From: | John McKown <jmckown(at)prodigy(dot)net> |
---|---|
To: | David Lloyd-Jones <david(dot)lloyd-jones(at)attcanada(dot)ca> |
Cc: | pgsql-sql(at)postgresql(dot)org |
Subject: | Re: Week of the Year? |
Date: | 2000-08-12 02:55:05 |
Message-ID: | Pine.LNX.4.21.0008112119390.15236-100000@linux2.johnmckown.net |
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Lists: | pgsql-sql |
Try using the function date_part such as:
select date_part('week',now());
"and the days that are in that week" I guess want to answer a question
such as:
Given a date, what is first date in that same week, and what is the last
date in that week. There are a couple of approaches to this. My first was:
select
to_date(date_part('year',now()),'YYYY')+(7*date_part('week',now()));
and the above +6 to the the last day of the week. Another approach for
this same question is much simplier (if the question is indeed what you
are asking)
select now()-date_part('dow',now());
This last select gives the Sunday for the current week. To get the
Saturday, simply:
select now()-date_part('dow',now())+6;
Of course, replace the now() with whatever contains the date or timestamp.
John McKown
> I'm probably staring right at it. (One of the difficulties with RTFMing, is
> having too many docs!)
>
> Is there anything in the API that produces the week of the year, from 1 to
> 52 or 53 depending on the week of the year, and the days that are in that
> week?
>
> Many thanks.
>
> -dlj.
>
>
>
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