需取出结果为： bh pm rq 1 32 20060908 3 40 20060911 5 50 20061111补充如果时间相同取出编号（bh）较大的 select * from tablename where bh in (select max(bh) from tablename group by pm) select * from 表名 a where a.rq=(select max(rq) from 表名 b where a.bh=b.bh group by pm ) 同意同意同意同意同意同意同意同意 上面的各种查询方法好像都漏了一点把，还要把PM分类查吧？ SQL code:
select * from @t a where not exists(select 1 from @t b where a.pm=b.pm and a.rq<b.rq)
