How to generate a range of dates in SQL Server

42,592

Solution 1

I would argue that for this specific purpose the below query is about as efficient as using a dedicated lookup table.

DECLARE @start DATE, @end DATE;
SELECT @start = '20110714', @end = '20110717';

;WITH n AS 
(
  SELECT TOP (DATEDIFF(DAY, @start, @end) + 1) 
    n = ROW_NUMBER() OVER (ORDER BY [object_id])
  FROM sys.all_objects
)
SELECT 'Bob', DATEADD(DAY, n-1, @start)
FROM n;

Results:

Bob     2011-07-14
Bob     2011-07-15
Bob     2011-07-16
Bob     2011-07-17

Presumably you'll need this as a set, not for a single member, so here is a way to adapt this technique:

DECLARE @t TABLE
(
    Member NVARCHAR(32), 
    RegistrationDate DATE, 
    CheckoutDate DATE
);

INSERT @t SELECT N'Bob', '20110714', '20110717'
UNION ALL SELECT N'Sam', '20110712', '20110715'
UNION ALL SELECT N'Jim', '20110716', '20110719';

;WITH [range](d,s) AS 
(
  SELECT DATEDIFF(DAY, MIN(RegistrationDate), MAX(CheckoutDate))+1,
    MIN(RegistrationDate)
    FROM @t -- WHERE ?
),
n(d) AS
(
  SELECT DATEADD(DAY, n-1, (SELECT MIN(s) FROM [range]))
  FROM (SELECT ROW_NUMBER() OVER (ORDER BY [object_id])
  FROM sys.all_objects) AS s(n)
  WHERE n <= (SELECT MAX(d) FROM [range])
)
SELECT t.Member, n.d
FROM n CROSS JOIN @t AS t
WHERE n.d BETWEEN t.RegistrationDate AND t.CheckoutDate;
----------^^^^^^^ not many cases where I'd advocate between!

Results:

Member    d
--------  ----------
Bob       2011-07-14
Bob       2011-07-15
Bob       2011-07-16
Bob       2011-07-17
Sam       2011-07-12
Sam       2011-07-13
Sam       2011-07-14
Sam       2011-07-15
Jim       2011-07-16
Jim       2011-07-17
Jim       2011-07-18
Jim       2011-07-19

As @Dems pointed out, this could be simplified to:

;WITH natural AS 
(
  SELECT ROW_NUMBER() OVER (ORDER BY [object_id]) - 1 AS val 
  FROM sys.all_objects
) 
SELECT t.Member, d = DATEADD(DAY, natural.val, t.RegistrationDate) 
  FROM @t AS t INNER JOIN natural 
  ON natural.val <= DATEDIFF(DAY, t.RegistrationDate, t.CheckoutDate);

Solution 2

I usually do this with a trick using row_number() on some table. So:

select t.name, dateadd(d, seq.seqnum, t.start_date)
from t left outer join
     (select row_number() over (order by (select NULL)) as seqnum
      from t
     ) seq
     on seqnum <= datediff(d, t.start_date, t.end_date)

The calculation for seq goes pretty fast, since no calculation or ordering is required. However, you need to be sure the table is big enough for all time spans.

Solution 3

If you have a "Tally" or "Numbers" table, life get's real simple for things like this.

 SELECT Member, DatePresent = DATEADD(dd,t.N,RegistrationDate)
   FROM @t 
  CROSS JOIN dbo.Tally t
  WHERE t.N BETWEEN 0 AND DATEDIFF(dd,RegistrationDate,CheckoutDate)
;

Here's how to build a "Tally" table.

--===================================================================
--      Create a Tally table from 0 to 11000
--===================================================================
--===== Create and populate the Tally table on the fly.
 SELECT TOP 11001
        IDENTITY(INT,0,1) AS N
   INTO dbo.Tally
   FROM Master.sys.ALL_Columns ac1
  CROSS JOIN Master.sys.ALL_Columns ac2
;
--===== Add a CLUSTERED Primary Key to maximize performance
  ALTER TABLE dbo.Tally
    ADD CONSTRAINT PK_Tally_N 
        PRIMARY KEY CLUSTERED (N) WITH FILLFACTOR = 100
;
--===== Allow the general public to use it
  GRANT SELECT ON dbo.Tally TO PUBLIC
;
GO

For more information on what a "Tally" table is in SQL and how it can be used to replace While loops and the "Hidden RBAR" of reursive CTEs that count, please see the following article.

http://www.sqlservercentral.com/articles/T-SQL/62867/

Share:
42,592
Daniel Cotter
Author by

Daniel Cotter

Updated on July 05, 2022

Comments

  • Daniel Cotter
    Daniel Cotter almost 2 years

    The title doesn't quite capture what I mean, and this may be a duplicate.

    Here's the long version: given a guest's name, their registration date, and their checkout date, how do I generate one row for each day that they were a guest?

    Ex: Bob checks in 7/14 and leaves 7/17. I want

    ('Bob', 7/14), ('Bob', 7/15), ('Bob', 7/16), ('Bob', 7/17) 
    

    as my result.

    Thanks!