我有一个扩展CRUDController
的抽象Controller
。为了使我的newAction
成功,我想使用showAction($slug)
方法重定向到redirect
:
return $this->redirect($this->generateUrl($route, $params));
但是
newAction
实际上是在子类UserController
中调用的,因此我无法在我的$route
中指定路由名称CRUDController
。class UserController extends CRUDController { }
abstract class CRUDController extends Controller
{
/** @Template */
public function showAction($slug) { }
/** @Template */
public function newAction(Request $request)
{
$model = $this->createModel();
$form = $this->createForm($this->createType(), $model);
if('GET' == $request->getMethod())
return array('form' => $form->createView());
$form->bindRequest($request);
if(!$form->isValid()) return array(
'errors' => $this->get('validator')->validate($model),
'form' => $form->createView()
);
$em = $this->getDoctrine()->getEntityManager();
$em->persist($model);
$em->flush();
// Success, redirect to showAction($slug)
}
}
路线示例:
users_show:
pattern: /users/show/{slug}
defaults: { _controller: AcmeSecurityBundle:User:show }
requirements:
_method: GET
users_new:
pattern: /users/new
defaults: { _controller: AcmeSecurityBundle:User:new }
requirements:
_method: GET
users_create:
pattern: /users/new
defaults: { _controller: AcmeSecurityBundle:User:new }
requirements:
_method: POST
最佳答案
您可以使用整个OO概念,并在抽象类中使用称为getRouteName()
的接口方法:
abstract public function getRoute();
然后,在您的具体类或子类UserController上,您只需重写并实现它:
public function getRoute()
{
return 'whatever:Route:YouWant';
}
因此,当在抽象类上调用实际的接口方法时,OO将处理所有类似魔术的事情:
public function newAction(Request $request)
{
...
return $this->redirect($this->generateUrl($this->getRouteName(), $params));
}
也许尝试一下,让我们知道工作是否正确。