Supplied argument is not a valid ....

manjeman

Warning: Supplied argument is not a valid MySQL-Link resource in /home/httpd/anteil/cphp/registration/include/DB_subs.inc on line 213

i've got this result when i am removing (&$conn) to ($conn). what is the meaning of "&" ? i've never seen $conn with "&". could anybody resolve for me??did i have to use &$conn instead of $conn?

Anon

this error usually comes when there is nothing in your resultset .. Please check your SQL Query and use "DIE" to break it..

and what do you mean by "(&$conn)" ??

Please clearify..

manjeman

i too never knows what is "&$conn". this is where i've found the variable

function DB_Open ($db,$host,$user,$passwd,&$conn) {

$conn = mysql_connect($host,$user,$passwd);
if ($conn) {
    if (! mysql_select_db($db,$conn)) {
        $status = mysql_error();
    } else {
        $status= 0;
            echo "succeessful connected to $db<br>";
    }
}  else {

    $status = mysql_error();
}
return ($status) ;

}

when i'm using the "&$conn" i've got error which is the table i'm refer to is not exist, but it is existed.

and when i'm remove "&" and just use "$conn" i've got msg error "Warning: Supplied argument is not a valid MySQL-Link resource in /home/httpd/anteil/cphp/registration/include/DB_subs.inc on line 213".

what should i do?

Anon

I don't know why are you using that function even you are unable to understand it properly.. why are you making it comlicated man ?? you can write simple statements for this ..

$db=mysql_connect("localhost") or die("Could not connect to the DB");
mysql_select_db("yourdatabase_name",$db) or die("Could not select to the DB");

and run the queries ..

$sql = "select * from yourtable_name";
$result=mysql_query($sql);

is it fine ??

Enjoy
Tabish

manjeman

okiee..okie...i've got itt...thanx for the help....🙂