是这样的，有一个表a,里面的数据如下：
其中id int,tatid varchar(18) workDate varchar(16)
id tatid workDate
1090 200403050000000066 2002-4--2003-11
1006 200403050000000066 2003-11--2004-11
1007 200403050000000066 2000-11--2002-4
1015 200403050000000075 2002-6--2003-10
1016 200403050000000075 NULL
1019 200403050000000075 2004-11--2006-10
......
我想查出结果,如:
id tatid workDate
1006 200403050000000066 2003-11--2004-11
1019 200403050000000075 2004-11--2006-10
......
就是workDate最靠后的,请问怎么写?
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使用时间倒序,top n不就行了?
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select * from a order by workDate desc
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说实话lz说的更让人迷糊了
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select id,tatid,max(workDate) from a where tatid in (select distinct tatid from a)
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select * from a as table1,
(select max(id) as maxid from a group by id) as table2
where table1.id = table1.id order by table1.id
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忘了加上group by
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select top 2 * from a order by workDate desc
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select * from a where 不存在workDate比当前行大的tatid
大概就这个意思，用EXISTS多试一下。
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select distinct tatid from a order by workDate desc
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不好意思，刚才写错了，修正如下：
select * from a as table1,
(select id from a where worddate in (select max(workDate) from a group by tatid)) as table2
where table1.id = table1.id order by table1.id