Help: Simple mysql_fetch_array error...

I'm getting an error with this statement...

$number = urldecode( $_GET['id'] );
$result = mysql_query("SELECT * FROM `Zombie` WHERE `id` = '$number'");

while($row = mysql_fetch_array($result)) {
echo $row['Username']. "<br>";
echo "Level:".$row['Level']."<br>";
echo "Exp:".$row['Exp']."<br>";
echo "Equipped Weapon:".$row['Equipped']."<br>";
}

What is the problem?

-The error is
'Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/mountgam/public_html/zombie/index.php on line 9'

Thanks,

James.

 

I also get the same error on another page, when I input...

$query = ("SELECT * FROM `Zombie` WHERE `Username` = 'Skillman13'");
$result = mysql_query($query);

while ($row = mysql_fetch_array($result)) {
echo "<a href= 'Username.php?" . "&id=". urlencode($row['id']) . "'>". $row['Username'] ."</a>". '<BR>'. "Level:". $row['Level']. '<BR>'. "Equipped Firearm:". $row['Equipped'];
echo '<BR>'. '<BR>';
}


Any ideas?

 

Working now. Thanks

 

Try this

mysql_query("SELECT * FROM `Zombie` WHERE id='".$number."'");