Mysql Syntax error

Hi--

I have the following script. It takes a variable from a link i.e.,
showsearch.php?id=103 and is supposed to simply display the record.

Instead I am getting the following error:

Warning : mysql_fetch_array(): supplied argument is not a valid MySQL
result resource in /usr/www/users/grndlvl/getdbtil.php on line 94

Where Line 94 is : if ($row = mysql_fetch_array($dbquery)) {

Here is the full script:

<?php


$id = addslashes($_GET['id']);

if( empty($id) ) {
echo "Sorry. No TIL Articles were found for your Question. Please try
again.";
}
else {
//connect to host

$dbServer = "host";
$dbDatabase = "db";
$dbUser = "user";
$dbPass = "password";


$sConn = mysql_connect($dbServer, $dbUser, $dbPass);
if (!$sConn)
echo 'Unable to connect to database server because ',mysql_error();
else {
$dConn = mysql_select_db($dbDatabase, $sConn);
if (!$dConn)
echo 'Unable to connect to database because ',mysql_error();
else {

//find which record has been sent by the GET
$dbQuery .= "SELECT * FROM articles WHERE id='" .$id ."'" ;

if ($row = mysql_fetch_array($dbquery)) {
echo "<b>Answer: </b>",$row[4];
echo "<b>Topic: </b>",$row[7];
echo "<b>Created: </b>",$row[5];
echo "<b>Last Modified: </b>",$row[6];
echo "<b>TIL No.: </b>",$row[8];
}
else {

echo "<b>Sorry, we couldn't find an answer to your Question.
Please contact Sales at: <a
href=\"sales@mysite.com\">sales@mysite.com</a></b>" ;
}
}
}
}


?>

Thank you!

All My Best,
Jeffrey

On Sat, 16 Aug 2003 22:22:24 -0700, Jeffrey Ellis <support@groundlevel.net>
wrote:

> $dbQuery .= "SELECT * FROM articles WHERE id='" .$id ."'" ;
>
> if ($row = mysql_fetch_array($dbquery)) {

You've not executed any queries. Look up mysql_query.

--
Andy Hassall (andy@andyh.co.uk) icq(5747695) (http://www.andyh.co.uk)
Space: disk usage analysis tool (http://www.andyhsoftware.co.uk/space)

in article rgsujvs8kvbnmog6bu05sd6pgcici7jhqm@4ax.com, Andy Hassall at
andy@andyh.co.uk wrote on 8/17/03 5:16 AM:

> On Sat, 16 Aug 2003 22:22:24 -0700, Jeffrey Ellis <support@groundlevel.net>
> wrote:
>
>> $dbQuery .= "SELECT * FROM articles WHERE id='" .$id ."'" ;
>>
>> if ($row = mysql_fetch_array($dbquery)) {
>
> You've not executed any queries. Look up mysql_query.
>
> --
> Andy Hassall (andy@andyh.co.uk) icq(5747695) (http://www.andyh.co.uk)
> Space: disk usage analysis tool (http://www.andyhsoftware.co.uk/space)

Right

I added the actual query and now it's working great. Thank you!

All My Best,
Jeffrey