I solved it. I had to pass the $conn when calling the function.
On Tue, 2002-01-29 at 14:09, Frank Bax wrote:
> I believe this is a problem with $conn, not your sql syntax. How is $conn
> defined? Is it passed as an argument to the function containing this code,
> or defined as a global? Are you using pg_pconnect or pg_connect()?
>
> At 01:34 PM 1/29/02 -0500, Jeff Self wrote:
> >I'm having trouble getting an SQL statement to use a variable. Here's
> >the statement:
> >
> > $sql = "SELECT emp_fname FROM employee WHERE username = '$username'";
> > $result = pg_exec($conn,$sql);
> > if (!$result) {
> > exit;
> > }
> >I get the following warning from this:
> >
> >Warning: Supplied argument is not a valid PostgreSQL link resource in
> >/var/www/personnel/include/functions.inc on line 25
> >
> >Is there another way to use a variable in an SQL statement. This format
> >works with MySQL.
> >
> >--
> >Jeff Self
> >Information Technology Analyst
> >Department of Personnel
> >City of Newport News
> >2400 Washington Ave.
> >Newport News, VA 23607
> >757-926-6930
> >
> >
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> >
> >http://www.postgresql.org/users-lounge/docs/faq.html
> >
>
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--
Jeff Self
Information Technology Analyst
Department of Personnel
City of Newport News
2400 Washington Ave.
Newport News, VA 23607
757-926-6930
