SQL Server not showing correct week numbers for given date - tsql

Question

SQL Server is showing week 53 for first week of 2011 except 1th of January, and needs to be week 1.
Below is the query and output:
declare #T table (dt datetime)
insert into #T values
('2010-12-26'),
('2010-12-27'),
('2010-12-28'),
('2010-12-29'),
('2010-12-30'),
('2010-12-31'),
('2011-01-01'),
('2011-01-02'),
('2011-01-03'),
('2011-01-04'),
('2011-01-05'),
('2011-01-06'),
('2011-01-07'),
('2011-01-08')
select dt,DATEPART(wk,dt) from #T
Output:
2010-12-26 00:00:00.000 53
2010-12-27 00:00:00.000 53
2010-12-28 00:00:00.000 53
2010-12-29 00:00:00.000 53
2010-12-30 00:00:00.000 53
2010-12-31 00:00:00.000 53
2011-01-01 00:00:00.000 1
2011-01-02 00:00:00.000 2
2011-01-03 00:00:00.000 2
2011-01-04 00:00:00.000 2
2011-01-05 00:00:00.000 2
2011-01-06 00:00:00.000 2
2011-01-07 00:00:00.000 2
2011-01-08 00:00:00.000 2
I want SQL Server to show week 1 from Dec 26th - Jan 1th. Does anybody know how to accomplish this?
Thanks and regards,
Aschwin.

Answer 1

It was alot harder than I first expected. I am comparing the end of last year to see if it is qualified to be part of the new year. If so i set the week as 1, otherwise i just use the normal week.
declare #T table (dt datetime)
insert into #T values
('2010-12-25'),
('2010-12-26'),
('2010-12-27'),
('2010-12-28'),
('2010-12-29'),
('2010-12-30'),
('2010-12-31'),
('2011-01-01'),
('2011-01-02'),
('2011-01-03'),
('2011-01-04'),
('2011-01-05'),
('2011-01-06'),
('2011-01-07'),
('2011-01-08'),
('2011-12-31'),
('2012-01-01')
select dt,
week = case when dt + 6 - datediff(day, -1, dt) % 7 = dateadd(year, datediff(year,-1, dt), 0)
then 1 else datepart(week, dt) end from #t
Proof:
https://data.stackexchange.com/stackoverflow/q/110527/

Answer 2

I am not sure it holds for all years (but it looks like it) but you could solve this using a CASE statement.
SELECT dt
, CASE WHEN DATEPART(wk, dt) <> 53
THEN DATEPART(wk, dt)
ELSE 1
END
FROM #T
The new ISO_WEEK datepart doesn't apply to your requested output.

Answer 3

I Created 2 functions to deal with this issue
1) to get First or last day of the week
2) to get the week number or year
function 1
CREATE FUNCTION [dbo].[fn_GetDayOf]
(
#Date datetime,
--#FirstDayOfWeek int = 7,
#Mode int =1
)
/*
Mode 1: First Day Of Week
Mode 2: Last Day Of Week
*/
RETURNS datetime
WITH EXECUTE AS CALLER
BEGIN
Declare #Return datetime
--SET DATEFIRST #FirstDayOfWeek
IF #Mode = 1
BEGIN
select #Return = dateadd(day,-(datepart(weekday,#date)-1),convert(date,#date))
END
ELSE IF #Mode = 2
BEGIN
select #Return = dateadd(SECOND,-1,convert(datetime,dateadd(day,(datepart(weekday,#date)),convert(date,#date))))
END
ELSE
BEGIN
SET #Return = #Date
END
--SET DATEFIRST 7
RETURN #Return
END
Function 2
CREATE FUNCTION [dbo].[fn_GetYearWeek]
(
#Date datetime,
--#FirstDayOfWeek int = 7,
#Mode int =1
)
/*
Mode 1 = Week Number
Mode 2 = Year
*/
RETURNS INT
BEGIN
declare #Return int
IF #Mode = 1
BEGIN
select #Return = case when datepart(week,[dbo].[fn_GetDayOf] (#Date,1)) <> datepart(week,[dbo].[fn_GetDayOf] (#Date,2)) then datepart(week,[dbo].[fn_GetDayOf] (#Date,1)) else datepart(week,[dbo].[fn_GetDayOf] (#Date,2)) end
END
ELSE IF #Mode = 2
BEGIN
select #Return = case when datepart(WEEK,[dbo].[fn_GetDayOf] (#Date,1)) <> datepart(week,[dbo].[fn_GetDayOf] (#Date,2)) then datepart(YEAR,[dbo].[fn_GetDayOf] (#Date,1)) else datepart(YEAR,[dbo].[fn_GetDayOf] (#Date,2)) end
END
ELSE
BEGIN
SET #Return = -1
END
Return #Return
END
Running Example
declare #T table (dt datetime)
insert into #T values
('2010-12-25'),
('2010-12-26'),
('2010-12-27'),
('2010-12-28'),
('2010-12-29'),
('2010-12-30'),
('2010-12-31'),
('2011-01-01'),
('2011-01-02'),
('2011-01-03'),
('2011-01-04'),
('2011-01-05'),
('2011-01-06'),
('2011-01-07'),
('2011-01-08'),
('2011-12-31'),
('2012-01-01'),
('2012-01-02'),
('2012-12-31'),
('2013-01-01')
select
dt,
datepart(week,dt),
--case when datepart(week,[dbo].[fn_GetDayOf] (dt,1)) <> datepart(week,[dbo].[fn_GetDayOf] (dt,2)) then datepart(week,[dbo].[fn_GetDayOf] (dt,1)) else datepart(week,[dbo].[fn_GetDayOf] (dt,2)) end
[dbo].[fn_GetYearWeek] (dt,1),
[dbo].[fn_GetYearWeek] (dt,2)
from #T
result:

Answer 4

Another way to retrieve the total number of weeks in current year:
DECLARE #LASTDAY DATETIME
DECLARE #weeks INT
SET #LASTDAY = DATEADD(ms,-3,DATEADD(yy,0,DATEADD(yy,DATEDIFF(yy,0,GETDATE())+1,0)))
SELECT #weeks = CASE DATEname(dw,#LASTDAY)
WHEN 'MONDAY' THEN DATEPART(WK, DATEADD(wk,DATEDIFF(wk,7,#LASTDAY),5))
WHEN 'TUESDAY' THEN DATEPART(WK, DATEADD(wk,DATEDIFF(wk,7,#LASTDAY),5))
WHEN 'WEDNESDAY' THEN DATEPART(WK, DATEADD(wk,DATEDIFF(wk,7,#LASTDAY),5))
ELSE DATEPART(WK, #LASTDAY)
END
select #weeks