What's wrong??

[NZ_Kiwis]

$result = mysql_query ("SELECT * FROM BET_Data WHERE `GameID` = $GID");
while ($row = mysql_fetch_array($result)){
$UID = $row['UserID'];

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/mysite/public_html/a/save.php on line 14

Line 14 is
while ($row = mysql_fetch_array($result)){
$GID = to 1 and GameID is a cell..... what is wrong???

[NogDog]

Usually that means the query failed to process for some reason, such as a SQL syntax error. Try this to get some debug info:

$sql = "SELECT * FROM BET_Data WHERE `GameID` = $GID";
$result = mysql_query ($sql);
if($result == false)
{
   die(mysql_error() . "<br />\n$sql");
}

Once debugged, change the die() to something more "graceful" that does not output database details to the browser.