我希望找到所有从
一起注册两个或更多部分的不同学生注册(sid *,年级,dname *,cno *,sectno *)
其中每个部分由(dname *,cno *,sectno *)
唯一标识答案 0 :(得分:3)
您可以自行加入标识某个部分的列,并指定e1.sid < e2.sid
以仅获取每对学生一次。
select e1.sid, e2.sid from enroll e1
join enroll e2 on e1.dname = e2.dname
and e1.cno = e2.cno
and e1.secno = e2.sectno
where e1.sid < e2.sid
group by e1.sid, e2.sid
having count(*) > 1
答案 1 :(得分:1)
/*
with enroll as (
select 1 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 1 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 2 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 3 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 1 sid, 'dname' dname, 2 cno, 2 secno from dual union all
select 2 sid, 'dname' dname, 2 cno, 2 secno from dual
)
*/
select e1.sid, e2.sid from enroll e1
join enroll e2 on e1.dname = e2.dname
and e1.cno = e2.cno
and e1.secno = e2.secno
where e1.sid < e2.sid
group by e1.sid, e2.sid
having count(*) >= 2;
(但如果一个sid在同一部分登记的时间超过一次,则此查询将无法正常工作)
答案 2 :(得分:1)
with enroll as (
select 1 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 1 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 2 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 3 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 3 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 1 sid, 'dname' dname, 2 cno, 2 secno from dual union all
select 2 sid, 'dname' dname, 2 cno, 2 secno from dual
)
select rn,sid from (
select row_number() over (partition by sid order by dname) as rn ,sid
from enroll )
where rn>1
使用row_number()
如果一个sid多次注册同一部分
,则此查询可正常工作distinct
with enroll as (
select 1 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 1 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 2 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 3 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 3 sid, 'dname' dname, 1 cno, 1 secno from dual union all
select 1 sid, 'dname' dname, 2 cno, 2 secno from dual union all
select 4 sid, 'dname' dname, 2 cno, 2 secno from dual union all
select 2 sid, 'dname' dname, 2 cno, 2 secno from dual
)
select distinct sid from (
select row_number() over (partition by sid order by dname) as rn ,sid
from enroll )
where rn>1
<强>输出强>
> SID
> ----------
> 1
> 2
> 3