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SQL query to return all dates between two date ranges for SQL Server 2005

Posted on 2008-10-18
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Last Modified: 2012-05-05
There is a post ID: 23398111 which shows how to do this in Oracle but I need a solution for SQL Server,
See link @ http://www.experts-exchange.com/Database/Oracle/Q_23398111.html

The only solution I could come up with is be using an existing table that already contains rows see query below:
SELECT CONVERT(char(10),dateadd(day,ROW_NUMBER() OVER (order by ColD),getdate()), 112) CurrDate
FROM Tbl
0
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Question by:HelmutP
  • 6
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13 Comments
 
LVL 39

Accepted Solution

by:
BrandonGalderisi earned 25 total points
ID: 22750307
The query would be:


declare @dtBegin datetime
     ,@dtEnd datetime
set @dtbegin = convert(datetime, '01/01/2008',101)
set @dtEnd = convert(datetime, '11/01/2008',101)

;with DateRange as
(
select @dtBegin theDAte
union all
select dateadd(d,n,@dtBegin) from vw_nums where n <= datediff(d,@dtBegin,@dtEnd)
)
select * from DateRange


and it uses the view below.
create view vw_Nums

as

with   cte0 as (select 1 as c union all select 1), 

       cte1 as (select 1 as c from cte0 a, cte0 b), 

       cte2 as (select 1 as c from cte1 a, cte1 b), 

       cte3 as (select 1 as c from cte2 a, cte2 b), 

       cte4 as (select 1 as c from cte3 a, cte3 b), 

       cte5 as (select 1 as c from cte4 a, cte4 b), 

       nums as (select row_number() over (order by c) as n from cte5)

       select n from nums 
 
 

go

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0
 
LVL 11

Assisted Solution

by:Louis01
Louis01 earned 25 total points
ID: 22939142
This is a function I use for exactly that purpose:
IF  EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'[dbo].[fnCalendar]') AND type in (N'FN', N'IF', N'TF', N'FS', N'FT'))

DROP FUNCTION [dbo].[fnCalendar]

GO
 

CREATE FUNCTION [dbo].[fnCalendar](@StartDate [smalldatetime], @EndDate [smalldatetime])

RETURNS @retMonth TABLE (

	[CalDate] [smalldatetime] NULL

) WITH EXECUTE AS CALLER

AS 

BEGIN

    IF @StartDate IS NOT NULL

      AND @EndDate IS NOT NULL 

      BEGIN

		DECLARE @Duration int

		SET @Duration = (SELECT DATEDIFF(day, @StartDate, @EndDate))

        SET @StartDate = CONVERT(DATETIME, CONVERT(CHAR(10), DATEADD(d, (DATEPART(d, @StartDate) * -1) + 1, @StartDate), 120))
 

        DECLARE @Counter INTEGER

        SET @Counter = 0

        WHILE @Counter < @Duration

          BEGIN

            INSERT INTO

              @retmonth

            VALUES

              (

               DATEADD(DAY, @Counter, @StartDate))

            SET @Counter = @Counter + 1

          END

      END  

    RETURN

   END

GO
 

select * from dbo.fnCalendar('01-Jan-2007', '11-Nov-2008')

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0
 
LVL 39

Expert Comment

by:BrandonGalderisi
ID: 23096127
I think that in the interest of "first come first serve" and considering I provided a solution almost a full month sooner, that I should get the full answer.
0
 
LVL 11

Expert Comment

by:Louis01
ID: 23102537
I posted a different solution because the one proposed by BrandonGalderisi was still running 20 minutes after I tried it. (It subsequently gave an Exception of type 'System.OutOfMemoryException'). In the interest of not splitting hairs though, I am happy if BrandonGalderisi has the 50 points.
0
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LVL 39

Expert Comment

by:BrandonGalderisi
ID: 23105006
20 minutes?
0
 
LVL 39

Expert Comment

by:BrandonGalderisi
ID: 23105078
The query I have posted above, which returns 306 records for Jan-1 to Nov-1 runs in 0 ms for me.  That means NO MEASURABLE TIME.


declare @dt datetime

set @dt=getdate()

declare @dtBegin datetime

     ,@dtEnd datetime

set @dtbegin = convert(datetime, '01/01/2006',101)

set @dtEnd = convert(datetime, '11/01/2008',101)
 

;with DateRange as

(

select @dtBegin theDAte

union all

select dateadd(d,n,@dtBegin) from vw_nums where n <= datediff(d,@dtBegin,@dtEnd)

)

select * from DateRange
 

select datediff(ms,@dt,getdate())

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0
 
LVL 39

Expert Comment

by:BrandonGalderisi
ID: 23126027
cleanup?  as in delete?
0
 
LVL 32

Expert Comment

by:Daniel Wilson
ID: 23129462
As in, 21 days from now I'm to make a recommendation again.
0
 
LVL 39

Expert Comment

by:BrandonGalderisi
ID: 24658054
I will re-state my earlier comment, but not object.


I think that in the interest of "first come first serve" and considering I provided a solution almost a full month sooner, that I should get the full answer.
0

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