SELECT * FROM rsl a, (SELECT CODE, max(time_key) time_key FROM rsl GROUP BY CODE ) b WHERE a. CODE = b. CODE AND a.time_key = b.time_key AND a. CODE IN ('HK.00700', 'HK.03888'); table :rsl 然后查询出根据每一种的code 中最新的一组数据…
declare @t table(name varchar(),qy varchar(),je int) insert into @t union all union all union all union all union all --select * from @t a where not exists --这是取表中的NAME相同的最大值 --( -- from @t where name=a.name and je>a.je --) --第一个答案: SELECT NAME,QY,JE…