Thread: Trouble with SQL statement using variable
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
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
>
>
>---------------------------(end of broadcast)---------------------------
>TIP 5: Have you checked our extensive FAQ?
>
>http://www.postgresql.org/users-lounge/docs/faq.html
>
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()?
$conn is defined as a global in a file called database.inc which is
here:
<?php
//
// database.inc
//
// Database Information
$dbname = "personnel";
$dbuser = "webuser";
$conn = pg_pconnect("dbname=$dbname user=$dbuser");
if (!$conn) {
echo "Could not connect to database.";
exit;
}
?>
As you can see, I'm using pg_pconnect.
>
> 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
> >
> >
> >---------------------------(end of broadcast)---------------------------
> >TIP 5: Have you checked our extensive FAQ?
> >
> >http://www.postgresql.org/users-lounge/docs/faq.html
> >
--
Jeff Self
Information Technology Analyst
Department of Personnel
City of Newport News
2400 Washington Ave.
Newport News, VA 23607
757-926-6930
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
> >
> >
> >---------------------------(end of broadcast)---------------------------
> >TIP 5: Have you checked our extensive FAQ?
> >
> >http://www.postgresql.org/users-lounge/docs/faq.html
> >
>
> ---------------------------(end of broadcast)---------------------------
> TIP 4: Don't 'kill -9' the postmaster
--
Jeff Self
Information Technology Analyst
Department of Personnel
City of Newport News
2400 Washington Ave.
Newport News, VA 23607
757-926-6930
Have you tried:
echo "CONN: $conn<br>";
echo "SQL: $sql<br>";
to see what is being passed to postgres?
Jeff
On Tuesday, January 29, 2002, at 10:34 AM, 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
>
>
> ---------------------------(end of broadcast)---------------------------
> TIP 5: Have you checked our extensive FAQ?
>
> http://www.postgresql.org/users-lounge/docs/faq.html
>