表:
CREATE TABLE Company_Info
(
?? id? smallint IDENTITY(1,1) ,
???username varchar(50) null,
???password varchar(50) null,
???PRIMARY KEY(id)
)
思路
------------顯示哪些人相同,相同的數(shù)量是多少?
select convert(int,SQRT(count(*))) as countU ,a.username from Company_Info a left join Company_Info b on
a.username=b.username and a.password=b.password
group by a.username having SQRT(count(*)) >1
?
----------取出所有相同的記錄到一個表MyRepeat
select c.* into MyRepeat from Company_Info c where c.username in(
select a.username from Company_Info a left join Company_Info b on
a.username=b.username and a.password=b.password
group by a.username having SQRT(count(*)) >1)
?
?
-----------在表MyRepeat中找出每個相同記錄的除最大id之外
select a.id from MyRepeat a where id not in(select max(b.id) from MyRepeat b group by b.username)--得出所有需要刪除的id
-----------
最后得出
delete Company_Info where id in (select h.id from (select c.* from Company_Info c where c.username in(select a.username from Company_Info a left join Company_Info b on a.username=b.username and a.password=b.password group by a.username having SQRT(count(*)) >1) ) h where h.id not in( select max(l.id) from (select k.id,username from (select f.* from Company_Info f where f.username in(select g.username from Company_Info g left join Company_Info o on g.username=o.username and g.password=o.password group by g.username having SQRT(count(*)) >1)) k ) l group by l.username))
?
?
posted on 2006-10-18 17:04
jackstudio 閱讀(1657)
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