From: Tom Lane <tgl@sss.pgh.pa.us>
To: Robert Creager <Robert_Creager@LogicalChaos.org>
Cc: Martin Marques <martin@bugs.unl.edu.ar>
Cc: pgsql-sql@postgresql.org
Subject: Re: inverse of "day of year"
Date: Fri, 19 Mar 2004 10:10:05 -0500
Message-ID: <24189.1079709005@sss.pgh.pa.us> (raw)
In-Reply-To: <20040319064714.405b304e.Robert_Creager@LogicalChaos.org>
References: <200403190906.17641.martin@bugs.unl.edu.ar>
<20040319064714.405b304e.Robert_Creager@LogicalChaos.org>
Robert Creager <Robert_Creager@LogicalChaos.org> writes:
> Martin Marques <martin@bugs.unl.edu.ar> confessed:
>> Is there a function that would give me the date for a given day of year?
> Something like:
> select date_trunc( 'year', now() ) + (extract( doy from now() ) - 1) *
> '1day'::interval;
timestamp + interval arithmetic is likely to give you problems at
daylight savings boundaries, since '1day' will be taken as '24hours'.
A more reliable way to get (what I assume is) the desired result is
to use the date + integer operator:
select date_trunc('year', now())::date + (extract(doy from now()) - 1)::integer;
regards, tom lane
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To: pgsql-sql@postgresql.org
Cc: tgl@sss.pgh.pa.us, Robert_Creager@LogicalChaos.org, martin@bugs.unl.edu.ar
Subject: Re: inverse of "day of year"
In-Reply-To: <24189.1079709005@sss.pgh.pa.us>
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