标题并没有完全表达我的意思,这可能是重复的。
这是长版本:给定客人的姓名、他们的注册日期和他们的结帐日期,我如何为他们作为客人的每一天生成一行?
例如:Bob 于 7/14 入住并于 7/17 离开。我想
('Bob', 7/14), ('Bob', 7/15), ('Bob', 7/16), ('Bob', 7/17)
作为我的结果。
谢谢!
标题并没有完全表达我的意思,这可能是重复的。
这是长版本:给定客人的姓名、他们的注册日期和他们的结帐日期,我如何为他们作为客人的每一天生成一行?
例如:Bob 于 7/14 入住并于 7/17 离开。我想
('Bob', 7/14), ('Bob', 7/15), ('Bob', 7/16), ('Bob', 7/17)
作为我的结果。
谢谢!
我会争辩说,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,所以这里有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);
我通常通过在某个表上使用 row_number() 的技巧来做到这一点。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 的计算非常快,因为不需要计算或排序。但是,您需要确保表对于所有时间跨度都足够大。
如果您有一个“Tally”或“Numbers”表,那么对于这样的事情来说,生活会变得非常简单。
SELECT Member, DatePresent = DATEADD(dd,t.N,RegistrationDate)
FROM @t
CROSS JOIN dbo.Tally t
WHERE t.N BETWEEN 0 AND DATEDIFF(dd,RegistrationDate,CheckoutDate)
;
这是构建“Tally”表的方法。
--===================================================================
-- Create a Tally table from 0 to 11000
--===================================================================
--===== Create and populate the Tally table on the fly.
SELECT TOP 11001
IDENTITY(INT,0,1) AS N
INTO dbo.Tally
FROM Master.sys.ALL_Columns ac1
CROSS JOIN Master.sys.ALL_Columns ac2
;
--===== Add a CLUSTERED Primary Key to maximize performance
ALTER TABLE dbo.Tally
ADD CONSTRAINT PK_Tally_N
PRIMARY KEY CLUSTERED (N) WITH FILLFACTOR = 100
;
--===== Allow the general public to use it
GRANT SELECT ON dbo.Tally TO PUBLIC
;
GO
有关什么是 SQL 中的“Tally”表以及如何使用它来替换 While 循环和重要的递归 CTE 的“隐藏 RBAR”的更多信息,请参阅以下文章。
这可能对你有用:
with mycte as
(
select cast('2000-01-01' as datetime) DateValue, 'Bob' as Name
union all
select DateValue + 1 ,'Bob' as Name
from mycte
where DateValue + 1 < '2000-12-31'
)
select *
from mycte
OPTION (MAXRECURSION 0)
我会创建一个触发器来创建额外的记录并在结帐时运行它。或者,您可以每天午夜做同样的工作(如果您需要数据库中的最新信息)。