我有一张桌子说:
id| AccID | Subject | Date
1 | 103 | Open HOuse 1 | 11/24/2011 9:00:00 AM
2 | 103 | Open HOuse 2 | 11/25/2011 10:00:00 AM
3 | 72 | Open House 3 | 11/26/2011 1:10:28 AM
4 | 82 | OPen House 4 | 11/27/2011 5:00:29 PM
5 | 82 | OPen House 5 | 11/22/2011 5:00:29 PM
从上表中,我需要Accid的所有唯一值。但是,如果有两个或更多列具有相同的Accid,那么我需要一个具有较小日期的列(在具有相同Accid的列中)
所以,从上表中,o / p应该是: 1 3 5
任何人都可以帮助我吗?感谢
答案 0 :(得分:1)
不仅仅是AccID,还有......
WITH SEL
AS
(
SELECT AccID, MIN(DATE)
FROM table
GROUP BY AccID
)
SELECT table.*
FROM table
JOIN SEL ON SEL.AccID = table.AccID
答案 1 :(得分:1)
SELECT t1.*
FROM [MyTable] t1
INNER JOIN
(
SELECT AccID, MIN(Date) Date
FROM [MyTable]
GROUP BY AccID
) t2 ON t1.AccID = t2.AccID AND t1.Date = t2.Date