Question

我在symfony2中有一个控制器，如下所示，如果用户表单有效，它将重定向到其他链接，但如果有错误，它将保留在同一页面并显示错误。只是在禁用客户端验证并且只有服务器端验证检查错误时的一般情况。

/**
 * Creates a new User entity.
 *
 * @Route("/create", name="admin_user_create")
 * @Method("post")
 * @Template("UserBundle:User:new.html.twig")
 */
public function createAction()
{
    $entity  = new User();
    $request = $this->getRequest();
    $form    = $this->createForm(new UserType() , $entity);

    $form->bindRequest($request);
    if($form->isValid())
    {
        // DO SOMETHING ... 
        return $this->redirect($this->generateUrl('some_link' , array( 'user_id' => $entity->getId() )));
    }

    // If the form is NOT valid, it will render the template and shows the errors.   
    return array(
        'entity' => $entity ,
        'form'   => $form->createView()
    );
}

场景如下所示：

用户输入表单
用户提交表单
控制器检查表单是否有效
由于它无效，它将呈现模板，在本例中为
```
@Template("UserBundle:User:new.html.twig")
```
浏览器中的路线为/create
如果浏览器链接上的用户click而不是post则会收到错误

我该如何解决这个问题？我必须再次重定向吗？由于方法是post，是否可以重定向？

Answer 1

不要指定@Method（“POST”）并在方法中执行此操作：

if ($request->getMethod() == 'POST')
{
    $form->bindRequest($request);
    if($form->isValid())
    {
        // DO SOMETHING ... 
        return $this->redirect($this->generateUrl('some_link' , array( 'user_id' => $entity->getId() )));
    }
}

Answer 2

您可以接受GET或POST并执行以下操作链接：

/**
 * Creates a new User entity.
 *
 * @Route("/create", name="admin_user_create")
 * @Method("GET|POST")
 * @Template("UserBundle:User:new.html.twig")
 */
public function createAction()
{
    $entity  = new User();
    $request = $this->getRequest();
    $form    = $this->createForm(new UserType() , $entity);

    // Is this POST? Bind the form...
    if('POST' == $request->getMethod()) $form->bindRequest($request);

    // GET or from not valid? Return the view...
    if('GET' == $request->getMethod() || !$form->isValid()) :
        return array(
            'entity' => $entity ,
            'form'   => $form->createView()
        );
    endif;

    // Success, then persist the entity and redirect the user
    return $this->redirect($this->generateUrl('some_link',
        array('user_id' => $entity->getId()))
    );
}

Answer 3

/**
 * Creates a new User entity.
 *
 * @Route("/create", name="admin_user_create")
 * @Method("GET|POST")
 * @Template("Use`enter code here`rBundle:User:new.html.twig")
 */
public function createAction()
{
    $entity  = new User();
    $request = $this->getRequest();
    $form    = $this->createForm(new UserType() , $entity);
// Is this POST? Bind the form...
if('POST' == $request->getMethod()) $form->bindRequest($request);

// GET or from not valid? Return the view...
if('GET' == $request->getMethod() || !$form->isValid()) :
    return array(
        'entity' => $entity ,
        'form'   => $form->createView()
    );
endif;

// Success, then persist the entity and redirect the user
return $this->redirect($this->generateUrl('some_link',
    array('user_id' => $entity->getId()))
);

}

如果提交的表单无效，带有@Method（“post”）的控制器应该如何处理请求？

3 个答案: