使用Jquery Mobile，PHP和MySQL的移动应用程序登录表单

我是移动开发的新手。我正在尝试开发一个登录表单，该表单将使用MySQL数据库检查用户名和密码。我已经构建了一些代码。我有两个问题。第一个是，提交表单后，在浏览器的 URL 栏中，我可以看到键入的用户名和密码，即使在我的 JQuery 脚本上使用 POST。第二个问题是页面不断返回登录.html而不是转到main.html。以下是我使用我在PHP，JQuery和MySQL的实际经验构建的内容。

网页表单

<div id="loginPage" data-role="page">
<div data-role="header">
<h1>App Authentication</h1>
</div>
<div data-role="content">   
    <div id="landmark-1" data-landmark-id="1">
    <form id="loginForm">
    <div data-role="fieldcontain" class="ui-hide-label">
        <label for="username">Username:</label>
        <input type="text" name="username" id="username" value="" placeholder="Username" />
    </div>
    <div data-role="fieldcontain" class="ui-hide-label">
        <label for="password">Password:</label>
        <input type="password" name="password" id="password" value="" placeholder="Password" />
    </div>
    <input type="submit" value="Login" id="submitButton">
    </form>
</div>  
</div>
<div data-role="footer">
    <h4>&copy; Silva's General Services</h4>
</div>

身份验证.php

<?php
$mysqli = new mysqli($mysql_hostname,$mysql_user, $mysql_password, $mysql_database);
header("access-control-allow-origin: *");
header("access-control-allow-methods: GET, POST, OPTIONS");
header("access-control-allow-credentials: true");
header("access-control-allow-headers: Content-Type, *");
header("Content-type: application/json");
// Parse the log in form if the user has filled it out and pressed "Log In"
if (isset($_POST["username"]) && isset($_POST["password"])) {
      CRYPT_BLOWFISH or die ('No Blowfish found.');
      //This string tells crypt to use blowfish for 5 rounds.
      $Blowfish_Pre = '$2a$05$';
      $Blowfish_End = '$';
      $username = mysql_real_escape_string($_POST['username']);
      $password = mysql_real_escape_string($_POST['password']);

      $sql = "SELECT id, password, salt, username FROM users WHERE username='$username' LIMIT 1";
      $result = $mysqli->query($sql) or die( $mysqli->error() );
      $row = mysqli_fetch_assoc($result);
      $hashed_pass = crypt($password, $Blowfish_Pre . $row['salt'] . $Blowfish_End);
      if ($hashed_pass == $row['password']) {
          $response_array['status'] = 'success'; 
      } else {
          $response_array['status'] = 'error'; 
      }
echo json_encode($response_array);
}
$mysqli->close();
?>

Jquery 脚本

<script>
$('#loginForm').submit(function(){
 e.preventDefault();
 jQuery.support.cors = true; 
 $.ajax({ 
     url: 'http://www.test.com/mobile/auth.php',
     crossDomain: true,
     type: 'post',
     data: $("#loginForm").serialize(), 
     success: function(data){
         if(data.status == 'success'){
             window.location.href = 'http://www.test.com/mobile/main.html'; 
         }else if(data.status == 'error'){
             alert("Authentication Invalid. Please try again!");
             return false;        
        }
     }
 }); 
 }); 
</script>

我真的很感激任何帮助。

谢谢。