Authentication CakePHP 3:尝试获取非对象的属性

我正在使用CakePHP 3.2，在

login（）

操作中，只允许状态为

verified=1的用户登录

public function login()
    {
      if ($this->request->is('post') || $this->request->query('provider')) {
        $user = $this->Auth->identify();
        if ($user) {
          if ($user->verified != 1) {   // LINE 6
            $this->Flash->error(__('You have not yet verified your account. Please check your email to verify your account before login'), [
              'params' => [
                'userId' => $user->id,   // LINE 9
              ],
              ['escape' => false]
            ]);
            $this->redirect(['action' => 'login']);
          }
          $this->Auth->setUser($user);
          $this->_setCookie();
          $this->Flash->success('Login Success');
          return $this->redirect($this->Auth->redirectUrl());
        }
        $this->Flash->error(__('Invalid username or password, try again'));
      }
    }

但这也带来了错误

Notice (8): Trying to get property of non-object [APP/Controller/UsersController.php, line 6]
Notice (8): Trying to get property of non-object [APP/Controller/UsersController.php, line 9]
Warning (2): Cannot modify header information - headers already sent by (output started at /var/www/html/argoSystems/projects/project_01/site/vendor/cakephp/cakephp/src/Error/Debugger.php:746) [CORE/src/Network/Session.php, line 572]
Warning (2): session_regenerate_id() [function.session-regenerate-id]: Cannot regenerate session id - headers already sent [CORE/src/Network/Session.php, line 576]

注意：行号已更改以满足代码段的要求

解决了我的问题。这是我更新的
login（）
方法

public function login()
    {
      if ($this->request->is('post') || $this->request->query('provider')) {
        $user = $this->Auth->identify();
        if ($user) {
          $this->Auth->setUser($user);
          if ($this->Auth->user('verified') != 1) { 
            $this->Flash->error(__('You have not yet verified your account. Please check your email to verify your account before login'), [
              'params' => [
                'userId' => $this->Auth->user('id'),
              ],
              ['escape' => false]
            ]);
            return $this->redirect($this->Auth->logout());
          }
          $this->_setCookie();
          $this->Flash->success('Login Success');
          return $this->redirect($this->Auth->redirectUrl());
        }
        $this->Flash->error(__('Invalid username or password, try again'));
      }
    }

解决了我的问题。这是我更新的
login（）
方法

public function login()
    {
      if ($this->request->is('post') || $this->request->query('provider')) {
        $user = $this->Auth->identify();
        if ($user) {
          $this->Auth->setUser($user);
          if ($this->Auth->user('verified') != 1) { 
            $this->Flash->error(__('You have not yet verified your account. Please check your email to verify your account before login'), [
              'params' => [
                'userId' => $this->Auth->user('id'),
              ],
              ['escape' => false]
            ]);
            return $this->redirect($this->Auth->logout());
          }
          $this->_setCookie();
          $this->Flash->success('Login Success');
          return $this->redirect($this->Auth->redirectUrl());
        }
        $this->Flash->error(__('Invalid username or password, try again'));
      }
    }

你本可以轻松解决这个问题的！例如，将第6行更改为

如果（$user->verified！=1）{//第6行

到

$user[“verified”！=1）{//第6行
您本可以轻松解决此问题！例如，将第6行从

如果（$user->verified！=1）{//第6行

到

$user[“已验证”]！=1）{//第6行
您是否更改了错误通知的行以匹配您的代码段？如果您这样做，请告诉人们，否则他们可能会忽略您，因为找出行号可能指向的位置是一项相当繁琐的任务。尽管如此，您确实需要开始更多地学习PHP和调试由于这是一个非常非常琐碎的问题，所以对基础知识进行分析，并找出其原因非常简单，只需阅读
AuthComponent:：identify（）的API文档即可
，并调试
$user
到底是什么。谢谢@ndm debugging$user提供了正确的参数和值。您是否更改了错误通知的行以匹配您的代码段？如果您这样做，请告诉人们，否则他们可能会忽略您，因为找出行号可能指向的位置是一个pr尽管如此，您确实需要开始学习更多关于PHP和调试基础知识的知识，因为这是一个非常非常琐碎的问题，找出它发生的原因就像阅读
AuthComponent:：Identifity（）的API文档一样简单
，并调试
$user
到底是什么。感谢@ndm debug$user提供正确的参数和值。