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Home/ Questions/Q 3624016
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T23:27:19+00:00 2026-05-18T23:27:19+00:00

I’m currently trying to write a stored procedure that can compute the biweekly periods

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I’m currently trying to write a stored procedure that can compute the biweekly periods when a date is passed in as a parameter.

The business logic: the first Monday of the year is first day of the biweekly period.

For example in 2010:

    period  period_start  period_end
    1       2010-01-04    2010-01-17
    2       2010-01-18    2010-01-31
    3       2010-02-01    2010-02-14
    ....
    26      2010-12-20    2011-01-02

Passing today’s date of 2010-12-31 will return 26, 2010-12-20 and 2011-01-02. How to achieve this?

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  1. Editorial Team
    Editorial Team
    2026-05-18T23:27:19+00:00Added an answer on May 18, 2026 at 11:27 pm

    @Lamak did the hard part, figuring out the first Monday in the year (upvoted). I revised his routine a bit to take a datetime value, like so:

    --  Return first Monday for the year being passed in
    CREATE FUNCTION dbo.FirstMonday (@TargetDay datetime)
    RETURNS DATE
    AS
    BEGIN
    DECLARE @Return DATE
    
    --  Set to first of its year
    SET @TargetDay = dateadd(dd, -datepart(dayofyear, @TargetDay) + 1, @TargetDay)
    
    ;WITH Dates AS
    (
      SELECT @TargetDay AS DateVal
      UNION ALL
      SELECT DATEADD(d, 1, DateVal) AS DateVal
      FROM Dates
      WHERE DATEADD(d, 1, DateVal) < DATEADD(m, 1, @TargetDay)
    )
    SELECT @Return = MIN(DateVal)
    FROM Dates
    WHERE DATENAME(WEEKDAY,DateVal) = 'Monday'
    
    RETURN @Return
    END 
    GO
    
    --  Test it
    print dbo.FirstMonday(getdate())
    print dbo.FirstMonday('Jan 1, 2010')
    

    From there, it’s just some datetime arithmatic:

    DECLARE
      @FirstMonday  datetime
     ,@TargetDay    datetime
     ,@BiWeek       int
    
    SET @TargetDay = getdate()
    --SET @TargetDay = 'Jan 17, 2010'
    
    --  Get the first Monday
    SET @FirstMonday = dbo.FirstMonday(@TargetDay)
    
    --  Calculate the bi-weekly period
    SET @BiWeek = datediff(dd, @FirstMonday, @TargetDay) / 14
    
    --  Print results
    PRINT @BiWeek + 1
    PRINT dateadd(dd, @BiWeek * 14, @FirstMonday)
    PRINT dateadd(dd, @BiWeek * 14 + 13, @FirstMonday)
    
    --  Or return them as a dataset
    SELECT
      @BiWeek + 1                                   Period
     ,dateadd(dd, @BiWeek * 14, @FirstMonday)       PeriodStart
     ,dateadd(dd, @BiWeek * 14 + 13, @FirstMonday)  PeriodEnd
    

    This fails when you pick a day in the year that falls before the first Monday, as it returns the first biweekly period after that date. (Or does it fail? You did not specify what the proper period is for such dates… 🙂

    However, as @HLGEM says, depending on what you are doing you are probably better off building and using some form of biweekly period lookup table, especially if you need proper handling of those early dates.

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