get date of first monday of the given year?

  • Hi,

    How to get date of first monday of the given year?

  • Check the below code

    Declare @year varchar(10)='2011' -- Input

    Declare @Dpart Tinyint

    set @year = @year+'-01-01'

    set @Dpart= Datepart(DW,@year)

    Select Case When @Dpart = 2 Then @year When @Dpart < 2 Then DateAdd(dd,2-@Dpart,@year) Else DateAdd(dd,(7-Datepart(DW,@year))+2,@year) End

  • The above code relies on the first day of the week being a Sunday, the US default. Your first day may be different (mine is), which will give an incorrect result using that code. Check out @@DATEFIRST and SET DATEFIRST in BOL if you want to use it.

  • Fast and accurate:

    SELECT [FirstMonday] = DATEADD(dd,CASE x.wd WHEN 'mo' THEN 0 WHEN 'su' THEN 1 WHEN 'sa' THEN 2 WHEN 'fr' THEN 3 WHEN 'th' THEN 4 WHEN 'we' THEN 5 WHEN 'tu' THEN 6 END,d.FirstDayOfYear)

    FROM (SELECT [FirstDayOfYear] = DATEADD(YEAR,DATEDIFF(YEAR,0,GETDATE()),0)) d

    CROSS APPLY(SELECT [wd] = LEFT(DATENAME(weekday,d.[FirstDayOfYear]),2)) x

    β€œWrite the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw

    For fast, accurate and documented assistance in answering your questions, please read this article.
    Understanding and using APPLY, (I) and (II) Paul White
    Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden

  • Fast and accurate, but dependent on the Language of the server.

    This code relies only on knowning the date of some known monday (but is perhaps a little clunky)...

    ;withDayOne as (select dateadd(year,dateDiff(year,0,@dt),0) as dt)

    ,KnownMonday as (select cast('1900-03-05' as date) as dt)

    ,DayOneOffset as (select DateDiff(day,KnownMonday.dt,dateadd(year,dateDiff(year,0,@dt),0))%7 as num from KnownMonday)

    selectdateadd(day,(7-DayOneOffset.num)%7,DayOne.dt) as FirstMonday

    fromDayOne, DayOneOffset;

  • Oh, what a can of worms date handling in SQL Server is!!

    declare @known_monday varchar(10) = '1900-03-05'

    set dateformat dmy

    select cast(@known_monday as date) as Date05Mar1900

    , cast(@known_monday as datetime) as DateTime05Mar1900

    , cast(@known_monday as datetime2) as DateTime05Mar1900

    set dateformat ymd

    select cast(@known_monday as date) as Date05Mar1900

    , cast(@known_monday as datetime) as DateTime05Mar1900

    , cast(@known_monday as datetime2) as DateTime05Mar1900

    Date05Mar1900DateTime05Mar1900DateTime05Mar1900

    1900-03-051900-05-03 00:00:00.0001900-03-05 00:00:00.0000000

    Date05Mar1900DateTime05Mar1900DateTime05Mar1900

    1900-03-051900-03-05 00:00:00.0001900-03-05 00:00:00.0000000

    Don't know what others get, but for me (as shwon above) the above code returns 5th March 1900 for five out of the six conversions above. However, when DateFormat is set to DMY (which being British is the "normal" setting for me), it returns 3rd May when converted to DateTime.

    DatePart and DateName also return 3rd May when DateFormat is set to DMY:

    declare @known_monday varchar(10) = '1900-03-05'

    set dateformat dmy

    selectdatepart(dw,@known_monday) as WeekDay05Mar1900, datepart(day,@known_monday) as Day05Mar1900, datepart(month,@known_monday) as Month05Mar1900, datepart(year,@known_monday) as Year05Mar1900

    ,datepart(dw,'1900-03-05') as WeekDay05Mar1900, datepart(day,'1900-03-05') as Day05Mar1900, datepart(month,'1900-03-05') as Month05Mar1900, datepart(year,'1900-03-05') as Year05Mar1900

    ,datename(dw,'1900-03-05') as WeekDay05Mar1900, datename(day,'1900-03-05') as Day05Mar1900, datename(month,'1900-03-05') as Month05Mar1900, datename(year,'1900-03-05') as Year05Mar1900

    set dateformat ymd

    selectdatepart(dw,@known_monday) as WeekDay05Mar1900, datepart(day,@known_monday) as Day05Mar1900, datepart(month,@known_monday) as Month05Mar1900, datepart(year,@known_monday) as Year05Mar1900

    ,datepart(dw,'1900-03-05') as WeekDay05Mar1900, datepart(day,'1900-03-05') as Day05Mar1900, datepart(month,'1900-03-05') as Month05Mar1900, datepart(year,'1900-03-05') as Year05Mar1900

    ,datename(dw,'1900-03-05') as WeekDay05Mar1900, datename(day,'1900-03-05') as Day05Mar1900, datename(month,'1900-03-05') as Month05Mar1900, datename(year,'1900-03-05') as Year05Mar1900

    WeekDay05Mar1900Day05Mar1900Month05Mar1900Year05Mar1900WeekDay05Mar1900Day05Mar1900Month05Mar1900Year05Mar1900WeekDay05Mar1900Day05Mar1900Month05Mar1900Year05Mar1900

    43519004351900Thursday3May1900

    WeekDay05Mar1900Day05Mar1900Month05Mar1900Year05Mar1900WeekDay05Mar1900Day05Mar1900Month05Mar1900Year05Mar1900WeekDay05Mar1900Day05Mar1900Month05Mar1900Year05Mar1900

    15319001531900Monday5March1900

  • Quite easy with a calendar table.

    Select top 1 * FROM dbo.Calendar WHERE Y = 2011 AND DW = <whatever fits your setting> ORDER BY dt.

  • Ninja's_RGR'us (8/19/2011)


    Quite easy with a calendar table.

    Select top 1 * FROM dbo.Calendar WHERE Y = 2011 AND DW = <whatever fits your setting> ORDER BY dt.

    Just as easy without. Make one for the first week of the year and pick the correct row from it πŸ˜‰

    β€œWrite the query the simplest way. If through testing it becomes clear that the performance is inadequate, consider alternative query forms.” - Gail Shaw

    For fast, accurate and documented assistance in answering your questions, please read this article.
    Understanding and using APPLY, (I) and (II) Paul White
    Hidden RBAR: Triangular Joins / The "Numbers" or "Tally" Table: What it is and how it replaces a loop Jeff Moden

  • Are you, by any chance, working on getting the iso week number? Because if you are, have a look at this article[/url] (remember to follow the discussions link, there are many examples in there, including one of myself recently) and if the end product is to run on SQL 2008, simply use datepart(isowk, <your date>)



    Posting Data Etiquette - Jeff Moden[/url]
    Posting Performance Based Questions - Gail Shaw[/url]
    Hidden RBAR - Jeff Moden[/url]
    Cross Tabs and Pivots - Jeff Moden[/url]
    Catch-all queries - Gail Shaw[/url]


    If you don't have time to do it right, when will you have time to do it over?

  • R.P.Rozema (8/19/2011)


    Are you, by any chance, working on getting the iso week number? Because if you are, have a look at this article[/url] (remember to follow the discussions link, there are many examples in there, including one of myself recently) and if the end product is to run on SQL 2008, simply use datepart(isowk, <your date>)

    R.P.Rozema I saw that thread, and actually added the IsoWeek to my calendar table because of it. Never know when or if it might be handy.

    Sure enough, less than a week later, a post that could benefit from it;

    for my specific calendarTable the WHERE statement would be:

    WHERE YearNumber = 2011 AND IsoWeek = 1 AND DayOfWeek = 'Monday'

    Lowell


    --help us help you! If you post a question, make sure you include a CREATE TABLE... statement and INSERT INTO... statement into that table to give the volunteers here representative data. with your description of the problem, we can provide a tested, verifiable solution to your question! asking the question the right way gets you a tested answer the fastest way possible!

  • ChrisM@Work (8/19/2011)


    Ninja's_RGR'us (8/19/2011)


    Quite easy with a calendar table.

    Select top 1 * FROM dbo.Calendar WHERE Y = 2011 AND DW = <whatever fits your setting> ORDER BY dt.

    Just as easy without. Make one for the first week of the year and pick the correct row from it πŸ˜‰

    Too many good uses to not have 1.

    Since my table is less than 2 MB and that I can query the table with an <clustered>index I don't really see the point of trying to go much faster than that. Not saying it's impossible, just never had that need! πŸ˜‰

  • Ninja's_RGR'us (8/19/2011)


    ChrisM@Work (8/19/2011)


    Ninja's_RGR'us (8/19/2011)


    Quite easy with a calendar table.

    Select top 1 * FROM dbo.Calendar WHERE Y = 2011 AND DW = <whatever fits your setting> ORDER BY dt.

    Just as easy without. Make one for the first week of the year and pick the correct row from it πŸ˜‰

    Too many good uses to not have 1.

    Since my table is less than 2 MB and that I can query the table with an <clustered>index I don't really see the point of trying to go much faster than that. Not saying it's impossible, just never had that need! πŸ˜‰

    I agree with Remi on this one. There are just too many uses for calendar tables not to have them. When properly done they give better performance than complicated date math in most cases. You can spell out a month name any way you want in any language and have it displayed correctly. Very simple aggregating and grouping. All sorts of uses.

    Todd Fifield

  • The code to find the first Monday of the year is fairly simple:

    First find the 7th day of the year (Jan 7):

    dateadd(yy,datediff(yy,0,a.DT),6)and then find the Monday on or before that date:

    dateadd(dd,(datediff(dd,0, JanuarySeventh )/7)*7,0)

    Does not depend on any setting of language or datefirst.

    select

    DT,

    FirstMonday =

    dateadd(dd,(datediff(dd,0,dateadd(yy,datediff(yy,0,a.DT),6))/7)*7,0)

    from

    ( -- Test Data

    select DT = getdate()union all

    select DT = '20111231'union all

    select DT = '20121231'union all

    select DT = '20131231'union all

    select DT = '20141231'union all

    select DT = '20110101'union all

    select DT = '20080229'

    ) a

    Results:

    DT FirstMonday

    ------------------------ -----------------------

    2011-08-19 14:42:26.310 2011-01-03 00:00:00.000

    2011-12-31 00:00:00.000 2011-01-03 00:00:00.000

    2012-12-31 00:00:00.000 2012-01-02 00:00:00.000

    2013-12-31 00:00:00.000 2013-01-07 00:00:00.000

    2014-12-31 00:00:00.000 2014-01-06 00:00:00.000

    2011-01-01 00:00:00.000 2011-01-03 00:00:00.000

    2008-02-29 00:00:00.000 2008-01-07 00:00:00.000

  • That's the same algorithm I posted further up the thread, although rather better explained.

    The only assumption that is made is that you know a Monday, which in your case is day 0.

    I rather unnecessarily tied myself in knots trying to make that assumption explicit and avoid dates before February 1900, to avoid leap year confusions.

  • paul_ramster (8/23/2011)


    That's the same algorithm I posted further up the thread, although rather better explained.

    The only assumption that is made is that you know a Monday, which in your case is day 0.

    I rather unnecessarily tied myself in knots trying to make that assumption explicit and avoid dates before February 1900, to avoid leap year confusions.

    If you are concerned about dates before 1900-01-01, then you could use 1753-01-01, which is the earliest possible datetime and also a Monday. Ask me how I knew that. πŸ™‚

    Start of Week Function

    http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=47307

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