[PHP][MYSQL]Warning: Invalid argument supplied for foreach() on line 25 Opcje

[marcin86s]

Kod wygląda tak:

[PHP] pobierz, plaintext 
    $g_link = false;
 
    function GetMyConnection()
    {
        global $g_link;
        if( $g_link )
            return $g_link;
        $g_link = mysql_connect( 'localhost', 'root', '') or die('Could not connect to server.' );
        mysql_select_db('baza', $g_link) or die('Could not select database.');
        return $g_link;
    }
 
    function CleanUpDB()
    {
        global $g_link;
        if( $g_link != false )
            mysql_close($g_link);
        $g_link = false;
    }
 
$results = mysql_query("SELECT * FROM tab", GetMyConnection() );
 
var_dump($results);
 
foreach ($results as $result){
	echo $result->date2;
}
[PHP] pobierz, plaintext 

Tablica tak:

[SQL] pobierz, plaintext 
CREATE TABLE IF NOT EXISTS `tab` (
  `id` int(10) UNSIGNED NOT NULL AUTO_INCREMENT,
  `date2` int(100) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=60 ;
 
INSERT INTO `tab` (`id`, `date2`) VALUES
(1, 1263855600),
(2, 1263250800);
[SQL] pobierz, plaintext 

Błąd jest taki:

Warning: Invalid argument supplied for foreach() on line 25

Co robię źle? Chce po prostu wyświetlić wszystkie wpisy z tabeli.

Ten post edytował marcin86s 13.11.2011, 14:33:55