SQL first of every month

Question

Supposing that I wanted to write table valued function in SQL that returns a table with the first day of every month between the argument dates, what is the simplest way to do this?

For example fnFirstOfMonths('10/31/10', '2/17/11') would return a one-column table with 11/1/10, 12/1/10, 1/1/11 , and 2/1/11 as the elements.

My first instinct is just to use a while loop and repeatedly insert first days of months until I get to before the start date. It seems like there should be a more elegant way to do this though.

Thanks for any help you can provide.

Answer 1

Something like this would work without being inside a function:

DECLARE @LowerDate DATE 
SET @LowerDate = GETDATE()

DECLARE @UpperLimit DATE
SET @UpperLimit = '20111231'

;WITH Firsts AS
(
    SELECT
        DATEADD(DAY, -1 * DAY(@LowerDate) + 1, @LowerDate) AS 'FirstOfMonth'

    UNION ALL

    SELECT
        DATEADD(MONTH, 1, f.FirstOfMonth) AS 'FirstOfMonth'
    FROM
        Firsts f
    WHERE
        DATEADD(MONTH, 1, f.FirstOfMonth)  <= @UpperLimit
)   
SELECT * 
FROM Firsts

It uses a thing called CTE (Common Table Expression) - available in SQL Server 2005 and up and other database systems.

In this case, I start the recursive CTE by determining the first of the month for the @LowerDate date specified, and then I iterate adding one month to the previous first of month, until the upper limit is reached.

Or if you want to package it up in a stored function, you can do so, too:

CREATE FUNCTION dbo.GetFirstOfMonth(@LowerLimit DATE, @UpperLimit DATE)
RETURNS TABLE 
AS 
   RETURN
      WITH Firsts AS
      (
          SELECT
             DATEADD(DAY, -1 * DAY(@LowerLimit) + 1, @LowerLimit) AS 'FirstOfMonth'
          UNION ALL
          SELECT
             DATEADD(MONTH, 1, f.FirstOfMonth) AS 'FirstOfMonth'
          FROM
             Firsts f
          WHERE
             DATEADD(MONTH, 1, f.FirstOfMonth)  <= @UpperLimit
       )    
       SELECT * FROM Firsts

and then call it like this:

SELECT * FROM dbo.GetFirstOfMonth('20100522', '20100831')

to get an output like this:

FirstOfMonth
2010-05-01
2010-06-01
2010-07-01
2010-08-01

PS: by using the DATE datatype - which is present in SQL Server 2008 and newer - I fixed the two "bugs" that Richard commented about. If you're on SQL Server 2005, you'll have to use DATETIME instead - and deal with the fact you're getting a time portion, too.

Answer 2

create function dbo.fnFirstOfMonths(@d1 datetime, @d2 datetime)
returns table as return
select dateadd(m,datediff(m,0,@d1)+v.number,0) as FirstDay
from master..spt_values v
where v.type='P' and v.number between 0 and datediff(m, @d1, @d2)
  and dateadd(m,datediff(m,0,@d1)+v.number,0) between @d1 and @d2
GO

Notes

• master..spt_values is a source for general purpose sequence numbers in SQL Server
• dateadd(m, datediff(m is a technique for working out the first day of month for any date
• +v.number is used to increase it by one month each time
• 0 and datediff(m, @d1, @d2) this condition gives us all the number s we need to generate a first-of-month date for each month between @d1 and @d2, inclusive of both months
• and dateadd(m,datediff(m,0,@d1)+v.number,0) between @d1 and @d2 the final filter to verify that the first-of-month date generated is between @d1 and @d2

---

Performance comparison against marc_s's code

Summary

declare @t table (dt datetime)
declare @d datetime
declare @i int
set nocount on

set @d = GETDATE()
set @i = 0
while @i < 10000
begin
insert @t select * from dbo.getfirstofmonth('20090102', '20100506')
delete @t
set @i = @i + 1
end

print datediff(ms, @d, getdate())

set @d = GETDATE()
set @i = 0
while @i < 10000
begin
insert @t select * from dbo.fnfirstofmonths('20090102', '20100506')
delete @t
set @i = @i + 1
end
print datediff(ms, @d, getdate())

Performante

Test

 declare @t table (dt datetime) declare @d datetime declare @i int set nocount on set @d = GETDATE() set @i = 0 while @i < 10000 begin insert @t select * from dbo.getfirstofmonth('20090102', '20100506') delete @t set @i = @i + 1 end print datediff(ms, @d, getdate()) set @d = GETDATE() set @i = 0 while @i < 10000 begin insert @t select * from dbo.fnfirstofmonths('20090102', '20100506') delete @t set @i = @i + 1 end print datediff(ms, @d, getdate()) Performante 

Answer 3

It will loop just between the months involved (4 times in the example):

set dateformat mdy;
declare @date1 smalldatetime,@date2 smalldatetime,@i int
set @date1= '10-31-2010'
set @date2= '02-17-2011'
set @i=1

while(@i<=DATEDIFF(mm,@date1,@date2))
begin
    select  convert(smalldatetime,CONVERT(varchar(6),DATEADD(mm,@i,@date1),112)+'01',112)
    set     @i=@i+1
end

Answer 4

I realize this isn't a function, but I'm going to throw this into the mix anyway.

select cal_date from calendar
where day_of_month = 1
and cal_date between '2011-01-01' and '2012-01-01'

This calendar table runs on a PostgreSQL server at work. I'll port it to SQL Server tonight, and run some speed comparisons. (Why? Because this stuff is fun, that's why.)

Answer 5

Just in case anybody is still reading this ... I cannot imaging that any of the aforementioned functions is faster than this:

declare @DatFirst date = '20101031', @DatLast date = '21110217';

declare @DatFirstOfFirstMonth date = dateadd(day,1-day(@DatFirst),@DatFirst);

select  DatFirstOfMonth = dateadd(month,n,@DatFirstOfFirstMonth)
from    (
        select  top (datediff(month,@DatFirstOfFirstMonth,@DatLast)+1)
                n=row_number() over (order by (select 1))-1
        from    (values (1),(1),(1),(1),(1),(1),(1),(1)) a (n)
        cross join (values (1),(1),(1),(1),(1),(1),(1),(1)) b (n)
        cross join (values (1),(1),(1),(1),(1),(1),(1),(1)) c (n)
        cross join (values (1),(1),(1),(1),(1),(1),(1),(1)) d (n)
        ) x