Can'；t使用CakePHP'登录；s身份验证登录

我知道这是一个吃多了的问题，但我无法在所有答案中找到答案。也许你能帮助我。我正在尝试登录一个用户，并且我得到了“用户名/密码无效”错误和正确的数据。这是我第一次使用cakePHP

遵守守则

型号：

App::uses('AppModel','Model');

Class User extends AppModel {

    public $useTable = 'Users';

    public $hasMany = array(
        'Costumer' => array(
            'className' => 'Costumer',
            'foreignKey' => 'users_id',
            'order' => 'Costumer.name ASC'
        )
    );

    //Suppressed the validation code, don't think it's important here

    public function beforeSave($options = array()){
        if (!empty($this->data['User']['pwd'])) {
            $this->data['User']['passwd'] = Security::hash($this->data['User']['pwd']);
        }
    }
}

控制器：

App::uses('AppController', 'Controller');

class UsersController extends AppController{

public $helpers = array('Html', 'Form');
public $name = 'Users';
public $components = array('Auth','Session');

public function beforeFilter(){
    parent::beforeFilter();
    $this->Auth->allow('add');
}

public function login(){
    //Tests
    $userEmail = $this->User->findByEmail($this->request->data['User']['email']);
    $userPass = $this->User->findByPasswd(Security::hash($this->request->data['User']['passwd']));
    die(var_dump($userEmail, $userPass));

    if ($this->request->is('post')) {
        $this->request->data['User']['passwd'] = Security::hash($this->request->data['User']['passwd']);
        if ($this->Auth->login()){
            return $this->redirect($this->Auth->redirectUrl());
        } else {
            $this->Session->setFlash(__('E-mail e/ou usuário incorretos, tente novamente.'));
        }
    }
}

视图：

}

疯狂的是，两条用户控制器的线路

$userEmail = $this->User->findByEmail($this->request->data['User']['email']);

及

返回我试图登录的用户，这样就不会出现数据错误

伙计们！我错过了什么

谢谢

编辑

因为我还没有找到任何方法以“优雅”的方式完成这项工作，所以我编写了一个虚拟解决方案。它根据数据库手动检查请求->数据值，并手动登录用户。这是一个暂时的解决办法，我稍后再谈

public function login(){
    if ($this->request->is('post')) {
        $user = $this->User->findByEmail($this->request->data['User']['email']);            
        if (!empty($user) && ($user['User']['passwd'] == Security::hash($this->request->data['User']['passwd']))){
            $this->Auth->login($this->request->data);
            return $this->redirect($this->Auth->redirectUrl());
        } else {
            $this->Session->setFlash(__('E-mail e/ou usuário incorretos, tente novamente.'));
        }
    }
}

---

我不熟悉Auth组件的passwordHasher属性，也不熟悉Blowfish

您应该使用Cake的内置密码哈希器

用户模型

public function beforeSave() {
    if (isset($this->data['User']['passwd'])) {
        $this->data['User']['passwd'] = AuthComponent::password($this->data['User']['passwd']);
    }
    return true;
}

AppController

public $components = array(
    'Session',
    'Auth' => array(
        'loginRedirect' => array('controller' => 'pages', 'action' => 'display', 'homemk'),
        'logoutRedirect' => array('controller' => 'pages', 'action' => 'display', 'homemk'),
        'authenticate' => array(
            'Form' => array(
                'fields' => array('username' => 'email', 'password' => 'passwd'),
            )
        ),
        'authError' => 'Para visualizar esta página, você precisa estar logado.'
    )
);

public function login() {
    if ($this->request->is('post')) {
        if ($this->Auth->login()){
            return $this->redirect($this->Auth->redirectUrl());
        } else {
            $this->Session->setFlash(__('E-mail e/ou usuário incorretos, tente novamente.'));
        } // end if cannot log in
    } // end if no form submitted
} // end login

用户控制器

public $components = array(
    'Session',
    'Auth' => array(
        'loginRedirect' => array('controller' => 'pages', 'action' => 'display', 'homemk'),
        'logoutRedirect' => array('controller' => 'pages', 'action' => 'display', 'homemk'),
        'authenticate' => array(
            'Form' => array(
                'fields' => array('username' => 'email', 'password' => 'passwd'),
            )
        ),
        'authError' => 'Para visualizar esta página, você precisa estar logado.'
    )
);

public function login() {
    if ($this->request->is('post')) {
        if ($this->Auth->login()){
            return $this->redirect($this->Auth->redirectUrl());
        } else {
            $this->Session->setFlash(__('E-mail e/ou usuário incorretos, tente novamente.'));
        } // end if cannot log in
    } // end if no form submitted
} // end login

---

这是一个很好的提示，bowlerae，但仍然显示出相同的错误。真倒霉这很奇怪，因为我遵循了2.x食谱和示例。快速转到您的数据库，将您的用户密码之一更改为“23235BBC17A56627A5BECF7A74CBFEA0B49FA5A”。这是单词“密码”的蛋糕密码哈希。然后尝试使用他们的电子邮件登录该用户并键入“密码”。它可能不起作用，因为密码是使用旧哈希保存在您的表中的。我认为这不是问题所在，因为我从users表中删除了所有用户，并使用AuthComponent:：password方法创建了一个新的用户。但我会试试的。不幸的是没有。正如您所建议的，我尝试直接在数据库中更改密码。同样疯狂的是，它不会让用户登录，但是测试查询“findByEmail”和“findByPasswd”返回的结果是正确的。我想知道这是否与使用电子邮件登录用户有关。我只使用过“用户名”，但我读过关于使用“电子邮件”的文档，它看起来很简单，而且你的代码看起来很正确。

public function login() {
    if ($this->request->is('post')) {
        if ($this->Auth->login()){
            return $this->redirect($this->Auth->redirectUrl());
        } else {
            $this->Session->setFlash(__('E-mail e/ou usuário incorretos, tente novamente.'));
        } // end if cannot log in
    } // end if no form submitted
} // end login