Re: Overhead of union versus union all

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
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