Salve ragazzi ho un problema, ho creato il seguente script:

Codice PHP:
        
        
include("class.php");
        
$db = new Database;
        
$db->connect();
        
        
$sql "SELECT * FROM news";
        
$result $db->query($sql) or die ("Errore query: " mysql_error());
        
        
$num_rows $db->numRows();
        
        for (
$k 0$num_rows $k$k++)
        {
              
            
// Ricaviamo i valori
            
$id mysql_result($result$k'id_news');
            
$testo mysql_result($result$k'testo');
            
$data mysql_result($result$k'data');
            
            if (
$k == 0)
                
$color "#F2F2F2";
            else
                
$color "FFFFFF";
                
            echo 
"<p style=\"background-color: ".$color."\">";
            echo 
"[b]<img src=\"img/freccia.png\" width=\"13\" height=\"9\" />".$data."[/b]
"
;
            echo 
$testo;
            echo 
"</p>";
        } 
Non riesco a capire perchè mi dà il seguente errore:

codice:
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in D:\Inetpub\webs\fitnutritionit\index.php on line 97

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in D:\Inetpub\webs\fitnutritionit\index.php on line 98

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in D:\Inetpub\webs\fitnutritionit\index.php on line 99
Ecco le righe 97,98 e 99:

Codice PHP:
$id mysql_result($result$k'id_news');
$testo mysql_result($result$k'testo');
$data mysql_result($result$k'data'); 
Grazie anticipatamente,
Gaten