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In-Reply-To: <4B636022.90804@kickstone.com>
References: <4B636022.90804@kickstone.com>
Date: Fri, 29 Jan 2010 16:50:31 -0700
Message-ID: <dcc563d11001291550j70867455q2f16fdf3a4a9cc4@mail.gmail.com>
Subject: Re: Partitioning by letter question
From: Scott Marlowe <scott.marlowe@gmail.com>
To: John Lister <john.lister-ps@kickstone.com>
Cc: pgsql-sql@postgresql.org
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On Fri, Jan 29, 2010 at 3:24 PM, John Lister
<john.lister-ps@kickstone.com> wrote:
> Hi, I was wondering if this was possible. I'm trying to partition a table=
,
> which is straightforward enough thanks to the great documentation, but i
> have a question:
>
> If I partition using something like a product_id for example and have che=
ck
> constraints such as (id>=3D1000 and id<2000) then everything is fine and =
the
> planner correctly uses the right subset of the tables. However I would li=
ke
> to partition by the first letter and using something like this
> substr(word,1,1)=3D'a' is ignored by the planner. From reading the docs I
> understand that complicated check constraints are ignored, but this doesn=
't
> seem overly complicated.
>
> Am i doing =A0something wrong or is there another better way to do this

Have you tried:

(word >=3D 'a' and word <'b')

?