is there a better way to do this?
From: Ben <benalvey_at_yahoo.com>
Date: Tue, 19 Feb 2008 12:32:36 -0800 (PST)
Message-ID: <faba7a38-01a2-4ac1-8735-714e119c2edf@s19g2000prg.googlegroups.com>
commit;
Date: Tue, 19 Feb 2008 12:32:36 -0800 (PST)
Message-ID: <faba7a38-01a2-4ac1-8735-714e119c2edf@s19g2000prg.googlegroups.com>
10.2.0.2 EE
Is there a more logical sql statement to get the same results than what I am running here?
create table t (a number, b number);
insert into t values (1, 1); insert into t values (1, 2); insert into t values (2, 3); insert into t values (3, 2); insert into t values (4, 4); insert into t values (4, 5); insert into t values (5, 1); insert into t values (6, 8);
commit;
select a.a, a.b
from t a, (
select b, count(distinct a)
from t
group by b
having count(distinct a) > 1) b
where a.b = b.b
A B C D ---------- ---------- -------------------- -------------------- 5 1 a a 1 1 a a 3 2 a a 1 2 a a
What I am wanting are the values for A where B is the same. Received on Tue Feb 19 2008 - 14:32:36 CST