如何在不使用JSON的情况下进行AJAX调用？ [关闭]

Question

I am trying to improve my jQuery and AJAX knowledge and I wanted to create a simple call to my database. However, I am having some issues with it and I can't really understand why. Here is the code:

function provaajax() {
    $.ajax({
        url: 'visualizza.php',           
        data: "",                      
        dataType: 'text',               
        complete: function(data) {
            var nome = data[0];
            var cognome = data[1];
            $('.provajax').html("<b>nome: </b>" + nome + "<b> cognome: </b>" + cognome);
        }
    });
}

And here is my PHP file:

<?
    Connection();
    $result = mysql_query("SELECT * FROM prova WHERE ID = 117");
    $array = mysql_result($result, 0, "Nome");
    $array = mysql_result($result, 0, "Cognome");
    echo $array;
?>

And a little code from my HTML:

<div>
    <button onclick="provaajax()">Prova Ajax</button>
    <p class="provajax"> Hello world </p>
</div>

But this code isn't working. I don't know how to use JSON, so it would be better not to use it. Any ideas?

Answer 1

To retrieve the nome and cognome values from your PHP it would probably be easiest to return JSON. To do that you can create an array with those values and encode it with json_encode:

<?
    Connection();
    $result = mysql_query("SELECT * FROM prova WHERE ID = 117");
    $arr = array('nome' => mysql_result($result, 0, "Nome"), 'cognome' => mysql_result($result, 0, "Cognome"));
    echo json_encode($arr);
?>

Then you can retrieve these values from the deserialised object provided to the success function by jQuery:

complete: function(data) {
    var nome = data.nome;
    var cognome = data.cognome;
    $('.provajax').html("<b>nome: </b>" + nome + "<b> cognome: </b>" + cognome);
}