


How to Efficiently Generate a Date Range for Multiple Guest Registrations in SQL Server?
Jan 10, 2025 am 11:56 AMEfficiently generate guest registration date range in SQL Server
While the title of the article is accurate, it does not fully convey the specific scenario of generating a row for each guest on each day based on the guest's check-in and check-out dates. This improved reply will delve into a technique for achieving efficiency using specialized lookup tables.
Use a dedicated lookup table
The solution using the ROW_NUMBER() function provides commendable efficiency for this particular task. The following optimized query utilizes a lookup table to generate the necessary date range:
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;
Expand to multiple guests
To extend this technique to multiple guests, the following query can be adapted:
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 d FROM [range]) ) SELECT t.Member, n.d FROM @t t INNER JOIN n ON n.d BETWEEN t.RegistrationDate AND t.CheckoutDate;
This modified query effectively generates the required date range for all members in the provided table.
The above is the detailed content of How to Efficiently Generate a Date Range for Multiple Guest Registrations in SQL Server?. For more information, please follow other related articles on the PHP Chinese website!

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