Date: 07/08/07 (PHP Community)    Keywords: php, mysql, html, database, sql

I'm sorry to disturb, I'm new in php and I have a trouble with this code:


    mysql_connect($host,$utente,$passwd) or die("Connessione non riuscita: " . mysql_error());
    mysql_select_db("news") or die("Selezione del database non riuscita");
    $np = "SELECT * FROM persona";
    if($riga = mysql_fetch_array($np,MYSQL_ASSOC)) {                        
                                echo "" . $riga["Nome"] . " - " . $riga["Cognome"] . " - " . $riga["Organizzazione"];
                                selected ($riga["id_persona"]);
                                    echo "";
                                  }
    while($riga = mysql_fetch_array($np,MYSQL_ASSOC)) {                        
                                echo "" . $riga["Nome"] . " - " . $riga["Cognome"] . " - " . $riga["Organizzazione"];
                                    echo "";
                                  }
?>
 

It has to extract a row from a mysql database and make a html select of the value id_persona displaying also the value of the other fields (Nome, Cognome, Organizzazione) but it don't works properly. Choices on select are displayed like in the attached photo instead with the correct data .

Source: http://community.livejournal.com/php/575883.html