Get a list of dates between two dates using a function

date, sql-server

Solution

Try something like this:

CREATE FUNCTION dbo.ExplodeDates(@startdate datetime, @enddate datetime)
returns table as
return (
with 
 N0 as (SELECT 1 as n UNION ALL SELECT 1)
,N1 as (SELECT 1 as n FROM N0 t1, N0 t2)
,N2 as (SELECT 1 as n FROM N1 t1, N1 t2)
,N3 as (SELECT 1 as n FROM N2 t1, N2 t2)
,N4 as (SELECT 1 as n FROM N3 t1, N3 t2)
,N5 as (SELECT 1 as n FROM N4 t1, N4 t2)
,N6 as (SELECT 1 as n FROM N5 t1, N5 t2)
,nums as (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 1)) as num FROM N6)
SELECT DATEADD(day,num-1,@startdate) as thedate
FROM nums
WHERE num <= DATEDIFF(day,@startdate,@enddate) + 1
);

You then use:

SELECT *
FROM dbo.ExplodeDates('20090401','20090531') as d;

Edited (after the acceptance):

Please note... if you already have a sufficiently large nums table then you should use:

CREATE FUNCTION dbo.ExplodeDates(@startdate datetime, @enddate datetime)
returns table as
return (
SELECT DATEADD(day,num-1,@startdate) as thedate
FROM nums
WHERE num <= DATEDIFF(day,@startdate,@enddate) + 1
);

And you can create such a table using:

CREATE TABLE dbo.nums (num int PRIMARY KEY);
INSERT dbo.nums values (1);
GO
INSERT dbo.nums SELECT num + (SELECT COUNT(*) FROM nums) FROM nums
GO 20

These lines will create a table of numbers containing 1M rows... and far quicker than inserting them one by one.

You should NOT create your ExplodeDates function using a function that involves BEGIN and END, as the Query Optimizer becomes unable to simplify the query at all.

Problem

My question is similar to this MySQL question, but intended for SQL Server: Is there a function or a query that will return a list of days between two dates? For example, lets say there is a function called ExplodeDates: ``` SELECT ExplodeDates('2010-01-01', '2010-01-13'); ``` This would return a single column table with the values: ``` 2010-01-01 2010-01-02 2010-01-03 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 2010-01-09 2010-01-10 2010-01-11 2010-01-12 2010-01-13 ``` I'm thinking that a calendar/numbers table might be able to help me here. Update I decided to have a look at the three code answers provided, and the results of the execution - as a % of the total batch - are: - Rob Farley's answer : 18% - StingyJack's answer : 41% - KM's answer : 41% Lower is better I have accepted Rob Farley's answer, as it was the fastest, even though numbers table solutions (used by both KM and StingyJack in their answers) are something of a favourite of mine. Rob Farley's was two-thirds faster. Update 2 Alivia's answer is much more succinct. I have changed the accepted answer.

Original source

Related problems