如何将数据从模型传递到视图（或控制器）

Question

问题很简单： 如何将数据从模型传递到视图（或返回控制器）以显示“密码太短”等错误

这是控制器

class UsersController extends Controller {

    private $username;
    private $password;

    function register()
    {
        if($_POST)
        {
            $this->User->username = $_POST['username'];
            $this->User->password = $_POST['password'];
            $this->User->register();
        }
    }



}

模型

class User extends Model {

    public $username;
    public $password;

    function register()
    {
        $username = $this->username;
        $password = $this->password;

        if (!empty($username) && !empty($password))
        {
            // registration process
        }
        else
        {
            // "you must provide a username and password" or something like that
        }
    }

Answer 1

只需将模型register中的return "PASSWORD";函数放到控制器中，让控制器从模型中返回并将其返回到视图中。让视图解释“PASSWORD”的错误输出是什么。

示例：

控制器

class UsersController extends Controller {

    private $username;
    private $password;

    function register()
    {
        if($_POST)
        {
            $this->User->username = $_POST['username'];
            $this->User->password = $_POST['password'];
            return $this->User->register();
        }
    }
}

模型

class User extends Model {

    public $username;
    public $password;

    function register()
    {
        $username = $this->username;
        $password = $this->password;

        if (!empty($username) && !empty($password))
        {
            // ...
            return "SUCCESS";
        }
        else
        {
            return "PASSWORD";
        }
    }
}

视图

$responses = array("SUCCESS" => "Registered Successfully!", "PASSWORD" => "You must provide a username and password!");

$result = $this->UsersController->register();
echo $responses[$result];

Answer 2

只需让你的模型方法返回一个值，或抛出异常，就像任何常规方法一样。然后在控制器中处理它。视图不应直接从模型触摸数据，这是控制器的工作。

Answer 3

public function addAction()
{
    $form = $this->_getForm();
    $this->view->form = $form;
    $this->render('add', null, true);
}

public function editAction()
{
    $id = $this->getRequest()->getParam(0);
    $Model = DI::get('yourclass_Model');
    $form = $this->_getForm();
    $data = $Model->getData();
    $form->populate($data);
    $this->view->flashMessages = $this->_helper->FlashMessenger->getMessages();
    $this->view->form = $form;
    $this->render('add', null, true);
}

public function saveAction()
{
    $form = $this->_getForm();
    $Model = DI::get('yourclass_Model');
    try{
    $saved = $Model->saveForm($form, $_POST);
    } catch (Exception $e) {
       echo "<pre>";
       print_r($e);
       exit;
    }
    if($saved)
    {
        $this->_helper->FlashMessenger('Record Saved!');
        $this->_redirect("edit".$form->id->getValue(), array('exit'=>true));
    } 
    $this->view->errorMessage = 'There were some errors';
    $this->view->form = $form;
    $this->render('add', null, true);
}

Answer 4

创建一个实现Singleton模式和ArrayAccess接口的类。 或者使用依赖注入创建类似的东西。

最终的解决方案是创建一些验证架构。 （模型验证其自身，并且视图中提供了错误状态。）