cakephp - 如何仅使用一个字段+ cakephp 2.x 进行登录

Question

我试图通过仅获取/授权一个输入 *user_number* （不是用户名 - 密码）来登录。

我用以下方式制作了我当前的登录页面： Cakephp2.x simple login

任何帮助请！

Answer 1

创建自定义身份验证对象

创建一个自定义身份验证对象，该对象仅对用户编号的使用进行身份验证；

创建自定义身份验证对象

app/Controller/Component/Auth/UserNumberAuthenticate.php

App::uses('BaseAuthenticate', 'Controller/Component/Auth');

class UserNumberAuthenticate extends BaseAuthenticate {
    public function authenticate(CakeRequest $request, CakeResponse $response) {
        $userModel = $this->settings['userModel'];
        list($plugin, $model) = pluginSplit($userModel);
        
        $fields = $this->settings['fields'];
        if (
            empty($request->data[$model])
            || empty($request->data[$model][$fields['username']])
        ) {
            return false;
        }

        return $this->_findUser($request->data[$model][$fields['username']]);
    }

    /**
     * Find a user record via his user-number/identifier
     *
     * @param string $usernumber The user-number/identifier.
     
     * @return Mixed Either false on failure, or an array of user data.
     */
    protected function _findUser($usernumber) {
        $userModel = $this->settings['userModel'];
        list($plugin, $model) = pluginSplit($userModel);
        $fields = $this->settings['fields'];
        
        $conditions = array(
            $model . '.' . $fields['username'] => $usernumber,
        );
        if (!empty($this->settings['scope'])) {
            $conditions = array_merge($conditions, $this->settings['scope']);
        }
        $result = ClassRegistry::init($userModel)->find('first', array(
            'conditions' => $conditions,
            'recursive' => $this->settings['recursive'],
            'contain' => $this->settings['contain'],
        ));
        if (empty($result) || empty($result[$model])) {
            return false;
        }
        $user = $result[$model];
        unset($result[$model]);
        return array_merge($user, $result);
    }
}

然后指定您要使用您的自定义身份验证对象

在您的 AppController 内部：

public $components = array(
    'Auth' => array(
        'authenticate' => array(
            'UserNumber' => array(
                'userModel' => 'User',
                'fields'    => array('username' => 'user_number')
            )
        )
    )
);

Answer 2

把事情简单化

如果您只有一种识别用户的方法，那么识别用户的最简单（因此也是推荐的）方法是定义您自己的登录函数。例如：

public function login() {
    if ($this->request->is('post')) {
        $number = $this->request->data['User']['user_number'];
        $user = $this->User->findByUserNumber($number);
        if ($user && $this->Auth->login($user)) {
            return $this->redirect($this->Auth->redirectUrl());
        } else {
            $this->Session->setFlash(__('User %d doesn\'t exist', $number), 'default', array(), 'auth');
        }
    }
}

请注意，这与使用 Cake 2.x登录用户的标准方式差别很小