Easy error

[Anonymous]

Should be for you pros. I have no clue

Heres the error on the page that I get:
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/teamlord/public_html/includes/sql_layer.php on line 301

Here is the code that pretains to that line and the adjacent lines.
function sql_fetch_row(&$res, $nr=0)
{
global $dbtype;
switch ($dbtype) {

case "MySQL":
$row = mysql_fetch_row($res); ****THIS IS LINE 301*****
return $row;
break;;

case "mSQL":
$row = mysql_fetch_row($res);
return $row;
break;;

case "PostgreSQL":
case "PostgreSQL_local":
if ( $res->get_total_rows() > $res->get_fetched_rows() ) {
$row = pg_fetch_row($res->get_result(), $res->get_fetched_rows() );
$res->increment_fetched_rows();
return $row;
} else {
return false;
}
break;;

case "ODBC":
case "ODBC_Adabas":
$row = array();
$cols = odbc_fetch_into($res, $nr, $row);
return $row;
break;;

case "Interbase":
$row = ibase_fetch_row($res);
return $row;
break;;

case "Sybase":
$row = sybase_fetch_row($res);
return $row;
break;;

default:
break;;
}
}

 

[Anonymous]

Heres the error on the page that I get:
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/teamlord/public_html/includes/sql_layer.php on line 301

Here is the code that pretains to that line and the adjacent lines.

function sql_fetch_row(&$res, $nr=0)
{
global $dbtype;
switch ($dbtype) {

    case "MySQL":
        $row = mysql_fetch_row($res); ****THIS IS LINE 301*****[code][/quote]

What's the value of $res that you're passing to sql_fetch_row()? Is it a valid MySQL result resource?