Bruno Wolff III <[EMAIL PROTECTED]> writes:
> On Thu, May 04, 2006 at 00:05:16 -0400,
> Greg Stark <[EMAIL PROTECTED]> wrote:
> > Bruno Wolff III <[EMAIL PROTECTED]> writes:
> >
> > > > Whereas it shouldn't be hard to prove that this is equivalent:
> > > >
> > > > stark=> explain select col1 from test group by upper(col1),col1 order
> > > > by upper(col1);
> > > > QUERY PLAN
> > > > ---------------------------------------------------------------------
> > > > Group (cost=88.50..98.23 rows=200 width=32)
> > > > -> Sort (cost=88.50..91.58 rows=1230 width=32)
> > > > Sort Key: upper(col1), col1
> > > > -> Seq Scan on test (cost=0.00..25.38 rows=1230 width=32)
> > > > (4 rows)
> > >
> > > I don't think you can assume that that will be true for any locale. If
> > > there
> > > are two different characters that both have the same uppercase version,
> > > this
> > > will break things.
> >
> > No it won't.
>
> Sure it will, because when you do the group by you will get a different
> number of groups. When grouping by the original characters you will get
> separate groups for characters that have the same uppercase character, where
> as when grouing by the uppercased characters you won't.
But grouping on *both* will produce the same groups as grouping on the
original characters alone.
--
greg
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