[RESOLVED] Error with mysql_fetch_array() in script

[Desbrina]

function normal() {
	$con = mysql_connect('localhost', '****', '****') or die('Could not connect to database!');
    mysql_select_db('****', $con);
	$sql = "(SELECT * FROM cards WHERE worth = 1 AND `deny`=0 ORDER BY RAND() LIMIT 1)"; 	   

	$result2 = mysql_query($sql, $con);
	if (!$result2) { echo( mysql_error()); }
	while ($row = mysql_fetch_array($result2)) {
	   echo "<img src=\"{$row[url]}\" alt=\"{$row[alttext]}\" /> " ; 
	   $alt = $row['alttext'];
		$collecting = "SELECT * FROM user_inventory WHERE alttext = '".$alt."' and name = '".$_COOKIE['dwtcg']['u']."' and status = 'collecting' LIMIT 1";
		$collect = mysql_query($collecting);
		while ($row = mysql_fetch_array($collect)) {
			if($row['status'] == 'collecting') {
				$sql = "SELECT * FROM `cards` WHERE alttext = '".$alt."'";		
				$result = mysql_query($sql, $con);
				if (!$result) { echo( mysql_error()); }		
				while ($row = mysql_fetch_array($result)) {
					$update = "UPDATE user_inventory set url = '".$row['url']."' WHERE alttext = '".$alt."' LIMIT 1";
					$result = mysql_query($update);
					if(!$result) { echo(mysql_error()); }
				}
			} elseif(mysql_num_rows($collect) == 0) {
				$sql1 = "SELECT * FROM `cards` WHERE alttext = '".$alt."'";		
				$result1 = mysql_query($sql1, $con);
				if (!$result1) { echo( mysql_error()); }		
				while ($row = mysql_fetch_array($result1)) {
					$add = "INSERT INTO `desbrina_sw`.`user_inventory` (`ID` ,`name` ,`url` ,`alttext` ,`deck` ,`user_id` ,`card_id` ,`status` ,`worth` ,`amount` ,`type`) VALUES (NULL , '".$_COOKIE['dwtcg']['u']."', '".$row['url']."', '".$row['alttext']."', '".$row['deck']."', '".$_COOKIE['dwtcg']['i']."', '".$row['ID']."', 'undefined', '".$row['worth']."', '".$row['amount']."', '".$row['type']."')";
					$result = mysql_query($add);
					if(!$result) { echo(mysql_error()); }
				}
			}
		}
	}
}

I seem to have a problem with the piece of code
It works correctly, ading and updating, but it gives this error when getting to one that needs to be updated not added

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/home/desbrina/public_html/games/insert.php on line 19

Whats causing it

Whoops, can this get moved to php forum please

[NogDog]

That error almost always means your query could not be parsed/processed by MySQL. You need to check the return values of your mysql_query() statements, and if they return false generate some debug output (to a log file, or using die(), etc.). In that debug output, include the query string variable (so you can see exactly what was sent to MySQL) along with a mysql_error() to see the most recent error message from MySQL.

[Desbrina]

solved it, it was because i was using $result and $sql twice. Soon as i changed them it started working