php - 坚持使用 Yii 框架表单教程

Question

我一直在一遍又一遍地阅读这篇 Yii 框架的教程
Yii Framework - Working with Form

我已经使用以下代码创建了我的模型

class LoginForm extends CFormModel{
    public $username;
    public $password;
    public $rememberMe = false;

    private $_identity;

    public function rules(){
        return array(
            /* array(<field>,<field>,<function to invoke>)
            * functions required and boolean are built-in validators of the yii framework.
            * you can invoke your own function by defining your own function
            */
            array('username','password','required'),
            array('rememberMe','boolean'),
            array('password','authenticate'),
        );
    }

    public function authenticate(){
        $this->_identity = new UserIdentity($this->username,$this->password);
        if(!$this->_identity->authenticate()){
            $this->addError("password","Incorrect Username or Password");
        }
    }

    public function attributeLabels(){
        return array(
            'username'=>"Username",
            'password'=>"Password",
            'rememberMe'=>"Remember Me",
        );
    }
}

以及我在 Controller 中使用此代码的 Action 函数

 public function actionLogin(){
        //calls the Login Model that will be used in this action
        $model = new LoginForm;
        if(isset($_POST["LoginForm"])){
            //collects user input
            $model->attributes = $_POST["LoginForm"];
            //validates user input using the model rules and redirects back to
            //previous page when user input is invalid
            if($model->validate()){
               $this->redirect(Yii::app()->user->returnUrl);   
            }
            //redisplay the login form
            $this->render('login',array('loginModel'=>$model));
        }
    }

最后在我看来

<div class="form">
<?php
    $formlogin = $this->beginWidget('CActiveForm');     
    echo $formlogin->errorSummary($model);
?>
       <div class="row">
       <?php
            $formlogin->label($model,'username');
            $formlogin->textField($model,'username');
       ?>
       </div>
       <div class="row">
       <?php
            $formlogin->label($model,'password');
            $formlogin->passwordField($model,'password');
       ?>
       </div>
       <div class="row rememberMe">
       <?php
            $formlogin->checkBox($model,'rememberMe');
            $formlogin->label($model,'rememberMe');
       ?>
       </div>
       <div class="row submit">
       <?php
            echo CHtml::submitButton('Login');
       ?>
       </div>
<?php 
$this->endWidget();
?>
</div>

在我看来，我总是出现这个错误
D:\xampp\htdocs\wiltalk\protected\views\sandbox\index.php(11)

Undefined variable: model

我想念什么吗？请让我知道...我知道这有点简单，但我是第一次使用这种基于组件的 MVC 框架....谢谢

Answer 1

public function actionLogin(){
        //calls the Login Model that will be used in this action
        $model = new LoginForm;
        if(isset($_POST["LoginForm"])){
            //collects user input
            $model->attributes = $_POST["LoginForm"];
            //validates user input using the model rules and redirects back to
            //previous page when user input is invalid
            if($model->validate()){
               $this->redirect(Yii::app()->user->returnUrl);   
            }            
        }
        //redisplay the login form
        $this->render('login',array('model'=>$model));
    }

您的代码不正确。
对您的代码进行这些更改。

您正在使用模型变量并传递 loginModel 变量。

您正在 POST 上呈现 View 文件。