try:
$sql = "select exhibition_name,to_char('YYYY',start_date) as start_date \
from exhibitions where exhibition_id = '$exhibition_id'";
----- Original Message -----
From: "Lynna Landstreet" <lynna@gallery44.org>
To: <pgsql-php@postgresql.org>
Sent: Wednesday, July 23, 2003 14:49
Subject: [PHP] Getting the year from a date column
> I'm having a bit of trouble with one part of the PHP front end of the art
> gallery database I'm working on - specifically, extracting the year from
the
> start date of an exhibition. I'm trying to print a list of exhibitions on
> the artist info pages, with the name, gallery space and year for each one,
> and everything's working except the year part.
>
> Here's what I have. First, near the beginning of the PHP portion of the
> page:
>
> $exh_query = "SELECT exhibition_name, start_date
> FROM exhibitions WHERE exhibition_id = '$exhibition_id'";
> $exhibition_info = pg_exec($db, $exh_query);
>
> Then further down, where I want the year to appear in the exhibition list:
>
> $date = getdate (pg_result($exhibition_info, 0, 'start_date'));
> $year = $date['year'];
> echo $year;
>
> (The 0 for row is because the way the code is structured, $exhibition_info
> only contains one row.) But I'm not sure if getdate() is the right
function
> to be using. It was the only function I could find in the PHP manual that
> was capable of breaking out the individual components of a date (day,
month,
> year, etc.), but it seems to be ignoring the actual date in the query
> results and returning '1969' for everything.
>
> Any idea what I'm doing wrong? Is there a different function I should be
> using instead for this? BTW, I'm using PHP 4.1 (and PostgreSQL 7.2), thus
> the old-style pg function names (pg_result(), etc.).
>
> Thanks,
>
> Lynna
> --
> Resource Centre Database Coordinator
> Gallery 44
> www.gallery44.org
>
>
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