从html解析返回数据到ajax

Question

I have some problem with the returned value of ajax.

this is the ajax code:

$(document).ready(function() {
    var request;
    $("#flog").submit(function(event) {
        if(request)
            request.abort();
        event.preventDefault();
        var form = $(this);
        var serializedData = form.serialize();
        var btnname = $('#log').attr('name');
        var btnval = $('#log').val();
        var btn = '&'+btnname+'='+btnval;
        serializedData += btn;      

        request = $.ajax({
            type: form.attr('method'),
            url: form.attr('action'),
            data: serializedData,
        });

        request.done(function(data, status, jdXHR) {
            alert(data);
        });

        request.fail(function(jdXHR, status, error) {

        });
    });
});

it takes data from a form and send it to another page.

this is the second page:

<?php include 'head.php'; ?>
    <?php
    if($_POST['login']) {

    session_regenerate_id(true);
    $con = mysqli_connect("localhost", "Alessandro", "ciao", "freetime")
        or die('Could not connect: ' . mysqli_error($con));

    $query = 'SELECT * FROM users WHERE username="' . $_POST['user'] . '"';
    $result = mysqli_query($con, $query) or die('Query failed: ' . mysqli_error($con));
    if(mysqli_num_rows($result) == 0) {
        mysqli_close($con);
        session_unset();
        session_destroy();
        $res = false;
        return $res;
    }

    $query = 'SELECT password FROM users WHERE username="' . $_POST['user'] . '"';
    $result = mysqli_query($con, $query) or die('Query failed: ' . mysqli_error($con));
    $line = mysqli_fetch_array($result, MYSQL_ASSOC);
    if(md5($_POST['password']) != $line['password']) {
        mysqli_close($con);
        session_unset();
        session_destroy();
        return false;
    }
?>
<?php include 'foot.php'; ?>

and in .done the returned data is all the html page. How can I retrieve only a data, like $res? I tried with json_encode() but with no results. If in the second page I delete the lines include 'head.php' and include 'foot.php' it works. But I need that the secon page is html, too. Somenone can help me?

Answer 1

Dont use the Data attribute from AJAX.

Replace

request.done(function(data, status, jdXHR) {
        alert(data);
    });

with

request.done(function(data, status, jdXHR) {
        alert(jdXHR.responseText);
    });

Answer 2

You could do it in a much simpler way. In PHP store the result of the attempted login into a variable, for instance $result =0; to start with If the login is valid change it to 1 and return it to ajax by doing an echo at the end of your PHP file. If you need other value returned such as the name you could add it to the variable with a separator such as || for instance.

in javascript collect your return and go data = data.split('||'); if (data[0] == 0){alert("Welcome back " + data[1]);}else{alert("wrong login...")}

Previous use is correct, you need to escape the user collected in your PHP script. Hope this helps.