Re: is there a better way to do this?

On Feb 19, 5:04 pm, "fitzjarr..._at_cox.net" <fitzjarr..._at_cox.net> wrote:
> On Feb 19, 2:32 pm, Ben <benal..._at_yahoo.com> wrote:
>
>
>
>
>
> > 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.
>
> What, exactly, does that last sentence mean?
>
> David Fitzjarrell- Hide quoted text -
>
> - Show quoted text -

in the original post my table definition doesn't have a column C or D but the output shows them. I removed those columns from my post to try to avoid confusion and forgot to remove them from the output of the query. Received on Wed Feb 20 2008 - 06:44:14 CST