蛋糕Php身份验证电子邮件或用户名登录

Question

我目前已经设置了下面的Auth组件，它工作得很好，但是我的客户现在希望能够为用户提供使用用户名或电子邮件地址登录的选项。

我当前的设置：

AuthComponent::$sessionKey = 'Auth.Customer';

        $this->Auth->allow();

        $this->Auth->authenticate = array(
            'Form' => array(
                'userModel' => 'Customer',
                'fields' => array(
                    'username' => 'email',
                    'password' => 'password',
                ),
                'scope' => array(
                    'Customer.guest' => 0,
                ),
                'passwordHasher' => 'Blowfish',
            )
        );

        $this->Auth->loginAction = array(
            'controller' => 'customers',
            'action' => 'login',
            'admin' => false,
            'plugin' => false,
        );
        $this->Auth->loginRedirect = array(
            'controller' => 'customers',
            'action' => 'my_account',
            'admin' => false,
            'plugin' => false,
        );

        if($this->Auth->user('id') == false){
            $this->Auth->authError = 'Please login to continue';
        }

        $this->Auth->authorize = array('Controller');

我如何才能更改此项，以便日志需要“电子邮件”或新字段“用户名”。

我会假设一个OR语句，但是下面的不工作：

$this->Auth->authenticate = array(
            'Form' => array(
                'userModel' => 'Customer',
                'fields' => array(

                    'password' => 'password',
                    'OR' => array(
                                'username' => 'email',
                                'username' => 'username'
                    )
                ),
                'scope' => array(
                    'Customer.guest' => 0,
                ),
                'passwordHasher' => 'Blowfish',
            )
        );

Answer 1

我已尝试更改配置，但无法执行此操作。我已自定义我的登录标识代码，如下所示：

 $findByEmail = $this->Users->findByEmail($this->request->data['username'])->first();
if (isset($findByEmail->username) && $findByEmail->username) {
  $this->request->data['username'] = $findByEmail->username;
 }
  $user = $this->Auth->identify(); 

无论用户名或电子邮件如何，这种方法都有效