[ djdejan @ 24.07.2004. 16:35 ] @
Da li neko vidi u cemu je problem? Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\web\testclass\class_mysql.php on line 24 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\web\testclass\class_mysql.php on line 27 PHP Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in D:\web\testclass\class_mysql.php on line 24 PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\web\testclass\class_mysql.php on line 27 Code: <?php error_reporting(E_ALL); class mysql { var $oKonekcija=''; var $oResult=''; var $aNiz=''; function db_connect($host, $username, $password) { $this->oKonekcija=mysql_connect($host='', $username='', $password='', $db=''); mysql_select_db($db, $this->oKonekcija); return True; } function db_query($query) { $this->oResult=mysql_query($query, $this->oKonekcija); return True; } function result2array() { if(mysql_num_rows($this->oResult)!=='0') { $i=0; while($aSlog=mysql_fetch_array($this->oResult)) { $this->aNiz[$i]['Naziv']=$aSlog['MailingEmailAdress']; $i++; } return $this->aNiz; } else { return False; } } } $veza=new mysql; $veza->db_connect("localhost", "root", "", "test"); $veza->db_query("select * from mailinglista"); $niz=$veza->result2array(); for ($i=0; $i<5; $i++) { echo $niz[$i]['Naziv']; } ?> |