登录在CakePHP的2.0版的电子邮件地址或用户名

Question

在CakePHP 2.0之前，你可以允许用户通过停止autoRedirect，然后在你的数据库的用户名数据的电子邮件列比较使用他们的电子邮件地址登录（显然蛋糕然后可以回退到用户名检查，如果不是电子邮件）。

在CakePHP 2.0这种情况已经改变，你登录使用$this->Auth->login()

我的问题是我怎么得到这个工作2.0手动？我有一些非常复杂的代码，可以处理各种各样的事情，例如处理ajax和回发请求，如果用户尝试登录太多次时锁定账户等，所以它非常长！

正如您将看到的那样，我将检查帐户是否实际存在，以便在通过身份验证过程前可以显示未找到的帐户消息（如果出现此情况，还可以使用此功能锁定该用户的帐户）尝试。

这里的主要问题是允许系统检查用户名和电子邮件地址以进行身份​​验证，如果您使用电子邮件地址进行验证，系统就会锁定用户，但它会始终因为身份验证无法处理它而失败。

希望有人能帮助，提供意见建议。由于

if ($this->request->is('post')) 
     {  
      $opts = array(
       'conditions'=>array(
        'OR'=>array(
         'User.username'=>$this->data['User']['username'], 
         'User.email'=>$this->data['User']['username'] 
        ) 
       ) 
      ); 

      $user = $this->User->find('first', $opts); 

      if(!empty($user)) 
      { 
       if($user['User']['status'] == 0) 
       { 
        if($this->request->is('ajax')) 
        { 
         $this->autoRender = false; 
         echo json_encode(array('authenticated'=>false,'error'=>__('Sorry your account is currently locked. Please reset your password.?'))); 
        } 
        else 
        { 
         $this->Session->setFlash(__('Sorry your account is currently locked. Please reset your password.'), 'default', array(), 'auth'); 
        } 
       } 
       else 
       { 
        if ($this->Auth->login()) 
        { 
         if ($this->request->is('ajax')) 
         { 
          $this->autoRender = false; 
          if(isset($this->params['url']['continue'])) 
          { 
           $pathtoredirect = $this->UrlEncode->base64url_decode($this->params['url']['continue']); 

           echo json_encode(array('authenticated'=>true,'redirect'=>$pathtoredirect,'base'=>false)); 
          } 
          else 
          { 
           $pathtoredirect = $this->Auth->redirect(); 

           echo json_encode(array('authenticated'=>true,'redirect'=>$pathtoredirect,'base'=>true)); 
          } 
         } 
         else 
         { 
          if(isset($this->params['url']['continue'])) 
          { 
           $pathtoredirect = $this->UrlEncode->base64url_decode($this->params['url']['continue']); 
          } 
          else 
          { 
           $pathtoredirect = $this->Auth->redirect(); 
          } 
          return $this->redirect($pathtoredirect); 
         }    
        } 
        else 
        {    
         if($this->Session->read('attempts')) 
         { 
          $attempts = $this->Session->read('attempts') + 1; 
         } 
         else 
         { 
          $attempts = 1; 
         } 

         $this->Session->write('attempts', $attempts); 

         if($attempts >= 5) 
         { 
          $this->User->id = $user['User']['id']; 
          $this->User->saveField('status', 0); 
          if ($this->request->is('ajax')) 
          { 
           $this->autoRender = false; 
           echo json_encode(array('authenticated'=>false,'error'=>__('Username or password is incorrect. For security reasons this account has now been locked and you must reset your password to unlock it.'))); 
          } 
          else 
          { 
           $this->Session->setFlash(__('Username or password is incorrect. For security reasons this account has now been locked and you must reset your password to unlock it.'), 'default', array(), 'auth'); 
          } 
         } 
         else 
         { 
          if ($this->request->is('ajax')) 
          { 
           $this->autoRender = false; 
           echo json_encode(array('authenticated'=>false,'error'=>__('Username or password is incorrect'))); 
          } 
          else 
          { 
           $this->Session->setFlash(__('Username or password is incorrect'), 'default', array(), 'auth'); 
          } 
         } 
        } 
       } 
      } 
      else 
      { 
       if ($this->request->is('ajax')) 
       { 
        $this->autoRender = false; 
        echo json_encode(array('authenticated'=>false,'error'=>__('Sorry that account does not exist.'))); 
       } 
       else 
       { 
        $this->Session->setFlash(__('Sorry that account does not exist.'), 'default', array(), 'auth'); 
       } 
      } 
     } 

Answer 1

我不知道，如果AuthComponent可配置为自动检查两个领域，但这里是一个另类：

/* 
* AppController 
*/ 
beforeFilter() 
{ 
    $this->Auth->authenticate = array('Form' => array('fields' => array('username' => 'email', 'password' => 'password'))); 
} 

/* 
* UsersController 
*/ 
function login() 
{ 
    if($this->request->is('post')) 
    { 
    $logged_in = false; 

    if($this->Auth->login()) 
    { 
     $logged_in = true; 
    } 
    else 
    { 
     $this->Auth->authenticate = array('Form' => array('fields' => array('username' => 'username', 'password' => 'password'))); 
     $this->Auth->constructAuthenticate(); 

     $this->request->data['User']['username'] = $this->request->data['User']['email']; 

     if($this->Auth->login()) 
     { 
      $logged_in = true; 
     } 
    } 

    if($logged_in) 
    { 
     /* 
     * Do what you want here 
     */ 
    } 
    else 
    { 
     /* 
     * Do what you want here 
     */ 
    } 
    } 
} 

然后，如果你希望能够执行的过程只有一个测试来检查这两个字段，您可以将此代码移动到Component中，而不是直接调用$this->Auth->login()方法。