只有一个实体的Symfony2表单实体OneToMany不起作用


Symfony2 Form Entity OneToMany with only one entity doesn't work

>我有两个表:会话和游戏(一个会话可以有多个游戏)

添加会话后,它会呈现我的游戏形式。但现在我有一个奇怪的行为:当我使用加载所有会话的实体类型表单时,游戏将正确保存在我的数据库中。但是当我只在表单中加载一个会话时,数据库中的session_id字段为 NULL。

游戏实体:

/**
 * @var int
 *
 * @ORM'Column(name="id", type="integer")
 * @ORM'Id
 * @ORM'GeneratedValue(strategy="AUTO")
 */
private $id;
/**
 * @ORM'ManyToOne(targetEntity="Session", inversedBy="game")
 * @ORM'JoinColumn(name="session_id")
 */
private $session;

会话实体:

/**
 * @var int
 *
 * @ORM'Column(name="id", type="integer")
 * @ORM'Id
 * @ORM'GeneratedValue(strategy="AUTO")
 */
private $id;
/**
 * @ORM'OneToMany(targetEntity="Game", mappedBy="session")
 */
private $game;

控制器:

/**
 * @param Request $request
 * @return 'Symfony'Component'HttpFoundation'Response
 * @Route("/session/new")
 */
public function addnewsessionAction(Request $request)
{
    $session = new Session();
    $form = $this->createForm(SessionType::class, $session);
    $form->handleRequest($request);
    if ($form->isSubmitted())
    {
        $em = $this->getDoctrine()->getManager();
        $em->persist($session);
        $em->flush();
        return $this->redirectToRoute('app_game_addnewsessiongame', array('sessionid' => $session->getId()));
    }
    return $this->render(':session:addsession.html.twig', array(
        'form' => $form->createView(),
    ));
}

/**
 * @param Request $request
 * @return 'Symfony'Component'HttpFoundation'Response
 * @Route("/session/{sessionid}/addnewgame")
 */
public function addnewsessiongameAction(Request $request, $sessionid)
{
    $game = new Game();
    $form = $this->createForm(GameType::class, $game, array('sessionid' => $sessionid));
    $form->handleRequest($request);
    if ($form->isSubmitted())
    {
        $em = $this->getDoctrine()->getManager();
        $em->persist($game);
        $em->flush();
        return $this->redirectToRoute('app_game_addnewsessiongame', array('sessionid' => $sessionid));
    }
    return $this->render(':game:addgame.html.twig', array(
        'form' => $form->createView(),
        'sessionid' => $sessionid,
    ));
}

游戏类型(手动选择会话时正确插入的游戏):

$builder
        ->add('session', 'entity', array(
            'class' => 'AppBundle'Entity'Session',
            'query_builder' => function(EntityRepository $er ) use ( $options ) {
                return $er->createQueryBuilder('w');
            },
        ))

游戏类型(当通过控制器只提供一个会话时):

        $builder
        ->add('session', 'entity', array(
            'class' => 'AppBundle'Entity'Session',
            'query_builder' => function(EntityRepository $er ) use ( $options ) {
                return $er->createQueryBuilder('w')
                    ->where('w.id = ?1')
                    ->setParameter(1, $options['sessionid']);
            },

在两个情况下,选择选项仅显示给定的实体,但以 NULL 保存在数据库中。即使我再次选择一个选项,NULL 也会保存在 db 中。

我想自动保存正确的会话,这样用户就不必选择会话。选择表单将显示:隐藏,附加"attr" =>数组("类" => "隐藏")

经过一夜的充分睡眠,我得到了解决方案。我获取了会话实体并将其通过表单扔到游戏实体中

控制器:

 /**
 * @param Request $request
 * @return 'Symfony'Component'HttpFoundation'Response
 * @Route("/session/{sessionid}/addnewgame")
 */
public function addnewsessiongameAction(Request $request, $sessionid)
{
    $game = new Game();
    $em = $this->getDoctrine()->getManager();
    $mysession = $em->getRepository('AppBundle:Session')->find($sessionid);
    $game->setSession($mysession);
    $form = $this->createForm(GameType::class, $game, array('sessionid' => $sessionid));
    $form->handleRequest($request);
    if ($form->isSubmitted())
    {
        $em = $this->getDoctrine()->getManager();
        $em->persist($game);
        $em->flush();
        return $this->redirectToRoute('app_game_addnewsessiongame', array('sessionid' => $sessionid));
    }
    return $this->render(':game:addgame.html.twig', array(
        'form' => $form->createView(),
        'sessionid' => $sessionid,
    ));
}

游戏类型:

$builder->add('session','entity', array('class' => 'AppBundle'Entity'Session','attr' => array('class' => 'hidden')))