Question

[设置]

Symfony 3.4
来自其他环境的POST数据

[问题]

我目前正在直接从外部POST数据持久化/创建新实体解析后，我刷新所有数据以创建我的新条目。

如果我的实体已存在于DB中，我应该如何在我的控制器中更改我的实体。

DeliveryBox实体

class DeliveryBox {
    /**
     * @var int
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;
    /**
     * @var string
     *
     * @ORM\Column(name="uuid", type="string", length=40, unique=true)
     */
    private $uuid;
    /**
     * @var string
     *
     * @ORM\Column(name="url", type="string", length=255, unique=true)
     */
    private $url;
}

POST数据

uuid=9a16dc07-902c-6115-6e9b-acb800968e25&url=https://sim9885.agni.lindenlab.com:12043/cap/4d6fe9c7-5fb7-77e8-8dbc-565dac4b0e49

控制器操作

/**
 * Creates a new deliveryBox entity.
 *
 * @Route("/registerbox/", name="register_box")
 * @Method({"POST"})
 *
 * @param Request $request
 * @return \Symfony\Component\HttpFoundation\RedirectResponse|\Symfony\Component\HttpFoundation\Response
 */
public function newAction(Request $request) {
    $deliveryBox=new Deliverybox();

    //Parse POST data and set DeliveryBox values
    $content=array();
    parse_str($request->getContent(), $content);
    $deliveryBox->setUuid($content['uuid']);
    $deliveryBox->setUrl($content['url']);

    $em=$this->getDoctrine()->getManager();
    $em->persist($deliveryBox);
    $em->flush();

    return $this->redirectToRoute('deliverybox_show', array('id'=>$deliveryBox->getId(),));
}

Answer 1

如果uuid是唯一的，它可能看起来像这样。

控制器操作

/**
 * Creates a new deliveryBox entity.
 *
 * @Route("/registerbox/", name="register_box")
 * @Method({"POST"})
 *
 * @param Request $request
 * @return \Symfony\Component\HttpFoundation\RedirectResponse|\Symfony\Component\HttpFoundation\Response
 */
public function newAction(Request $request) {
    // Parse POST data
    $content = array();
    parse_str($request->getContent(), $content);

    // Get entity manager instance
    $em = $this->getDoctrine()->getManager();

    // Find existing delivery box by UUID
    $deliveryBox = $em->getRepository(DeliveryBox::class)->findOneByUuid($content['uuid']);
    if (null === $deliveryBox) {
        // If delivery box with current UUID is not found, create one
        $deliveryBox = new DeliveryBox();
        $deliveryBox->setUuid($content['uuid']);
    }

    // Update URL
    $deliveryBox->setUrl($content['url']);

    $em->persist($deliveryBox);
    $em->flush();

    return $this->redirectToRoute('deliverybox_show', array('id'=>$deliveryBox->getId(),));
}

Answer 2

 /**
  *  This is sample code, which receives id of entity
  *  to edit and create the edit form and do the update,
  *  if form is submitted and valid. See this for more info 
  *  https://symfony.com/doc/3.4/controller.html

  *  Edit an existing entity.
  *  
  * @Route("/registerbox-edit/{id}", name="register_box_edit")
  * @Method({"POST"})
  *
  * @param Request $request
  * @return \Symfony\Component\HttpFoundation\RedirectResponse|\Symfony\Component\HttpFoundation\Response
  */
   public function editAction(Request $request,  Deliverybox $deliverybox)
   {
     $editForm = $this->createForm('AppBundle\Form\DeliveryboxType', $deliverybox);
     $editForm->handleRequest($request);

     if ($editForm->isSubmitted() && $editForm->isValid()) {
        $this->getDoctrine()->getManager()->flush();

         return $this->redirectToRoute('deliverybox_show', array('id'=>$deliveryBox->getId(),));
      }        

   }

一切顺利

更新实体（如果已存在）

2 个答案: