Question

I get this error message when I try to use mysql_fetch_array():

while( $deffefgpSfet->fetch() ) {
   $roefe= mysql_fetch_array($ffefe, MYSQL_NUM);

Warning:  mysql_fetch_array(): supplied argument is not a valid 
         MySQL result resource in <b>/opt/lampp/htdocs/index.php on line 157

Answer 1

You don’t show the code for the object $defeft but I can assume what you’re trying to do…

$blah = mysql_query("SELECT 1 FROM information_schema.tables");

if(!$blah) { // check for mysql errors
  echo mysql_error();
  exit;
}

while( $defeft = mysql_fetch_array($blah)) {

 echo $defeft['row'];

}

Defeft holds the rows data in a array, and while loops through each row.