mysql_close(): supplied argument is not a valid MySQL-Link resource

Question

Here's what I'm trying to do: I've got a db.php file that does all the db manipulation.

It has 2 static methods, connect and deconnect.

In my other file i simply use db::connect() and db::deconnect(). The mysql_close($con) in the deconnect method just doesn't know who $con is.

Since I don't want to instantiate my class static is the only way to go.

Declaring 'private $con' in class db doesn't seem to have an effect.

Any ideas?

class db {

    public static function connect() {
        $dbData = parse_ini_file($_SERVER['DOCUMENT_ROOT'].'/config.ini');

        $con = mysql_connect($dbData['host'],$dbData['user'],$dbData['pass']);
        $db = mysql_select_db($dbData['db']);
        if ((!$con) || (!$db))
            return 0;
        else return 1;
    }

    public static function deconnect() {
        mysql_close($con);
    }

}

Answer 1

The message makes sense as $con is out of scope in your deconnect() method (deconnect...?).

Use a static data member

class db { static $con; }

Access it through self::$con.

Answer 2

In deconnect, $con is out of scope.

You should make it a static member, like this:

class db {
    static $con;

    public static function connect() {
        $dbData = parse_ini_file($_SERVER['DOCUMENT_ROOT'].'/config.ini');

        self::$con = mysql_connect($dbData['host'],$dbData['user'],$dbData['pass']);
        $db = mysql_select_db($dbData['db']);
        if ((!self::$con) || (!$db))
            return 0;
        else return 1;
    }

    public static function deconnect() {
        if( !isset( self::$con ) ) return;
        mysql_close( self::$con );
    }

}

Answer 3

Well, it seems to me that $con is a local variable (local to the method connect). Thus, when you call mysql_close, most likely $con is undefined. Try declaring $con as a private variable in your class. Look here for more info on how to do that.