sql查询去重
if not object_id('tempdb..#t') is null
drop table #t
go
create table #t([id] int,[name] nvarchar(1),[memo] nvarchar(2))
insert #t
select 1,n'a',n'a1' union all
select 2,n'a',n'a2' union all
select 3,n'a',n'a3' union all
select 4,n'b',n'b1' union all
select 5,n'b',n'b2'
go
--i、name相同id最小的记录(推荐用1,2,3),方法3在sql05时,效率高于1、2
方法1:
select * from #t a where not exists(select 1 from #t where name=a.name and id<a.id)
方法2:
select a.* from #t a join (select min(id)id,name from #t group by name) b on a.name=b.name and a.id=b.id
方法3:
select * from #t a where id=(select min(id) from #t where name=a.name)
方法4:
select a.* from #t a join #t b on a.name=b.name and a.id>=b.id group by a.id,a.name,a.memo having count(1)=1
方法5:
select * from #t a group by id,name,memo having id=(select min(id)from #t where name=a.name)
方法6:
select * from #t a where (select count(1) from #t where name=a.name and id<a.id)=0
方法7:
select * from #t a where id=(select top 1 id from #t where name=a.name order by id)
方法8:
select * from #t a where id!>all(select id from #t where name=a.name)
方法9(注:id为唯一时可用):
select * from #t a where id in(select min(id) from #t group by name)
--sql2005:
方法10:
select id,name,memo from (select *,min(id)over(partition by name) as minid from #t a)t where id=minid
方法11:
select id,name,memo from (select *,row_number()over(partition by name order by id) as minid from #t a)t where minid=1
生成结果:
/*
id name memo
----------- ---- ----
1 a a1
4 b b1
(2 行受影响)
*/
--ii、name相同id最大的记录,与min相反:
方法1:
select * from #t a where not exists(select 1 from #t where name=a.name and id>a.id)
方法2:
select a.* from #t a join (select max(id)id,name from #t group by name) b on a.name=b.name and a.id=b.id order by id
方法3:
select * from #t a where id=(select max(id) from #t where name=a.name) order by id
方法4:
select a.* from #t a join #t b on a.name=b.name and a.id<=b.id group by a.id,a.name,a.memo having count(1)=1
方法5:
select * from #t a group by id,name,memo having id=(select max(id)from #t where name=a.name)
方法6:
select * from #t a where (select count(1) from #t where name=a.name and id>a.id)=0
方法7:
select * from #t a where id=(select top 1 id from #t where name=a.name order by id desc)
方法8:
select * from #t a where id!<all(select id from #t where name=a.name)
方法9(注:id为唯一时可用):
select * from #t a where id in(select max(id) from #t group by name)
--sql2005:
方法10:
select id,name,memo from (select *,max(id)over(partition by name) as minid from #t a)t where id=minid
方法11:
select id,name,memo from (select *,row_number()over(partition by name order by id desc) as minid from #t a)t where minid=1
生成结果2:
/*
id name memo
----------- ---- ----
3 a a3
5 b b2
(2 行受影响)
*/