Php Zend登录表单身份验证AJAX错误捕获

Havin in LoginController.php ajax操作如下：

public function ajaxAction()
{
    $form = $this->getServiceLocator()->get('LoginForm');
    $form->setData($post);
    $post = $this->request->getPost();
    $response   = $this->getResponse();
    if (!$form->isValid()){
        // email is invalid; print the reasons
        $json= $form->getMessages();
        $response->setContent(\Zend\Json\Json::encode($json));
        return $response;
    }

    $this->getAuthService()->getAdapter()->setIdentity(
        $this->request->getPost('email'))->setCredential(
        $this->request->getPost('password'));
    $result = $this->getAuthService()->authenticate();

    switch ($result->getCode()) {
        case Result::FAILURE_IDENTITY_NOT_FOUND:
            $json = 'No such email found';      
            $response->setContent(\Zend\Json\Json::encode($json));
            return $response;
            break;

        case Result::FAILURE_CREDENTIAL_INVALID:
            $json =  'Invalid password';
            $response->setContent(\Zend\Json\Json::encode($json));
            return $response;
            break;
    }

    $dbTableAuthAdapter = $this->getServiceLocator()->get('AuthService')[1];
    if($result->isValid()) {
        $result =  $this->getAuthService()->getStorage();
        $result->write($dbTableAuthAdapter->getResultRowObject(array(
                                                        'email',
                                                        'name',
                                                    )));; // Writes email and name to the storage
        $result->write($dbTableAuthAdapter->getResultRowObject(
            null,
            'password'
        ));

        $user_session = new Container('user');
        $user_session->user_name = $this->getAuthService()->getStorage()->read()->name;
        $user_session->user_email = $this->getAuthService()->getStorage()->read()->email; // gets email from storage
        $user_session->login_session = true;
    }
}

在script.js中，为loginForm编写以下脚本

var urlformLogin = "login/ajax";
$("#Login").submit( function() {
    return false;    
});
$("#btnLogin").click( function() {
    $.ajax({
        url: urlformLogin,
        type: 'POST',
        dataType: 'json',
        async: true,
        data: $("#Login").serialize(),
        success: function (data) {              
                var msgs = $.map(data, function (fieldObj, key) 
                {     return [$.map(fieldObj, function (msg, key) {         return msg;     })] 
                });
              $('#lCheck').html(msgs.join('<hr>'));
            console.log(data);
        },
        error: function () {
            location.href = "auth";
        }
    }); 
});

---

如果您遇到

内部服务器错误

，则会在操作中出错并引发错误

$post = $this->request->getPost();
$form->setData($post);

您切换了此行，并且从未初始化过

$post

。 如果您想实现json响应，我建议您使用zf2本身的

ViewJsonStrategy

。在module.config.php中进行以下更改

'view_manager' => array(
    /** OTHER SETTINGS **/

    /** ADD THIS **/
    'strategies' => array(
        'ViewJsonStrategy',
    ),
),

在您的操作中，返回一个

Zend\View\JsonModel

，您的输出是干净的json

$result = new JsonModel(array(
    'some_parameter' => 'some value',
    'success'=>true,
));
return $result;

---

以及如何使用jQuery文件将“some_参数”传递给AJAX？请看一下

$result = new JsonModel(array(
    'some_parameter' => 'some value',
    'success'=>true,
));
return $result;