如何在SQL Server中生成一系列日期

Question

标题并没有完全体现我的意思,这可能是重复的.

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

作为我的结果.

谢谢!

Answer 1

我认为,对于这个特定目的,下面的查询与使用专用查找表一样有效.

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, 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!

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);

Answer 2

我通常在某些表上使用row_number()来处理这个问题.所以:

select t.name, dateadd(d, seq.seqnum, t.start_date)
from t left outer join
     (select row_number() over (order by (select NULL)) as seqnum
      from t
     ) seq
     on seqnum <= datediff(d, t.start_date, t.end_date)

seq的计算速度非常快,因为不需要计算或排序.但是,您需要确保该表足够大以适应所有时间跨度.