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Generate date range in SQL Server
While the title implies generating a range of dates, the main problem is creating multiple rows for each day a guest stayed at the facility. Given a guest name, check-in date, and check-out date, the goal is to output a row for each day of the stay.
The following query effectively solves this task:
<code class="language-sql">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;</code>
Executing this query will produce the following results (based on the example provided):
<code>Bob 2011-07-14 Bob 2011-07-15 Bob 2011-07-16 Bob 2011-07-17</code>
For situations where multiple guests need to be accommodated, the query can be adapted into a more comprehensive form:
<code class="language-sql">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 ), 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;</code>
This adapted query produces the following results, which contain data for multiple guests:
<code>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</code>
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