【发布时间】:2014-08-02 07:08:36
【问题描述】:
我们已经集成了 yii 身份验证访问规则。在登录页面,提交后 表单,它会显示以下错误信息显示
致命错误:在第 13 行的 D:\wamp\www\onlinetest\protected\components\UserIdentity.php 中调用未定义的方法 LoginForm::model()
这是控制器代码
public function actionLogin()
{
$model=new LoginForm;
// if it is ajax validation request
if(isset($_POST['ajax']) && $_POST['ajax']==='login-form')
{
echo CActiveForm::validate($model);
Yii::app()->end();
}
// collect user input data
if(isset($_POST['LoginForm']))
{
$model->attributes=$_POST['LoginForm'];
// validate user input and redirect to the previous page if valid
if($model->validate() && $model->login())
$this->redirect(Yii::app()->user->returnUrl);
}
// display the login form
$this->render('login',array('model'=>$model));
}
这里是登录表单模型
class LoginForm extends CFormModel
{
public $username;
public $password;
public $rememberMe;
private $_identity;
public function tableName()
{
return 'tbl_login';
}
public function authenticate($attribute,$params)
{
if(!$this->hasErrors()) // we only want to authenticate when no input errors
{
$identity=new UserIdentity($this->username,$this->password);
$identity->authenticate();
switch($identity->errorCode)
{
case UserIdentity::ERROR_NONE:
Yii::app()->user->login($identity);
break;
case UserIdentity::ERROR_USERNAME_INVALID:
$this->addError('username','Username is incorrect.');
break;
default: // UserIdentity::ERROR_PASSWORD_INVALID
$this->addError('password','Password is incorrect.');
break;
}
}
}
public function login()
{
if($this->_identity===null)
{
$this->_identity=new UserIdentity($this->username,$this->password);
$this->_identity->authenticate();
}
if($this->_identity->errorCode===UserIdentity::ERROR_NONE)
{
$duration=$this->rememberMe ? 3600*24*30 : 0; // 30 days
Yii::app()->user->login($this->_identity,$duration);
return true;
}
else
return false;
}
}
这是组件中的 useridentity.php
class UserIdentity extends CUserIdentity
{
private $_id;
public function authenticate()
{
$record=LoginForm::model()->findByAttributes(array('VarUser_type'=>$this->username)); // here I use Email as user name which comes from database
if($record===null)
{
$this->_id='user Null';
$this->errorCode=self::ERROR_USERNAME_INVALID;
}
else if($record->E_PASSWORD!==$this->password) // here I compare db password with passwod field
{ $this->_id=$this->username;
$this->errorCode=self::ERROR_PASSWORD_INVALID;
}
else
{
$this->_id=$record['VarUser_type'];
$this->setState('title', $record['VarUser_type']);
$this->errorCode=self::ERROR_NONE;
}
return !$this->errorCode;
}
public function getId() // override Id
{
return $this->_id;
}
}
我该如何解决这个问题?如果你知道帮助我
【问题讨论】:
-
这里的 LoginForm 是 CFormModel,你将它用作 ActiveRecord 模型,所以它是错误的......
-
感谢 jaimin 的即时回复。我们如何改变模型格式来解决这个问题?
-
您从哪个表验证用户...?
-
你必须从那个表中创建一个 CActiveRecord 模型
-
你必须使用 Gii 来创建该模型...