取分组 日期最大的数据
将数据 分组，取得日期最大的记录，如果日期最大有相同 就一起取
如下：
WITH test (ID,[Date])
AS
(
SELECT 1,'2012-3-10'
UNION ALL SELECT 1,'2012-3-10'
UNION ALL SELECT 1,'2012-3-4'
UNION ALL SELECT 2,'2012-2-10'
UNION ALL SELECT 2,'2012-2-10'
UNION ALL SELECT 2,'2012-3-8'
UNION ALL SELECT 2,'2012-3-9'
UNION ALL SELECT 3,'2012-5-12'
UNION ALL SELECT 3,'2012-4-12'
UNION ALL SELECT 3,'2012-1-2'
UNION ALL SELECT 4,'2012-5-5'
UNION ALL SELECT 5,'2012-3-7'
UNION ALL SELECT 6,'2012-4-2'
UNION ALL SELECT 7,'2012-2-1'
)
SELECT id,MIN ([Date])[Date]
FROM Test
GROUP BY id
/*
id Date
----------- ---------
1 2012-3-10
2 2012-2-10
3 2012-1-2
4 2012-5-5
5 2012-3-7
6 2012-4-2
7 2012-2-1
(7 行受影响)
*/
UNION ALL SELECT 3, '2012-4-12'
UNION ALL SELECT 3, '2012-1-2'
UNION ALL SELECT 4, '2012-5-5'
UNION ALL SELECT 5, '2012-3-7'
UNION ALL SELECT 6, '2012-4-2'
UNION ALL SELECT 7, '2012-2-1'
SELECT * FROM #TB T WHERE DATE=(SELECT MAX(DATE) FROM #TB WHERE ID=T.ID)
/*
ID DATE
----------- ------------------------------------------------------
7 2012-02-01 00:00:00.000
6 2012-04-02 00:00:00.000
5 2012-03-07 00:00:00.000
4 2012-05-05 00:00:00.000
3 2012-05-12 00:00:00.000
2 2012-03-09 00:00:00.000
1 2012-03-10 00:00:00.000
1 2012-03-10 00:00:00.000
（所影响的行数为 8 行）
[解决办法]
