How to create a row for every day in a date range using a stored procedure?

Question

I would like to create a stored procedure that will create a row in a table for every day in a given date range. The Stored Procedure accepts two inputs - A start date and end date of the date range desired by the user.

So, let's say I have a table like so:

SELECT Day, Currency
FROM ConversionTable

Day is a DateTime, and Currency is just an integer.

To keep things simple, let's just say I always want the Currency column to be 1 for each of these inserted rows. So, if someone inputs 'March 5, 2017' as the start date and 'April 11, 2017' as the end date, I would like the following rows created:

2017-03-05 00:00:00, 1
2017-03-06 00:00:00, 1
...
2017-04-11 00:00:00, 1

What's the best way to code the stored procedure to do this? I am using SQL Server 2008 R2 in my test environment, but our real environment uses SQL Server 2012, so I can upgrade my test machine if there is new functionality introduced in 2012 that makes this task easier.

RDFozz · Accepted Answer · 2018-06-27T17:20:21.340

One option is a recursive CTE:

DECLARE @StartDate datetime = '2017-03-05'
       ,@EndDate   datetime = '2017-04-11'
;

WITH theDates AS
     (SELECT @StartDate as theDate
      UNION ALL
      SELECT DATEADD(day, 1, theDate)
        FROM theDates
       WHERE DATEADD(day, 1, theDate) <= @EndDate
     )
SELECT theDate, 1 as theValue
  FROM theDates
OPTION (MAXRECURSION 0)
;

(MAXRECURSION hint added thanks to Scott Hodgin's comment, below.)

score 9 · Answer 2 · answered Mar 02 '17 at 23:06

Another option is to use a Table-Valued-Function. This approach is very fast, and offers a little more flexibility. You supply the Date/Time Range, DatePart and Increment. Also offers the advantage of including it in a CROSS APPLY

For Example

Select * from [dbo].[udf-Range-Date]('2017-03-05','2017-04-11','DD',1)

Returns

RetSeq  RetVal
1   2017-03-05 00:00:00.000
2   2017-03-06 00:00:00.000
3   2017-03-07 00:00:00.000
4   2017-03-08 00:00:00.000
5   2017-03-09 00:00:00.000
...
36  2017-04-09 00:00:00.000
37  2017-04-10 00:00:00.000
38  2017-04-11 00:00:00.000

The UDF if Interested

CREATE FUNCTION [dbo].[udf-Range-Date] (@R1 datetime,@R2 datetime,@Part varchar(10),@Incr int)
Returns Table
Return (
    with cte0(M)   As (Select 1+Case @Part When 'YY' then DateDiff(YY,@R1,@R2)/@Incr When 'QQ' then DateDiff(QQ,@R1,@R2)/@Incr When 'MM' then DateDiff(MM,@R1,@R2)/@Incr When 'WK' then DateDiff(WK,@R1,@R2)/@Incr When 'DD' then DateDiff(DD,@R1,@R2)/@Incr When 'HH' then DateDiff(HH,@R1,@R2)/@Incr When 'MI' then DateDiff(MI,@R1,@R2)/@Incr When 'SS' then DateDiff(SS,@R1,@R2)/@Incr End),
         cte1(N)   As (Select 1 From (Values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) N(N)),
         cte2(N)   As (Select Top (Select M from cte0) Row_Number() over (Order By (Select NULL)) From cte1 a, cte1 b, cte1 c, cte1 d, cte1 e, cte1 f, cte1 g, cte1 h ),
         cte3(N,D) As (Select 0,@R1 Union All Select N,Case @Part When 'YY' then DateAdd(YY, N*@Incr, @R1) When 'QQ' then DateAdd(QQ, N*@Incr, @R1) When 'MM' then DateAdd(MM, N*@Incr, @R1) When 'WK' then DateAdd(WK, N*@Incr, @R1) When 'DD' then DateAdd(DD, N*@Incr, @R1) When 'HH' then DateAdd(HH, N*@Incr, @R1) When 'MI' then DateAdd(MI, N*@Incr, @R1) When 'SS' then DateAdd(SS, N*@Incr, @R1) End From cte2 )

    Select RetSeq = N+1
          ,RetVal = D 
     From  cte3,cte0 
     Where D<=@R2
)
/*
Max 100 million observations -- Date Parts YY QQ MM WK DD HH MI SS
Syntax:
Select * from [dbo].[udf-Range-Date]('2016-10-01','2020-10-01','YY',1) 
Select * from [dbo].[udf-Range-Date]('2016-01-01','2017-01-01','MM',1) 
*/

score 4 · Answer 3 · answered Mar 02 '17 at 21:59

Using Aaron Bertrand's post on how to create a date dimension table as an example, I came up with this:

DECLARE @StartDate DATE ='2017-03-05 00:00:00'
DECLARE @EndDate DATE ='2017-04-11 00:00:00'

Declare @DateTable table ([date]       DATE PRIMARY KEY);

-- use the catalog views to generate as many rows as we need

INSERT @DateTable ([date])
SELECT d
FROM (
    SELECT d = DATEADD(DAY, rn - 1, @StartDate)
    FROM (
        SELECT TOP (DATEDIFF(DAY, @StartDate, @EndDate)) rn = ROW_NUMBER() OVER (
                ORDER BY s1.[object_id]
                )
        FROM sys.all_objects AS s1
        CROSS JOIN sys.all_objects AS s2
        -- on my system this would support > 5 million days
        ORDER BY s1.[object_id]
        ) AS x
    ) AS y;

SELECT *
FROM @DateTable
ORDER BY [date]

You should be able to put this type of logic in your stored procedure and add whatever else you need.

score 2 · Answer 4 · answered Apr 24 '18 at 16:27

I had to solve a similar problem recently on Redshift where I only had read access and thus needed a purely SQL-based solution (no stored procedures) to getting a row for every hour in a date range as a starting point for my result set. I'm sure others can make this more elegant and modify it for their purposes, but for those in need, here is my hacked solution:

with hours as
   (select 0 clockhour union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 10 union select 11 union select 12 
    union select 13 union select 14 union select 15 union select 16 union select 17 union select 18 union select 19 union select 20 union select 21 union select 22 union select 23)
, days as
   (select *
    from 
       (select to_number(n0.number || n1.number, '99') daynum
        from
           (select 0 as number union select 1 union select 2 union select 3) as n0
           cross join
           (select 1 as number union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) as n1)
    where daynum between 1 and 31)
, months as
   (select 1 as monthnum, 'jan' as themonth, 31 as numdays union select 2, 'feb', 28 union select 3, 'mar', 31 union select 4, 'apr', 30 union select 5, 'may', 31 union select 6, 'jun', 30 
    union select 7, 'jul', 31 union select 8, 'aug', 31 union select 9, 'sep', 30 union select 10, 'oct', 31 union select 11, 'nov', 30 union select 12, 'dec', 31)
, years as
   (select century || decade || yr as yr
    from 
       (select 19 century union select 20) 
    cross join
       (select 0 decade union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) 
    cross join
       (select 0 yr union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9))
select cast(daynum || '-' || themonth || '-' || yr || ' ' || clockhour || ':00:00' as timestamp) dayhour
from hours
cross join days
cross join months
cross join years
where cast(daynum || '-' || themonth || '-' || yr || ' ' || clockhour || ':00:00' as timestamp) 
between date_trunc('month', dateadd('month', -$MONTHS_AGO, getdate()))
and     date_trunc('month', dateadd('month', $MONTHS_AHEAD, getdate()))
and   daynum <= numdays
order by yr, monthnum, daynum, clockhour;

score 0 · Answer 5 · answered Jul 08 '22 at 13:51

Another option is to insert the date records into the table using a loop.

The code below adds them to a temp table but this could easily be adapted to run in a stored proc with a permanent table.

DECLARE @d DATE
DECLARE @StartDate DATE = '2020-01-01'
DECLARE @EndDate DATE = '2022-12-31'
DROP TABLE IF EXISTS #d
CREATE TABLE #d (Id INT IDENTITY(1,1) NOT NULL PRIMARY KEY, Date DATE)
SET @d = @StartDate
WHILE @d <= @EndDate BEGIN
INSERT INTO #d (Date) VALUES (@d)

SET @d = DATEADD(DAY, 1, @d)


END

score -2 · Answer 6 · answered Nov 10 '20 at 02:45

-2

Fill about 10 records in Item & 3 in Category tables, then fill Ledger table with historical data from 01.01.10 to 31.12.20 as 10 records per day using a stored procedure.

answered Nov 10 '20 at 02:45

Samer

1

How to create a row for every day in a date range using a stored procedure?

6 Answers6