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通過ajax提交我的表單後，即使使用正確的登錄信息，郵件也會始終以失敗告終！我已經編寫了非OOP風格的代碼，它完美的工作，但是當我使用這種風格的代碼時，它會掛起。活動網站是http://andyholmes.me/demo/summersproperty/OOP/login.php和用戶名是[email protected]和密碼是adminOOP和AJAX表單提交問題
的login.php -
<?PHP
session_start();
include('includes/class.login.php');
$login = new Login();
$token = $_SESSION['token'] = md5(uniqid(mt_rand(), true));
if ($_POST['ajax']) {
exit($login->getStatus());
}
?>
<style>
#message { display: none; cursor: pointer; }
.loader { display: none; }
</style>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#loginForm").submit(function(e) {
$(this).fadeOut(300);
$('.loader').delay(300).fadeIn(100);
$.post("<?=$_SERVER['PHP_SELF'];?>", { username: $('#username').val(), password: $('#password').val(), ajax: true }).done(function(data) {
if (data.logged_in == true) {
// Redirect with javascript
$('.loader').delay(2000).fadeOut(100);
$('#message').html('<p>Success! We\'ll redirect you in a minute...</p>').delay(2200).fadeIn(200);
} else {
$('.loader').delay(2000).fadeOut(100);
$('#message').html('<p>Failed... Click to try again!').delay(2200).fadeIn(200);
$('#message').on('click', function(){
$(this).fadeOut(200);
$('#loginForm').delay(350).fadeIn(200);
});
}
});
return false;
});
});
</script>
<form id="loginForm" method="POST" action="">
<table>
<tr><td>Username:</td><td><input type="text" name="username" id="username"/></td></tr>
<tr><td>Password:</td><td><input type="password" name="password" id="password"/></td></tr>
</table>
<input type="hidden" name="token" value="<?=$token;?>"/>
<input type="submit" name="login" value="Log In"/>
</form>
<div class="loader">
<img src="loader.gif"/>
</div>
<div id="message"></div>
，登錄類 -
<?PHP
class Login
{
private $_id;
private $_username;
private $_password;
private $_passmd5;
private $_errors;
private $_access;
private $_login;
private $_token;
public function __construct()
{
$this->_errors = array();
$this->_login = isset($_POST['login']) ? 1 : 0;
$this->_access = 0;
$this->_token = $_POST['token'];
$this->_id = 0;
$this->_username = ($this->_login) ? $this->filter($_POST['username']) : $_SESSION['username'];
$this->_password = ($this->_login) ? $this->filter($_POST['password']) : '';
$this->_passmd5 = ($this->_login) ? md5($this->_password) : $_SESSION['password'];
}
public function isLoggedIn()
{
($this->_login) ? $this->verifyPost() : $this->verifySession();
return $this->_access;
}
public function filter($var)
{
return preg_replace('/[^a-zA-Z0-9]/','',$var);
}
public function verifyPost()
{
try
{
if(!$this->isTokenValid())
throw new Exception('Invalid form submission');
if(!$this->isDataValid())
throw new Exception('Invalid form data entered');
if(!$this->verifyDatabase())
throw new Exception('Invalid username/password combination');
$this->_access = 1;
$this->registerSession();
}
catch(Exception $e)
{
$this->_errors[] = $e->getMessage();
}
}
public function verifySession()
{
if($this->sessionExist() && $this->verifyDatabase())
$this->_access = 1;
}
public function verifyDatabase()
{
include('dbConfig.php');
$data = mysql_query("SELECT user_id FROM users WHERE user_username = '{$this->_username}' AND user_password = '{$this->_passmd5}'");
if(mysql_num_rows($data))
{
list($this->_id) = @array_values(mysql_fetch_assoc($data));
return true;
}
else
{
return false;
}
}
public function isDataValid()
{
return (preg_match('/^[a-zA-Z0-9]/', $this->_username) && preg_match('/^[a-zA-Z0-9]/', $this->_password)) ? 1 : 0;
}
public function isTokenValid()
{
return (!isset($_SESSION['token']) || $this->_token != $_SESSION['token']) ? 0 : 1;
}
public function registerSession()
{
$_SESSION['id'] = $this->_id;
$_SESSION['username'] = $this->_username;
$_SESSION['password'] = $this->_passmd5;
}
public function sessionExist()
{
return (isset($_SESSION['username']) && isset($_SESSION['password'])) ? 1 : 0;
}
public function showErrors()
{
echo "<h3>Errors</h3>";
foreach($this->_errors as $key=>$value)
echo $value."<br>";
}
public function getStatus()
{
return json_encode(array('logged_in' => $this->isLoggedIn(), 'errors' => $this->showErrors()));
}
}
?>
順便說一句，我知道我需要使用PDO等，但我只想讓腳本達到一個點，它在我更改數據庫連接數據之前很好地工作。我知道我很接近，但它非常令人沮喪！
如果你能幫助我，我將不勝感激！
編輯備註：此代碼已更新爲想出了使用建議之後，從user1781670
不是所有使用類的東西都是oop :-) – PeeHaa 2013-04-24 20:18:34
但是這是OOP：S – 2013-04-24 20:21:48
對不起，但是:-) – PeeHaa 2013-04-24 20:24:22