如何在SQL Server中生成一系列日期

这是长版本：给定一个客人的姓名,他们的注册日期和结账日期,我如何为他们作为客人的每一天生成一行？

例如：鲍勃在7月14日检查并离开7月17日.我想要

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

作为我的结果.

谢谢！

解决方法

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;

结果：

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

大概你需要这个作为一个集合,而不是单个成员,所以这是一种适应这种技术的方法：

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,(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!

结果：

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

正如@Dems指出的那样,这可以简化为：

;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);

（编辑：李大同）

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