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

Posted on 2010-08-18
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Last Modified: 2012-05-10
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.
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Question by:dgerler
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  • +4
10 Comments
 
LVL 25

Expert Comment

by:Lee Savidge
Comment Utility
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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LVL 16

Expert Comment

by:vdr1620
Comment Utility
SELECT DATEADD(week,34,'2009-01-01')
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LVL 10

Expert Comment

by:Jini Jose
Comment Utility


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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Author Comment

by:dgerler
Comment Utility
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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LVL 16

Expert Comment

by:vdr1620
Comment Utility
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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LVL 11

Expert Comment

by:Larissa T
Comment Utility
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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LVL 11

Assisted Solution

by:Larissa T
Larissa T earned 200 total points
Comment Utility
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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Author Comment

by:dgerler
Comment Utility
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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LVL 69

Expert Comment

by:ScottPletcher
Comment Utility
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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LVL 58

Accepted Solution

by:
cyberkiwi earned 300 total points
Comment Utility

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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