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Sådan genereres en række datoer i SQL Server

Jeg vil påstå, at nedenstående forespørgsel til dette specifikke formål er omtrent lige så effektiv som at bruge en dedikeret opslagstabel.

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;
 

Resultater:

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

Formentlig skal du bruge dette som et sæt, ikke for et enkelt medlem, så her er en måde at tilpasse denne teknik på:

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!
 

Resultater:

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
 

Som @Dems påpegede, kunne dette forenkles til:

;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);

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