Question

There is a datepart iso_week returning the iso_week for a selected day.

SELECT DATEPART(iso_week, getdate())

This will return the european standard for week current. It is not nessasarily the same as week.

Here comes the tricky part:

SELECT DATEPART(iso_week, '2011-01-01')

Returns

52 

so it belongs to last year
In order to group by year and iso_week, I need to consider that iso_week is not from the same year.

Iso_week starts on monday and belongs to the year where most days overlap. So 4 days in 2010 and 3 days in 2011, the week is 52 and all days belong to iso_year 2010. However TSQL doesn’t have the detepart iso_year.

declare @t table(col1 datetime)

insert @t 
ALL SELECT '2010-12-28'
UNION ALL SELECT '2010-12-29'
UNION ALL SELECT '2010-12-30'
UNION ALL SELECT '2010-12-31'
UNION ALL SELECT '2011-01-01'
UNION ALL SELECT '2011-01-02'
--UNION ALL SELECT '2011-01-03' Stackexchange is bugged and can't handle this line showing 53
UNION ALL SELECT '2011-01-04'

I need something like (iso_year doesn’t exist):

SELECT DATEPART(ISO_WEEK, col1) WEEK, DATEPART(iso_YEAR, col1) YEAR, COUNT(*) COUNT 
FROM @t
GROUP BY DATEPART(ISO_WEEK, col1), DATEPART(iso_YEAR, col1)
ORDER BY  2,1

Expected result

WEEK  YEAR          COUNT
52    2010           6
1     2011           2

Answer 1

The same ISO week’s Thursday will unambiguously provide you with the correct year. And this answer can provide you with the idea how to obtain the right Thursday from the given date.

SELECT
  Week  = DATEPART(ISOWK, TheThursday),
  Year  = DATEPART(YEAR, TheThursday),
  Count = COUNT(*)
FROM (
  SELECT
    TheThursday = DATEADD(
      DAY,
      3 - (DATEPART(DW, col1) + @@DATEFIRST - 2) % 7,
      col1
    )
  FROM @t
) s
GROUP BY
  TheThursday