Saturday: SET DATEFIRST 6

DECLARE @restore AS int
SET @restore = @@DATEFIRST
SET DATEFIRST 1

DECLARE @d AS datetime
SET @d = '1/15/2009'
PRINT @d
PRINT DATEADD(day, 8 - DATEPART(weekday, @d), @d)
SET @d = '1/19/2009'
PRINT @d

PRINT DATEADD(day, 8 - DATEPART(weekday, @d), @d)
SET DATEFIRST @restore

Answer 2

Today's day-of-week:
SELECT @dow = DATEPART(d, GETDATE()) where 1 = Sunday, 7 = Saturday

You want to add enough days to get the next Sunday.

If today is 1 = Sunday, add 7
If today is 2 = Monday, add 6
If today is 3 = Tuesday, add 5 etc.

so you are always adding 8 - today's day-of-week value.

SELECT DATEADD(d, GETDATE(), 8 - @dow(GETDATE))

EDIT: But Cade wins!

Answer 3

Thanks for the SET DATEFIRST hint, thats just helped me out with a query I had.