jquery .ajax always returns error - data being added to database

Question

I am trying to add users to a database using jquery ajax calls. The users get added just fine to the database, but the ajax always returns with error. I'm not sure how to retrieve the specific error either. Below is my code, form, php, and jquery.

Here is the jquery

$(document).ready(function() {

    //ajax call for all forms. 
    $('.button').click(function() {
        var form = $(this).closest('form');
        $.ajax({
            type: "POST",
            url: form.attr('data'),
            dataType: 'json',
            data: form.serialize(),
            success: function (response) {
                alert('something');
            },
            error: function() {
                alert('fail');
            }
        });
    });
  });

Here is the PHP

<?php
include 'class_lib.php';

if(isset($_POST['username'])) {
    $user = new Users;
    $user->cleanInput($_POST['username'], $_POST['password']);
    if($user->insertUser()) {
        echo json_encode('true');
    } else {
        echo json_encode('false');
    }
}

Here is the HTML

<div id='newUser' class='tool'>
    <h3>New User</h3>
    <form method='post' name='newUser' data='../php/newUser.php'>
        <span>Username</span><input type='text' name='username'><br>
        <span>Password</span><input type='password' name='password'>
        <input type='submit' name='submit' class='button' style='visibility: hidden'>
    </form>
    <span class='result'> </span>
</div>

Answer 1

@Musa, above you mentioned

My guess is its a parsing error, try removing dataType: 'json', and see if it works

You absolutely solved the problem I was having! My ajax post request was similar to above and it just kept returning to the 'error' section. Although I checked using firebug, the status was 200(ok) and there were no errors.

removing 'dataType:json' solved this issue for me. Thanks a lot!

Answer 2

Turns out I had to add async: false to the $.ajax function. It wasn't getting a response back from the php.