From: | Mark Fox <mark(dot)fox(at)gmail(dot)com> |
---|---|
To: | Dann Corbit <DCorbit(at)connx(dot)com> |
Cc: | pgsql-general(at)postgresql(dot)org |
Subject: | Re: Days in month query |
Date: | 2005-03-30 23:45:43 |
Message-ID: | bca49a9f0503301545687bee23@mail.gmail.com |
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Thread: | |
Lists: | pgsql-general |
Greetings,
Thanks Dan, but I searched for, and scoured, that page before asking
my question. It helped with some of the details, but not on the
general approach. I'll try to restate my problem in a better way:
What I want is SELECT statement that references no tables but returns
the days in a given month. I'm now thinking that I might be able to
come up with something using an IN clause and using EXTRACT, but
haven't figured it out yet.
Mark
On Wed, 30 Mar 2005 15:16:19 -0800, Dann Corbit <DCorbit(at)connx(dot)com> wrote:
> The online documentation has a search function. It would lead you to
> this:
> http://www.postgresql.org/docs/8.0/static/functions-datetime.html
>
> -----Original Message-----
> From: pgsql-general-owner(at)postgresql(dot)org
> [mailto:pgsql-general-owner(at)postgresql(dot)org] On Behalf Of Mark Fox
> Sent: Wednesday, March 30, 2005 3:07 PM
> To: pgsql-general(at)postgresql(dot)org
> Subject: [GENERAL] Days in month query
>
> Greetings,
>
> This is more of an SQL question, but since each database server seems
> to use it's own syntax for handling dates...
>
> Is there a way to query for the days in a month? For example,
> querying for the days in January of this year? Listing the days
> between two dates would be useful as well.
>
> I'm sure I saw a query like this somewhere, but I can't track it down.
> Just to be clear, there were no tables involved. Just a SELECT
> statement that returned all the days in a given month.
>
> Basically, I have a table of "events" and I'd like to generate a
> histogram of how many events occur on the days of a particular month.
> What I do now is create a temporary table, fill it with the
> appropriate days, and then do a cross join and summation to generate
> what I need. This works, but seems messy to me.
>
> Mark
>
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