如何在 SQL Server 中生成多个客人入住的每日日期范围？

在SQL Server中生成日期范围

虽然标题暗示生成一系列日期，但主要问题是为客人入住机构的每一天创建多行。给定客人姓名、入住日期和退房日期，目标是为入住的每一天输出一行。

下面的查询有效地解决了这个任务：

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;

执行此查询后，将生成以下结果（根据提供的示例）：

<code>Bob     2011-07-14
Bob     2011-07-15
Bob     2011-07-16
Bob     2011-07-17</code>

对于需要容纳多个客人的情况，可以将查询改编成更全面的形式：

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;

此改编后的查询会生成以下结果，其中包含多个客人的数据：

<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>

以上是如何在 SQL Server 中生成多个客人入住的每日日期范围？的详细内容。更多信息请关注PHP中文网其他相关文章！