TA貢獻(xiàn)1796條經(jīng)驗(yàn) 獲得超4個(gè)贊
我認(rèn)為，為此特定目的，以下查詢與使用專用查找表一樣有效。
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;
結(jié)果：
Bob 2011-07-14
Bob 2011-07-15
Bob 2011-07-16
Bob 2011-07-17
大概您需要將此作為一個(gè)集合，而不是單個(gè)成員，因此這里是一種適應(yīng)此技術(shù)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!
結(jié)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指出的那樣，這可以簡化為：
;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);
TA貢獻(xiàn)1765條經(jīng)驗(yàn) 獲得超5個(gè)贊
我通常在某些表上使用row_number（）來達(dá)到目的。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的計(jì)算非?？?，因?yàn)椴恍枰?jì)算或排序。但是，您需要確保該表對(duì)于所有時(shí)間段都足夠大。
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