<![CDATA[problema mysql_result]]>sul sito mi esce qst errore: Warning: mysql_result() expects parameter 1 to be resource, object given in

codice da me usato :

[PHP]<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>[/PHP]

nella riga 5 e 6 ho aggiunto la ' i ' alla funzione mysql e funziona, ma nella mysql_result nn va..
sapete darmi una mano?
grazie a tutti!!

]]>https://connect.gt/topic/215629/problema-mysql_resultRSS for NodeTue, 10 Mar 2026 10:28:23 GMTTue, 07 Jul 2015 08:38:57 GMT60<![CDATA[Reply to problema mysql_result on Tue, 07 Jul 2015 14:51:06 GMT]]>a ki dovesse servire ho risolto cosi:

[PHP]<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; cursorointer;" width="120" align="center" valign="top"><img src="<?php echo $row["NOME"]; ?>" height="90" /></td>
                            <td style="padding:15px 15px 15px 10px; cursorointer;" 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>[/PHP]
]]>https://connect.gt/post/1203282https://connect.gt/post/1203282Tue, 07 Jul 2015 14:51:06 GMT<![CDATA[Reply to problema mysql_result on Tue, 07 Jul 2015 12:24:07 GMT]]>provato e riprovato ma niente...
scusa riusciresti a farmi un esempio?
grazie mille

]]>https://connect.gt/post/1203281https://connect.gt/post/1203281Tue, 07 Jul 2015 12:24:07 GMT<![CDATA[Reply to problema mysql_result on Tue, 07 Jul 2015 09:42:21 GMT]]>Non funziona perché anche alla fine dovresti usare questa

http://php.net/manual/en/class.mysqli-result.php

]]>https://connect.gt/post/1203263https://connect.gt/post/1203263Tue, 07 Jul 2015 09:42:21 GMT