Using T-SQL Need to return first day of the 34th week last year.

I need to calculate what the date was for the first day of the 34th week of last year. This so I can compare week 34 last year to this week (also week 34).

I am using Sql Server 2005.
dgerlerAsked:
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Lee SavidgeCommented:
Well, as the 34th week is 238 days in try this:

Lee
declare @dt datetime
select @dt = cast('1 Jan 2009' as datetime) + 238
select @dt

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vdr1620Commented:
SELECT DATEADD(week,34,'2009-01-01')
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Jini JoseSenior .Net DeveloperCommented:


declare @d datetime
declare @numbers table (n int)

set @d = '5/13/2005'

insert into @numbers(n)
select 0 union all
select 1 union all
select 2 union all
select 3 union all
select 4 union all
select 5 union all
select 6 union all
select -1 union all
select -2 union all
select -3 union all
select -4 union all
select -5 union all
select -6

select min(d) AS WeekBegin, max(d) AS WeekEnd
from
(
      select dateadd(d, n, @d) as d, datepart(week, dateadd(d, n, @d)) as w
      from @numbers
) t
where datepart(week, @d) = w




You might want to consider making the numbers table a permanent table so you don't have to build it each time.

EDIT: modified the numbers table for the values that we need.

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dgerlerAuthor Commented:
My week should always start on Monday and week one would be the first full week of the year. In other words, week one starts with the first Monday of the year.
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vdr1620Commented:
DECLARE @YearBegin DATE = (SELECT DATEADD(yy, DATEDIFF(yy,0,getdate()), 0))
DECLARE @WKBegin DATE  = (SELECT DATEADD(wk, DATEDIFF(wk,0, dateadd(dd,6-datepart(day,@YearBegin),@YearBegin)), 0))

SELECT DATEADD(week,34,@WKBegin)
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Lara FEACommented:
You may want to redefine what is current week, According to your week numbering this week is not 34 week of the year, but 33

declare @lastYearFM datetime, @thisYearFM datetime, @oneOff int, @thisWeek int
select @thisYearFM=getdate() - datepart(dy,getdate())+1
select  @lastYearFM= dateadd(year,-1,@thisYearFM), @oneOff = 0

select @thisYearFM, @lastYearFM

if  datepart(dw,@lastYearFM) > 2       select @lastYearFM = @lastYearFM - datepart(dw,@lastYearFM), @oneOff=1

select @thisWeek = datepart(week,getdate())
select @thisYearFM, @lastYearFM, @thisWeek

select @lastYearFM+7*@thisWeek
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Lara FEACommented:
Sorry,  +/- 1 errors.
 I am in US - so week is Sunday (day 1) - Saturday (day 7)
declare @lastYearFM datetime, @thisYearFM datetime, @oneOff int, @thisWeek int
select @thisYearFM=getdate() - datepart(dy,getdate())+1
select  @lastYearFM= dateadd(year,-1,@thisYearFM), @oneOff = 0

select @thisYearFM, @lastYearFM, @thisWeek

select @thisYearFM,datepart(dw,@thisYearFM), @lastYearFM,datepart(dw,@lastYearFM)

if  datepart(dw,@lastYearFM) > 2       select @lastYearFM = @lastYearFM + (8-datepart(dw,@lastYearFM)), @oneOff=1
if  datepart(dw,@thisYearFM) > 2       select @thisYearFM = @thisYearFM + (8-datepart(dw,@thisYearFM))

select @thisWeek = datepart(week,getdate())-@oneOff

select @thisYearFM, @lastYearFM, @thisWeek

select @lastYearFM+7*(@thisWeek-1)
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dgerlerAuthor Commented:
Could be. I got the week number from what vbscript gave me and it's probably using "System default"

What I really need is a script that will return a Monday at the beginning of a numbered week. this year and last.

The ultimate goal is to be able to compare this week to date to last year's week to date for the corresponding week.

Week 1 of a given year will be the first full week of the year. This allows us to compare Monday and Tuesday of this week to a Monday and Tuesday last year in the corresponding week whether this week is 33 or 34.
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Scott PletcherSenior DBACommented:
You can get the first Monday of any year with the computation shown below.  
To get a specific week-starting date, we just need to add (wk#  - 1) * 7 days to that starting date.  
For example:
DECLARE @date datetime
DECLARE @date2 datetime
DECLARE @weekno tinyint

SET @date = DATEADD(YEAR, -0, GETDATE()) --adjust -0 to test for different years
SET @date2 = DATEADD(YEAR, -1, GETDATE()) --get prior yr of @date year
SET @weekno = 34

SELECT DATEADD(DAY, (@weekno - 1) * 7, 
    --determine first Monday of the year
    DATEADD(WEEK, (DATEDIFF(DAY, 0, CAST(YEAR(@date) AS char(4)) + '0101') + 6) / 7, 0)) AS WeekNoMondayDate,
       DATEADD(DAY, (@weekno - 1) * 7, 
    --determine first Monday of the year
    DATEADD(WEEK, (DATEDIFF(DAY, 0, CAST(YEAR(@date2) AS char(4)) + '0101') + 6) / 7, 0)) AS WeekNoMondayDate2

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cyberkiwiCommented:

drop function dbo.weekXMonday
go
create function dbo.weekXMonday(@year int, @weekno int)
returns datetime as
begin
declare @d datetime
set @d=dateadd(yy, @year-1900, 0)
return (dateadd(wk,@weekno-1,@d)-((datepart(dw,@d-2)+@@datefirst)%7))
return @d
end
GO
select dbo.weekXMonday(2009,34)
select dbo.weekXMonday(2010,34)

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Microsoft SQL Server 2005

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