我在使用新的Auth组件测试用户是否处于活动状态时遇到了麻烦。我有3种用户状态:0未激活(默认),1激活,2去激活。我尝试在登录函数中实现这个,这样我就可以返回他们是没有注册还是已经被禁止。
登录名:
public function login() {
if ($this->request->is('post')) {
if($this->Auth->login()) {
$results = $this->User->find('all', array(
'conditions' => array('User.email' => $this->Auth->user('email')),
'fields' => array('User.is_active')
));
if ($results['User']['is_active'] == 0) {
// User has not confirmed account
$this->Session->setFlash('Your account has not been activated. Please check your email.');
$this->Auth->logout();
$this->redirect(array('action'=>'login'));
}
// not working atm
else if ($results['User']['is_active'] == 2) {
// User has been deactivated
$this->Session->setFlash('Your account has been deactivated. Contact site admin if you believe this is in error.');
$this->Auth->logout();
$this->redirect(array('action'=>'login'));
}
else if ($results['User']['is_active'] == 1) {
// User is active
$this->redirect($this->Auth->redirect());
}
} else {
$this->Session->setFlash(__('Your email/password combination was incorrect'));
}
}
}
不知道我哪里出错了。拥有管理员权限的用户和激活的用户仍然会收到未激活帐户错误。
更新
决定删除User.is_active字段并在角色中处理它。我在AppController中处理它,它现在几乎可以工作了。在isAuthorized函数中,如果用户被禁止或未激活,它现在会抛出错误,但我也需要它来注销他们。
public function isAuthorized($user) {
// This isAuthorized determines what logged in users are able to see on ALL controllers. Use controller
// by controller isAuthorized to limit what they can view on each one. Basically, you do not want to allow
// actions on all controllers for users. Only admins can access every controller.
if (isset($user['role']) && $user['role'] === 'admin') {
return true; //Admin can access every action
}
elseif (isset($user['role']) && $user['role'] === 'unactivated') { // Account has not been activated
$this->Session->setFlash("You haven't activated your account yet. Please check your email.");
return false;
}
elseif (isset($user['role']) && $user['role'] === 'banned') { // Your account has been banned
$this->Session->setFlash("You're account has been banned. If you feel this was an error, please contact the site administrator.");
return false;
}
return false; // The rest don't
}
2条答案
按热度按时间j8ag8udp1#
如果用户登录,可以使用
$this->Auth->user()
访问用户模型信息。因此,您应该能够执行以下操作:您可以在
login()
之后使用debug($this->Auth->user());
来查看为什么用户一直显示为未激活。ojsjcaue2#
现在,在2021年,对于CakePHP 4.x,这是正确的方法: