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Calcuate the number of weeks in a month

    Question

  • I need a query to return the # of weeks ina  month for eg

    June has MAY has 4 weeks, ie 5-1-07 thru 5-7-07 is week 1,

    '5-6-07' and '5-10-07' is week 2 ETc.. hence i need my results to show as follows, but it needs to be automated for every month, ie the number of weeks in a month.

    EG

    when date between '5-1-07' and '5-5-07' then 'Week1'
    when date between  '5-6-07' and '5-12-07' then week2'
    when date between '5-13-07' and '5-19-07' then 'week3'

     

    I need the results, above to be automated for each month..

     

     

    Monday, June 25, 2007 5:00 PM

Answers

  • Try:

     

    declare @d datetime

     

    set @d = '20070515'

     

    select datediff(week, convert(varchar(6), @d, 112) + '01', @d) + 1

    go

     

     

    AMB

    Tuesday, June 26, 2007 12:51 PM

All replies

  • You should consider the use of a Calendar table. (See this reference.)

     

    A Calendar table is a very useful support object in most databases where it is often necessary to handle date related data.

    Monday, June 25, 2007 5:22 PM
  • I do have the calendar tbl. But that does not ahve the week of the month in it. I also looked at the reference u listed, I need the week of the month, ie between 1-5, not the week of the year.
    Monday, June 25, 2007 6:08 PM
  • I do have the calendar tbl. But that does not ahve the week of the month in it. I also looked at the reference u listed, I need the week of that  month, ie in May we have 5 weeks, so I need it sho show as week 1, week2 etc..ie between 1-5, not the week of the year.
    Monday, June 25, 2007 6:13 PM
  •  

     

    Without knowing quite how you want to use it...here's an approach to start.

    Code Snippet

     

    declare @date datetime

    set @date = getdate()

     

    select 1 + (W + first.first_week_nbr) as MoWeekNbr

    from dbo.Calendar

    inner join

    (

        select @date as seldate, first_week_nbr

        from dbo.calendar where dt = dateadd(d, -1*(day(@date)-1), @date)

    ) first

    on first.seldate = @date

    where dt = @date

     

    Monday, June 25, 2007 6:27 PM
  • It is easy to add additional columns to the Calendar table for your specific needs.

     

    If you need a 'Week of the Month', then add a column and use an UPDATE to populate the data. It's not a complex algorithm...

    Monday, June 25, 2007 7:21 PM
  • Does the first week of each month ALWAYS start on the first?
    Monday, June 25, 2007 7:31 PM
  • Yes,  the first week of the month, always starts on the FIRST.
    Monday, June 25, 2007 7:48 PM
  • This code gives me erros such as Invalid column name 'first_week_nbr'. etc.. we dont have the same calendar table
    Monday, June 25, 2007 7:51 PM
  • IF you add a 'computed' WeekOfMonth column to your Calendar table, the following will create the proper values.

     

    Code Snippet


    ALTER TABLE Calendar
       ADD WeekOfMonth AS datepart( wk, dateadd( day, 0, datediff( day, dateadd( month, datediff( month, 0, [date] ), 0 ), [date] )))

     

    Now you can use a JOIN with the Calendar table to find the WeekOfMonth and use that value in groupings, etc.

    Monday, June 25, 2007 8:10 PM
  • THis code does not return the rite result, for eg if the date is 2007-02-05, it falls under week 2.-BUT The code returns week1

    select datepart( wk, dateadd( day, 0, datediff( day, dateadd( month, datediff( month, 0, '20070205' ), 0 ),
    '20070205' )))

    Monday, June 25, 2007 9:14 PM
  • I asked you specifically, if

    Does the first week of each month ALWAYS start on the first?
    .

     

    And you replied,

    Yes,  the first week of the month, always starts on the FIRST.

     

    Following that logic, the first through the 7th is in week one, 8th through 14th in week two, etc. Therefore '2007-02-05' would correctly fall in week one.

     

    However, it seems like now your previous response may have mislead me. (And of course, I take responsibility for not asking my question in a more exacting form. I clearly see how the confusion exists.)

     

    It appears that you may have meant that a week is from Sunday-Saturday (Calendar). And that any number of days falling within a calendar week (even if that calendar week contains the days of a different month) constitutues the 'first week of the month'.

     

    My assumption was that a week was seven days.

     

    Please clarify. What determines a Week? (Numerical or Calendar)

    Monday, June 25, 2007 10:43 PM
  • select day(getdate()) / 7 as [WeekOfMonth]

    , convert(char(1),day(getdate()) / 7 ) + 'Week' as [MonthWeek]

    --

    select *

    ,day(<DateField>) / 7 as [WeekOfMonth]

    , convert(char(1),day(<DateField>) / 7 ) + 'Week' as [MonthWeek]

    from <TableName>

     

     

     

    Tuesday, June 26, 2007 2:16 AM
  • Rusag2,

     

    Somehow, this just doesn't seem right... (using your suggested algorithms)

     


    Code Snippet

    Select
       Today = getdate(),
       WeekOfMonth = day(getdate()) / 7,
       MonthWeek = convert(char(1),day(getdate()) / 7 ) + 'Week'

     

    Today                      WeekOfMonth MonthWeek
    --------------------------- ----------- ---------
    2007-06-25 19:46:26.130     3           3Week

     

    It seems like today 'should be' in the forth (or fifth) week of the month...


     

    Tuesday, June 26, 2007 2:50 AM
  • here's another alternative, you can also make use of a calendar udf, here's my favorite calendar udf

     

     [edited]

    CREATE  FUNCTION dbo.GetCalendarDates
    (
     @StartDate smalldatetime
     , @EndDate smalldatetime
    )
    RETURNS @CalendarDates TABLE (
     Row int IDENTITY(1,1)
     , CalendarDate smalldatetime
     , [Day]   AS day( [CalendarDate] )
            , [Month] AS month( [CalendarDate] )
            , [Year]  AS year( [CalendarDate] )
            , YearDay AS datepart( dayofyear, [CalendarDate] )
     , DayOfWeek AS datepart( weekday, [CalendarDate] )
     , WeekOfYear AS datepart( week, [CalendarDate] )
     , DateFirst AS @@DATEFIRST
     , WeekOfMonth int
     , NumOfWeeks int
     , WeekStart smalldatetime
     , WeekEnd smalldatetime
    )
    AS
    BEGIN
     DECLARE @StartOfMonth smalldatetime
     DECLARE @EndOfMonth smalldatetime

     SET @StartOfMonth = CONVERT(varchar(2),MONTH(@StartDate)) + '/1/' + CONVERT(varchar(4),YEAR(@StartDate))
     SET @EndOfMonth = CONVERT(varchar(2),MONTH(@EndDate)) + '/1/' + CONVERT(varchar(4),YEAR(@EndDate))
     SET @EndOfMonth = DATEADD(day,-1,DATEADD(month,1,@EndOfMonth))

     WHILE @StartOfMonth <= @EndOfMonth
      BEGIN
       
       INSERT
       INTO @CalendarDates
       SELECT CONVERT(varchar(10),@StartOfMonth,101), 1, 1, CONVERT(varchar(10),@StartOfMonth,101), CONVERT(varchar(10),@StartOfMonth,101)

       SET @StartOfMonth = DATEADD(d,1,@StartOfMonth)
      END
     
     

     UPDATE a
     SET WeekOfMonth = WeekOfYear - MinWeek
     FROM @CalendarDates a INNER JOIN
      (
       SELECT [Year]
        , [Month]
        , MIN(WeekOfYear) - 1 AS MinWeek
       FROM @CalendarDates
       GROUP BY
        [Year]
        , [Month]
      ) b ON a.[Year] = b.[Year]
       AND a.[Month] = b.[Month]
     
     UPDATE a
     SET WeekStart = b.WeekStart
      , WeekEnd = b.WeekEnd
      , NumOfWeeks = b.NumOfWeeks
     FROM @CalendarDates a INNER JOIN
      (
       SELECT [Year]
        , [Month]
        , WeekOfMonth
        , MIN(CalendarDate) AS WeekStart
        , MAX(CalendarDate) AS WeekEnd
        , COUNT(Row) AS NumOfWeeks
       FROM @CalendarDates
       GROUP BY
        [Year]
        , [Month]
        , WeekOfMonth
      ) b ON a.[Year] = b.[Year]
       AND  a.[Month] = b.[Month] 
       AND a.WeekOfMonth = b.WeekOfMonth

     

     UPDATE a
     SET NumOfWeeks = b.NumOfWeeks
     FROM @CalendarDates a INNER JOIN
      ( SELECT [Year]
        , [Month]
        , Max(WeekOfMonth) AS NumOfWeeks
        FROM @CalendarDates
       GROUP BY
        [Year]
        , [Month]
      ) b ON a.[Year] = b.[Year]
       AND  a.[Month] = b.[Month] 

     

     DELETE 
     FROM @CalendarDates
     WHERE CalendarDate NOT BETWEEN @StartDate AND @EndDate

    RETURN
    END

     

    ex.

     

    declare @to smalldatetime
    declare @from smalldatetime
    declare @thedate smalldatetime

     

    set @to = '06/01/2007'
    set @from = '07/30/2007'
    set @thedate = '06/25/2007'

     

    [edited] 

    select distinct @thedate, a.WeekOfMonth, a.WeekStart, a.WeekEnd, 'Week ' + convert(varchar(1),a.WeekOfMonth)
    from dbo.GetCalendarDates(@to,@from) a

    Tuesday, June 26, 2007 7:17 AM
  •  

     

    Oops, buggered that.

    Try this:

    Code Snippet

     

    declare @date datetime

    set @date = getdate()

     

    select 1 + (W + first.first_week_nbr) as MoWeekNbr

    from dbo.Calendar

    inner join

    (

        select @date as seldate, W as first_week_nbr

        from dbo.calendar where dt = dateadd(d, -1*(day(@date)-1), @date)

    ) first

    on first.seldate = @date

    where dt = @date

     

    Tuesday, June 26, 2007 9:20 AM
  • Yes Arnie, you are rite. TOday ie 2007-6-26, should really be in the 5th week. I applogize for not clarifying earlier.. Sorry abt that.
    Tuesday, June 26, 2007 12:46 PM
  • Yes , thats rite- the week is from Sunday thru Saturday.. ie eg 2007-06-01, though falls on a Friday, it should show as WEEK1 , For June.
    Tuesday, June 26, 2007 12:48 PM
  • Try:

     

    declare @d datetime

     

    set @d = '20070515'

     

    select datediff(week, convert(varchar(6), @d, 112) + '01', @d) + 1

    go

     

     

    AMB

    Tuesday, June 26, 2007 12:51 PM
  • Tarana,

     

    It appears that Alejandro's (hunchback/AMB) suggested algorithm works as you desire.

     

    Try using it to populate your calendar table.

     

    Alejandro -very nice, simple and elegant. Counts the number of week 'boundaries' between the first of the month and the supplied date.

     

    (Note: Check the DATEFIRST setting to verify what is the first day of the week.)

     

     

    Tuesday, June 26, 2007 2:30 PM
  • this code is very good if the week starts with sunday. I need the same thing, but i need the week to start with monday. can u help me? thanks

    Monday, January 14, 2013 10:04 AM