Please help Quick fix needed

When i try to run this script i get an error (You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1)

include'mysql.php';

$result = mysql_query("SELECT * FROM cat_gambling")or die(mysql_error());
$id = $row['id'];

while($row = mysql_fetch_array( $result )) {

$result2 = mysql_query("SELECT * FROM listings WHERE id = $id")or die(mysql_error());

while($row2 = mysql_fetch_array( $result2 )) {

echo $row['site'];
};};

Can anyone help me find a way to fix it please ?

 

hi
i can help for $10 by paypal

pm me
I am online now


Regards
Alex

 

the purpose of the script is to get all id's (int) from table cat_gambling

Then use the id's to grab information from the listings table

eventually i will display the info in seperate tables using (while) function

 

Thanks for the offer but i would rather work it out myself or go to phpfreaks forum

 

Fixed it myself using tizag.com

// Get all the data from the "example" table
$result = mysql_query("SELECT * FROM cat_gambling")or die(mysql_error());

// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
$id = $row['poker'];
$result2 = mysql_query("SELECT * FROM listings WHERE id = $id")or die(mysql_error());
$row2 = mysql_fetch_array($result2) or die(mysql_error());
echo $row2['site'];
};