CakePHP的验证：问题 [英] CakePhp Auth : Questions

问题描述

我已经做出CakePHP的鉴别，我想用验证的组成部分。我想，看它是否满足我的要求：

I've to make a cakePhp authentification, and I wish to use the "Auth" component. I'm trying to see if it fill my requirement:

• 我需要与他们的电子邮件或他们的客户ID来验证用户身份，（当然的加密码）。我找不到它是否可能有两个字段（或多个），其上的认证可以做到

• I need to authenticate users with their email OR their customerId, (with an addition password of course). I can't find if it is possible to have two fields(or more) on which the authentication can be done

我已经将我需要验证几个部分。但我需要的粒度型动物：

I've several parts on which I need to be authenticated. But I need differents granularity:

• 对于一些事情，这是整个控制器，它不应该访问（如果可能的异常（例如，所有的用户控制，除登​​录/注册动作）其他我真的需要它的整个控制器（如购物车控制器）


• 有时候，我需要的，只有一些动作不可用而不被记录


• 有时候，我需要的，只有视图的一部分不显示（不显示例如登录元素）

非常感谢你的帮助。

推荐答案

简短的回答是，是的，你可以做这些事情，但在我看来，ACL的可能是矫枉过正满足您的需要（但我也倾向于避免了ACL，如果有在所有任何开口这样做）。为了你的观点：

The short answer is that yes, you can do these things, but it seems to me that the ACL might be overkill for your needs (but I also tend to avoid the ACL if there's any opening at all to do so). To your points:

• 由于伊沃建议，你需要一个自定义的 UsersController ::登录（）方法，通过多个字段进行身份验证（如果你的权威性模型并不用户，然后使用适当的控制器）。如果验证组件的登录方法失败，则控制传递给您的自定义登录（）方法。下面是从一个项目，我一直工作在一个片段：

• As Ivo suggests, you'll need a custom UsersController::login() method to authenticate by multiple fields (If your auth model isn't User, then use the appropriate controller). If the Auth component's login method fails, it passes control to your custom login() method. Here's a snippet from a project I've been working on:

function login() {
  # Allow login by either username (legacy) or email.
  # If no authenticated user exists at this point then the Auth
  # component's login() method has failed and control has been passed
  # here for any further handling. Since the expected credentials
  # (email and password) have failed we're going to check for
  # username and password.
  $user = $this->Auth->user();
  if( empty( $user ) && !empty( $this->Auth->data['User']['email'] ) && !empty( $this->Auth->data['User']['password'] ) ) {
    $user = $this->User->find(
      'first',
      array(
        'recursive'  => -1,
        'conditions' => array(
          'User.username' => $this->Auth->data['User']['email'],
          'User.password' => $this->Auth->data['User']['password'],
        )
      )
    );

    if( empty( $user ) || !$this->Auth->login( $user ) ) {
      # Set the configured flash message b/c login failed after both checks.
      $this->Session->setFlash( $this->Auth->loginError, null, array(), 'auth' );
    }
  }

  $this->redirect( $this->Auth->redirect(), null, true );
}

有关行动访问，只需要使用 $这个 - ＆GT; Auth-＆GT;允许（）和 $这个 - ＆GT; Auth-＆GT;否认（）在每个相关负责人的 beforeFilter（）回调方法。例如，在 UsersController ，您可能需要做这样的事情：

For action access, just use the $this->Auth->allow() and $this->Auth->deny() methods in each relevant controller's beforeFilter() callback. For example, in the UsersController, you may want to do something like this:

public function beforeFilter() {
  parent::beforeFilter();

  $this->Auth->deny('*');
  $this->Auth->allow( 'login', 'logout' );
}

在景色，那里确定用户是否通过测试 Auth.User 值验证，以确定哪些显示来自匿名/认证/隐藏：

In views, just determine whether the user is authenticated by testing the Auth.User value to determine what to display/hide from anonymous/authenticated:

if( $this->Session->check( 'Auth.User' ) ) { ... }

如果密码更改，可以透明地通过调用重新验证用户 $这个 - ＆GT; Auth-＆GT;登录（$ USER_DATA）。我这样做，例如，当用户寄存器。我不想让他/她不得不然后去登录，所以我就自动登录。

If the password changes, you can re-authenticate the user transparently by calling $this->Auth->login( $user_data ). I do this, for example, when a user registers. I don't want him/her to have to then go login, so I just login automatically.

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