如何 - Symfony 2執行CRUD而不將ID傳遞給url

Question

經過很長時間的這些板子上的行事之後，我終於決定做我的第一篇文章。我最近開始玩Symfony（2.4，請不要吼我:)）。使用原則的終端命令，我生成了CRUD事件。現在，這是偉大的，除了你必須在URL中傳遞ID，例如：www.mydomain.com/account/16/。這將使用來自mysql中id爲16的行的數據預填充表單。我的問題是，如何操作預先創建的CRUD（僅對更新感興趣），以便我不必將id傳遞給url，而是根據登錄用戶的id呈現表單與他們的賬戶有關聯？

這裏是我的代碼：

class AccountController extends Controller 
{ 
/** 
* Displays a form to edit an existing Event entity. 
* @Route("/account/{id}", name="post_login_account_edit") 
* @PreAuthorize("isFullyAuthenticated()") 
*/ 
public function editAction($id) 
{ 
    $em = $this->getDoctrine()->getManager(); 

    $entity = $em->getRepository('UserBundle:User')->find($id); 

    if (!$entity) { 
     throw $this->createNotFoundException('Unable to find Event entity.'); 
    } 

    $editForm = $this->createEditForm($entity); 

    return $this->render('UserBundle:Account:account.html.twig', array(
     'entity'  => $entity, 
     'edit_form' => $editForm->createView() 
    )); 
} 

/** 
* Creates a form to edit a Event entity. 
* 
* @param User $entity The entity 
* 
* @return \Symfony\Component\Form\Form The form 
*/ 
private function createEditForm(User $entity) 
{ 
    $form = $this->createForm(new UserType(), $entity, array(
     'action' => $this->generateUrl('post_login_account_update', array('id' => $entity->getId())), 
     'method' => 'PUT', 
    )); 

    $form->add('submit', 'submit', array('label' => 'Update')); 

    return $form; 
} 
/** 
* Edits an existing User entity. 
* @Route("/account/{id}/update", name="post_login_account_update") 
* @PreAuthorize("isFullyAuthenticated()") 
*/ 
public function updateAction(Request $request, $id) 
{ 
    $em = $this->getDoctrine()->getManager(); 

    $entity = $em->getRepository('UserBundle:User')->find($id); 

    if (!$entity) { 
     throw $this->createNotFoundException('Unable to find Event entity.'); 
    } 

    $editForm = $this->createEditForm($entity); 
    $editForm->handleRequest($request); 

    if ($editForm->isValid()) { 
     $em->flush(); 

     return $this->redirect($this->generateUrl('post_login_account')); 
    } 

    return $this->render('UserBundle:Account:account.html.twig', array(
     'entity'  => $entity, 
     'edit_form' => $editForm->createView() 
    )); 
} 

}

Answer 1

簡單地得到登陸用戶控制器：

$entity = $this->get('security.context')->getToken()->getUser();