在我目前的系统中,我需要使用用户名或电子邮件和密码登录。 任何人都知道如何实现这个目标吗?
我的表格:
<?php echo $this->Form->create('User', array('action' => 'login'));
echo $this->Form->input('username', array('class' => 'TextField js-user-mode'));
echo $this->Form->input('password', array('class' => 'TextField'));
?>
MY AppController:
public $components = array(
'Email'=>array(),
'Auth' => array(
'loginAction' => array(
'admin' => false,
'controller' => 'users',
'action' => 'login'
),
'authError' => 'Your session has ended due to inactivity. Please login to continue.',
'authenticate' => array(
'Form' => array(
'fields' => array('username' => array('username','email')),
),
'all' => array(
'userModel' => 'User',
'scope' => array('User.status' =>array('active'))
)
)
)
);
让我知道我还需要做什么.. ??
答案 0 :(得分:5)
不确定在发布旧问题的答案时礼仪是什么,但这就是我为此所做的。
在我的登录功能
$username = $this->data['User']['username'];
$password = $this->request->data['User']['password'];
$user = $this->User->findByUsername($username);
if (empty($user)) {
$user = $this->User->findByEmail($username);
if (empty($user)) {
$this->Session->setFlash(__('Incorrect Email/Username or Password'));
return;
}
$this->request->data['User']['username'] = $user['User']['username'];
}
答案 1 :(得分:4)
我们可以在您的Auth登录操作之前执行此操作:
$emailUsername = @$this->request->data['User']['email'];
if (!filter_var($emailUsername, FILTER_VALIDATE_EMAIL)) {
$emailFromUsername = $this->User->find('first', array('conditions' => array('User.username' => $emailUsername), 'recursive' => -1, 'fields' => array('email')));
//pr($emailFromUsername );
if (!empty($emailFromUsername)) {
$emailFromUsernameDB = $emailFromUsername['User']['email'];
} else {
$emailFromUsername = '';
}
$this->request->data['User']['email'] = $emailFromUsername;
}
答案 2 :(得分:2)
您可以使用MultipleColumn Auth适配器:
更新: 新版本@ https://github.com/dereuromark/cakephp-tools/blob/master/src/Auth/MultiColumnAuthenticate.php
答案 3 :(得分:2)
我从url找到了以下代码。我认为这是最简单的感觉。在登录操作中使用以下代码:
public function login() {
if($this->request->is('post')&&!empty($this->request->data)) {
App::Import('Utility', 'Validation');
if( isset($this->data['User']['username']) && Validation::email($this->request->data['User']['username'])) {
$this->request->data['User']['email'] = $this->request->data['User']['username'];
$this->Auth->authenticate['Form'] = array('fields' => array('username' => 'email'));
}
if($this->Auth->login()) {
/* login successful */
$this->redirect($this->Auth->redirect());
} else {
/* login unsuccessful */
}
}
}
并使用以下代码 login.ctp :
<?php
echo $this->form->create('User');
echo $this->form->input('username');
echo $this->form->input('password');
echo $this->form->end('Submit');
?>
答案 4 :(得分:2)
假设您的用户表格中有用户名和电子邮件这两个字段
在你的AppController.php
中public function beforeFilter() {
if ($this->request->is('post') && $this->action == 'login') {
$username = $this->request->data['User']['username'];
if (filter_var($username, FILTER_VALIDATE_EMAIL)) {
$this->Auth->authenticate['Form']['fields']['username'] = 'email';
$this->request->data['User']['email'] = $username;
unset($this->request->data['User']['username']);
}
}
}
此代码适用于CakePHP 2.x,未在3.x版本上测试过,您的用户表中应该有电子邮件字段。
答案 5 :(得分:0)
我发现此解决方案很有用。 我创建了两个扩展FormAuthenticate的类:
app / Controller / Component / Auth / ClassNameAuthenticate.php 和
<?php
App::uses('FormAuthenticate', 'Controller/Component/Auth');
class ClassNameAuthenticate extends FormAuthenticate {
}
应用/控制器/组件/认证/ ClassNameEmailAuthenticate.php 强>
<?php
App::uses('FormAuthenticate', 'Controller/Component/Auth');
class ClassNameEmailAuthenticate extends FormAuthenticate {
}
然后在我的Controller中将Auth Component添加到$ components
public $components = array(
'Session',
'Auth' => array(
'authenticate' => array(
'ClassName' =>array(
'userModel'=>'ClassName',
'fields' => array(
'username' => 'username',
),
'scope' => array('ClassName.active' => 1)
),
'ClassNameEmail' =>array(
'userModel'=>'ClassName',
'fields' => array(
'username' => 'email',
),
'scope' => array('ClassName.active' => 1)
)
)
),
);
登录视图:login.ctp
<div class="form">
<?php echo $this->Form->create('ClassName'); ?>
<fieldset>
<legend><?php echo __('Login'); ?></legend>
<?php
echo $this->Form->input('username');
echo $this->Form->input('password');
?>
</fieldset>
<?php echo $this->Form->end(array('label'=>__('Login'))); ?>
</div>
和login()操作:
public function login(){
if ($this->Auth->loggedIn()) {
return $this->redirect($this->Auth->redirect());
}
if ($this->request->is('post')) {
//Need to duplicate field email for ClassNameEmail Auth
$this->request->data['ClassName']['email'] = $this->request->data['ClassName']['username'];
if ($this->Auth->login()) {
return $this->redirect($this->Auth->redirect());
}
$this->Session->setFlash(__('Invalid username/email or password, try again'));
}
}
我希望有人会觉得这很有用。
答案 6 :(得分:0)
这是解决该问题的解决方案
public function login(){
if($this->request->is('post')){
$this->User->set($this->request->data);
if($this->User->validates()){
if(Validation::email($this->data['User']['username'])){
$this->Auth->authenticate['Form'] = array_merge($this->Auth->authenticate['Form'],array('fields'=>array('username'=>'email')));
$this->request->data['User']['email'] = $this->request->data['User']['username'];
unset($this->request->data['User']['username']);
}
if($this->Auth->login()){
$this->User->id = $this->Auth->user('id');
$this->User->saveField('last_login',time());
if($this->data['User']['remember']){
unset($this->request->data['User']['remember']);
$this->request->data['User']['password'] = Security::hash($this->request->data['User']['password'],'blowfish');
$this->Cookie->write('rememberMe',$this->request->data['User'],true,'2 days');
}
$this->redirect($this->Auth->loginRedirect);
}
$this->Session->setFlash('Invalid Username or Password entered, please try again.','default',array('class'=>'alert alert-warning'),'warning');
}
}
$this->set('title','Login Page');
}