I don't see? You?

[djdejan]

I am just learning to write php classes and I have problems:

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

<?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'];
}
?>

[Mordecai]

The problem isn't with the class (or at least, appears not to be), but rather with the query itself.

Whatever query you are sending it appears to be failing.

[Weedpacket]

Have a look at the first two tips in this thread. They'll help you solve your problem, as will using functions like [man]mysql_error[/man].