Received: from localhost (maia-1.hub.org [200.46.204.191])
	by postgresql.org (Postfix) with ESMTP id 66F009FB3BA
	for <pgsql-sql-postgresql.org@postgresql.org>;
	Sat, 17 Feb 2007 14:40:03 -0400 (AST)
Received: from postgresql.org ([200.46.204.71])
	by localhost (mx1.hub.org [200.46.204.191]) (amavisd-new, port 10024)
	with ESMTP id 33333-08 for <pgsql-sql-postgresql.org@postgresql.org>;
	Sat, 17 Feb 2007 14:40:00 -0400 (AST)
X-Greylist: from auto-whitelisted by SQLgrey-1.7.4
Received: from zenon.apartia.fr (zenon.apartia.fr [82.66.93.83])
	by postgresql.org (Postfix) with ESMTP id 538209FB3B1
	for <pgsql-sql@postgresql.org>; Sat, 17 Feb 2007 14:40:00 -0400 (AST)
Received: from styx.apartia.fr (styx.apartia.fr [10.0.3.109])
	(using TLSv1 with cipher DHE-RSA-AES256-SHA (256/256 bits))
	(Client CN "styx.apartia.fr", Issuer "ca.apartia.fr" (verified OK))
	by zenon.apartia.fr (Postfix) with ESMTP id D693CF4B8A32F
	for <pgsql-sql@postgresql.org>; Sat, 17 Feb 2007 19:39:58 +0100 (CET)
Received: by styx.apartia.fr (Postfix, from userid 1000)
	id 20078B2AE50B; Sat, 17 Feb 2007 19:40:00 +0100 (CET)
Date: Sat, 17 Feb 2007 19:40:00 +0100
From: Louis-David Mitterrand <vindex+lists-pgsql-sql@apartia.org>
To: pgsql-sql@postgresql.org
Subject: Re: sub-limiting a query
Message-ID: <20070217183959.GA29756@apartia.fr>
Mail-Followup-To: pgsql-sql@postgresql.org
References: <20070217125634.GA20589@apartia.fr> <45D747B4.5050900@gmail.com>
MIME-Version: 1.0
Content-Type: text/plain; charset=us-ascii
Content-Disposition: inline
In-Reply-To: <45D747B4.5050900@gmail.com>
User-Agent: Mutt/1.5.13 (2006-08-11)
X-Virus-Scanned: Maia Mailguard 1.0.1
X-Spam-Status: No, hits=1.509 tagged_above=0 required=5 tests=AWL, BAYES_50,
	RCVD_IN_NJABL_DUL, SPF_HELO_PASS, SPF_PASS
X-Spam-Level: *
X-Archive-Number: 200702/188
X-Sequence-Number: 27522

On Sat, Feb 17, 2007 at 07:21:40PM +0100, M.P.Dankoor wrote:
> Louis-David Mitterrand wrote:
> >  
> I thought of another solution, actually it's of those top n query tricks 
> that I picked up somewhere, can't remember
> where.
> Assuming that your table is called shows, the following query should 
> give you the results you want (hope so)
> 
> SELECT *
> FROM shows a
> WHERE 3 > (SELECT COUNT(*)
>           FROM shows b
>           WHERE b.created_on >= a.created_on
>           and a.show_type = b.show_type)

This is stunning and it works!

I can barely understand the query: it's so terse it hurts :)

/me goes back studying it


Thanks a lot!