| From: | Simon Riggs <simon(at)2ndQuadrant(dot)com> |
|---|---|
| To: | Bruce Momjian <bruce(at)momjian(dot)us> |
| Cc: | Scott Marlowe <scott(dot)marlowe(at)gmail(dot)com>, Scott Bailey <artacus(at)comcast(dot)net>, pgsql-general(at)postgresql(dot)org |
| Subject: | Re: Overhead of union versus union all |
| Date: | 2009-07-10 14:00:47 |
| Message-ID: | 1247234447.11347.598.camel@ebony.2ndQuadrant |
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| Thread: | |
| Lists: | pgsql-general |
On Fri, 2009-07-10 at 09:46 -0400, Bruce Momjian wrote:
> Simon Riggs wrote:
> > or a query like this
> >
> > Select '1', ...
> > ...
> > union
> > Select status, ...
> > ...
> > where status != '1';
> > ;
> >
> > then it is clear that we could automatically prove that the the distinct
> > step is redundant and so we could either hash or sort. This is the same
> > as replacing the UNION with UNION ALL.
>
> In the last example, how do you know that status != '1' produces unique
> output?
You don't. I was assuming that you could already prove that each
subquery was distinct in itself.
It's one for the TODO, that's all. I see it often, but I'm not planning
to work on the code for this myself.
--
Simon Riggs www.2ndQuadrant.com
PostgreSQL Training, Services and Support
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