Re: inverse of "day of year"
| От | Tom Lane |
|---|---|
| Тема | Re: inverse of "day of year" |
| Дата | |
| Msg-id | 24189.1079709005@sss.pgh.pa.us обсуждение |
| Ответ на | Re: inverse of "day of year" (Robert Creager <Robert_Creager@LogicalChaos.org>) |
| Ответы |
Re: inverse of "day of year"
|
| Список | pgsql-sql |
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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