使用jQuery从PHP文件返回JSON数据

Question

Long time reader, first time poster. I'm very new to the world of jQuery and JSON and have been seeing an issue with a login script I'm running.

The end goal is to capture data from a form, pass that data to a PHP file for processing via jQuery.ajax() post, compare the data against a MySQL database for authentication and return a data for either a success of failure.

My problem is that I cannot get the JSON formatted data to be passed from the PHP script back to the jQuery. When viewing the processing with Chrome's Developer Tools, I see that 'Login Failure'. I've double checked the array $rows by throwing it to my error_log file and it returns properly formatted JSON, but I just can't for the life of me get it to return to the jQuery file. Any help is appreciated.

My form input:

<!-- BEGIN: Login Page -->
    <section data-role="page" id="login">
        <header data-role="header" data-theme="b">
            <a href="#landing" class="ui-btn-left">Back</a>
        <h1>Please Log In</h1>
        </header>
        <div data-role="content" class="content" data-theme="b">
            <form id="loginForm" action="services.php" method="post">
                <div data-role="fieldcontain">
                    <label for="schoolID">School ID</label>
                    <input type="text" name="schoolID" id="schoolID" value=""  />

                    <label for="userName">Username</label>
                    <input type="text" name="userName" id="userName" value=""  />

                    <label for="password">Password</label>
                    <input type="password" name="password" id="password" value=""  />

                    <h3 id="notification"></h3>
                    <button data-theme="b" id="submit" type="submit">Submit</button>
            <input type="hidden" name="action" value="loginForm" id="action">
                </div>
            </form>
        </div>
        <footer data-role="footer" data-position="fixed" data-theme="b">
            <h1>Footer</h1>
        </div>
    </section>
    <!-- END: Login Page -->

My jQuery Handler:

// Listen for the the submit button is clicked, serialize the data and send it off
    $('#submit').click(function(){
        var data = $("#loginForm :input").serializeArray();
    var url = $("#loginForm").attr('action');

    $.ajax({
        type: 'POST',
        url: url,
        cache: false,
        data: data,
        dataType: 'json',
            success: function(data){
            $.ajax({
                type: 'GET',
                url: "services.php",
                success: function(json){
                alert(json);
                $('#notification').append(json);
                }
            });
            }
        });
    });

And here is my PHP processing:

if (isset($_POST['action'])) {
    $schoolID = $_POST['schoolID'];
    $userName = $_POST['userName'];
    $password = $_POST['password'];

    $sql = "SELECT FirstName, LastName, FamilyID, StudentID, UserID ";
    $sql .= "FROM Users ";
    $sql .= "WHERE SchoolID = '$schoolID' ";
    $sql .= "AND Username = '$userName' ";
    $sql .= "AND Password = '$password'";
    $rs = mysql_query($sql);

    $rows = array();
    while($row = mysql_fetch_array($rs, MYSQL_ASSOC)) {
        $row_array['firstName'] = $row['FirstName'];
        $row_array['lastName'] = $row['LastName'];
        $row_array['familyID'] = $row['FamilyID'];
        $row_array['studentID'] = $row['StudentID'];
        $row_array['userID'] = $row['UserID'];
        array_push($rows, $row_array);
    }

    header("Content-type: application/json", true);
    echo json_encode(array('rows'=>$rows));
    exit;

    }else{

    echo "Login Failure";
    }

Answer 1

Look what mistake you are making here it is very simple

 $.ajax({
    type: 'POST',
    url: 'services.php',
    cache: false,
    data: $('#loginForm').serialize(),
    dataType: 'json',
        success: function(data){
            alert(data);
            $('#notification').append(data);
        }
    });
});

Use serialize function of jquery. And when you have a parameter in success function it is not that 'data' you have in this instruction

data : data,

It is returned from php end it is new data returned on success. To avoid confliction use some thing else like new_data

    success: function(new_data){
        alert(new_data);
        $('#notification').append(new_data);
    }

New data is in json format check it. Use console.log to see in firebug if you are using firefox.

    success: function(new_data){
        console.log(new_data);
        alert(new_data);
        $('#notification').append(new_data);
    }

Answer 2

Try changing:

var data = $("#loginForm :input").serializeArray();

to:

var data = $("#loginForm").serialize();

This should work as I'm pretty sure "data" in the .ajax request is supposed to be a serialized string, serializeArray returns an array of objects.

http://api.jquery.com/serialize/

http://api.jquery.com/serializeArray/

EDIT 1:

This looks like an event bubbling problem. Try the following:

Change your javascript to:

function processForm(formObj) {
    var $form = $(formObj);
    var data = $form.serializeArray();
    var url = $form).attr('action');

    $.ajax({
        type: 'POST',
        url: url,
        cache: false,
        data: data,
        dataType: 'json',
        success: function(data){
            $('#notification').append(data);
        }
    });

    return false;
}

And your form to:

<form id="loginForm" action="services.php" method="post" onsubmit="return processForm(this)">
    <div data-role="fieldcontain">
        <label for="schoolID">School ID</label>
        <input type="text" name="schoolID" id="schoolID" value=""  />

        <label for="userName">Username</label>
        <input type="text" name="userName" id="userName" value=""  />

        <label for="password">Password</label>
        <input type="password" name="password" id="password" value=""  />

        <h3 id="notification"></h3>
        <button data-theme="b" id="submit" type="submit">Submit</button>
        <input type="hidden" name="action" value="loginForm" id="action">
    </div>
</form>

Answer 3

Simple answer: user error. I didn't truly understand the concept of the getJSON() function. It is meant to send the data to the PHP script and also handle the return data on success. I assumed that it would require one function to send info and one function to retrieve info.

Thanks for all your help!