The "\" character is commonly used as a line continuation character.
The PostgreSQL manual (which you should read) shows to_char working like
this (on this page:
http://www.postgresql.org/docs/7.3/static/functions-formatting.html) :
to_char([data],[format]). I guess I had the parameters reversed, but AFAICT
to_char doesn't work with date anyways. So now the PHP manual shows many
useful date functions (http://www.php.net/manual/en/ref.datetime.php). Of
note is date() (http://www.php.net/manual/en/function.date.php) and
strtotime() (http://www.php.net/manual/en/function.strtotime.php). So
perhaps this would work:
$x=date('r', strtotime(pg_result($exhibition_info, 0, 'start_date')));
/B
----- Original Message -----
From: "Lynna Landstreet" <lynna@gallery44.org>
To: "David Busby" <busby@pnts.com>
Sent: Wednesday, July 23, 2003 15:46
Subject: Re: [PHP] Getting the year from a date column
> on 7/23/03 5:59 PM, David Busby at busby@pnts.com wrote:
>
> > try:
> > $sql = "select exhibition_name,to_char('YYYY',start_date) as start_date
\
> > from exhibitions where exhibition_id = '$exhibition_id'";
>
> That got me a whole slew of error messages - originally it choked on the
\,
> and when I took that out I got this:
>
> Warning: PostgreSQL query failed: ERROR: Function 'to_char(unknown, date)'
> does not exist Unable to identify a function that satisfies the given
> argument types You may need to add explicit typecasts in /[path to my home
> directory]/db/artist_exhibitions.php on line 104
>
> Followed by about five other error messages, but those seemed to stem from
> this query failing. Not sure what that first message means...
>
>
> Lynna
>
>
> > ----- Original Message -----
> > From: "Lynna Landstreet" <lynna@gallery44.org>
> > To: < >
> > 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
> >>
> >>
> >> ---------------------------(end of
broadcast)---------------------------
> >> TIP 9: the planner will ignore your desire to choose an index scan if
your
> >> joining column's datatypes do not match
> >
> >
> > ---------------------------(end of broadcast)---------------------------
> > TIP 8: explain analyze is your friend
> >
>
> --
> Resource Centre Database Coordinator
> Gallery 44
> www.gallery44.org
