problema mysql_result

[dj89]

sul sito mi esce qst errore: Warning: mysql_result() expects parameter 1 to be resource, object given in

codice da me usato :

<div>  
            <? 
            $query = "SELECT c.CAS_ID, c.TITOLO, i.NOME,c.DES_BREVE,c.DESC_ESTESA,c.OFFERTA_SPECIALE FROM `CASE` c LEFT JOIN `IMMAGINI` i ON c.CAS_ID = i.CAS_ID where VISIBLE = 1 and DEF = 1 and HOME = 1 order by c.CAS_ID DESC"; 
            $result = mysqli_query($mysqli , $query); 
            $num = mysqli_num_rows($result); 
            $i=0; 
            $count = 1; 
            while ($i < $num && $i <5) { 
                $id_case[] = mysqli_result($result,$i,"CAS_ID"); 
                ?> 
                    <a style="border:1px solid #e2001a; margin-bottom:10px; display:block;" class="box_case" href="javascript:;" onClick="showCasa('bigbox_<? echo mysql_result($result,$i,"CAS_ID");?>'); updateStatCasa(<? echo mysql_result($result,$i,"CAS_ID");?>)">  
                        <table border="0" cellpadding="0" cellspacing="0" width="100%"> 
                            <tr> 
                                <td style="padding:15px 0 15px 15px; cursor:pointer;" width="120" align="center" valign="top"><img src="<?php echo mysql_result($result,$i,"NOME"); ?>" height="90" /></td> 
                                <td style="padding:15px 15px 15px 10px; cursor:pointer;" valign="top"> 
                                    <span style="font-weight:bold; font-size:12px; color:#13007d;"><?php echo mysql_result($result,$i,"TITOLO"); ?></span>  
                                    <br /> 
                                    <?php echo mysql_result($result,$i,"DES_BREVE"); ?> 
                                </td> 
                            </tr> 
                        </table> 
                    </a> 
                <? $i++;$count++; 
            } ?> 
        </div>

nella riga 5 e 6 ho aggiunto la ' i ' alla funzione mysql e funziona, ma nella mysql_result nn va..
guardato anke il manuale php, ma nn riesco..
potreste darmi una mano?
grazie a tutti!!

[.Kurt]

mysql e mysqli sono due api differenti.

[dj89]

<div> 
            <?
            $query = "SELECT c.CAS_ID, c.TITOLO, i.NOME,c.DES_BREVE,c.DESC_ESTESA,c.OFFERTA_SPECIALE FROM `CASE` c LEFT JOIN `IMMAGINI` i ON c.CAS_ID = i.CAS_ID where VISIBLE = 1 and DEF = 1 and HOME = 1 order by c.CAS_ID DESC";
            $result = mysqli_query($mysqli , $query);
            $num = mysqli_num_rows($result);
            $i=0;
            $count = 1;
             while ($row = mysqli_fetch_assoc($result)) {?>
                    <a style="border:1px solid #e2001a; margin-bottom:10px; display:block;" class="box_case" href="javascript:;" onClick="showCasa('bigbox_<? echo $row["CAS_ID"];?>'); updateStatCasa(<? echo $row["CAS_ID"];?>)"> 
                        <table border="0" cellpadding="0" cellspacing="0" width="100%">
                            <tr>
                                <td style="padding:15px 0 15px 15px; cursor:pointer;" width="120" align="center" valign="top"><img src="<?php echo $row["NOME"]; ?>" height="90" /></td>
                                <td style="padding:15px 15px 15px 10px; cursor:pointer;" valign="top">
                                    <span style="font-weight:bold; font-size:12px; color:#13007d;"><?php echo $row["TITOLO"]; ?></span> 
                                    <br />
                                    <?php echo $row["DES_BREVE"]; ?>
                                </td>
                            </tr>
                        </table>
                    </a>
                <? $i++;$count++;
            } ?>
        </div>