问题

Question

我的目的是使用Symfony表单允许用户选择多个兴趣并将其与用户相对保存。

我有三个实体来映射和存储此数据：

User
 - id
 - name

Interest
 - id
 - name

UserInterest
 - id
 - user_id (FK ManyToOne user.id)
 - interest_id (FK ManyToOne interest.id)

我正在努力寻找最动态的方式来处理Interest的{{1}}并将其保存到User实体。

InterestsController.php

UserInterest

UserInterestsType.php

public function interests(Request $request)
{
    $error = null;

    $userInterests = new UserInterest();
    $userInterests->setUser($this->getUser());

    $form = $this->createForm(UserAccountInterests::class, $userInterests);
    $form->handleRequest($request);

    if ($form->isSubmitted() && $form->isValid()) {
        try {
            $this->entityManager->persist($userInterests);
            $this->entityManager->flush();
    } catch (\Exception $e) {
            $error = $e->getMessage();
        }
    }

    $parameters = [
        'error' => $error,
        'user_interests_form' => $form->createView()
    ];

    return $this->render('user/interests.html.twig', $parameters);
}

问题

此刻我面临的问题是，在提交表单时，这会创建链接到class UserInterestsType extends AbstractType { /** * @param FormBuilderInterface $builder * @param array $options */ public function buildForm(FormBuilderInterface $builder, array $options) { $builder ->add('interest', EntityType::class, [ 'class' => Interest::class, 'choice_label' => 'name', 'expanded' => true, 'multiple' => true ]) ->add('update', SubmitType::class); } }的{{1}}实体的实例化，但会选择多个UserInterest作为数组，并且不是多个user。

我该如何以正确的Symfony方式执行此操作，以允许一种形式和简单的控制器逻辑，以便用户可以加载页面，从多个复选框类型中选择他们感兴趣的内容，单击“保存”，然后在重新加载表单时自动填充为先前选择的内容？

Answer 1

对于您要实现的目标，您的第一个解决方案有点太复杂了。如果只希望用户选择一些兴趣，则不需要UserInterest实体。

对于一个专用于该关系的新实体，我唯一想到的用途是该关系带有一些您要保留的额外属性，例如关系开始的日期，顺序或优先级或您需要的任何东西。如果您要存储的唯一数据是userId和interestId，那么一个新的实体将不会给您带来很多麻烦，而是会带来很多麻烦。您也不需要CollectionType（无论如何也不会在此表单上产生新的兴趣）。在用户和兴趣之间建立简单的ManyToMany关系

 User:
  type: entity
  manyToMany:
    interests:
      targetEntity: Interest
      inversedBy: users
      joinTable: 
        ## name will be the name of the table storing your entities 's relations
        ## Beware not to give it a name already defined for an entity table (like your UserInterest wich you don't need anymore)
        name: users_interests 
        joinColumns:
          user_id:
            referencedColumnName: id
        inverseJoinColumns:
          interest_id:
            referencedColumnName: id

Interest:
  type: entity
  manyToMany:
    users:
      targetEntity: User
      mappedBy: interests

然后，您可以使用以下形式的简单多个EntityType：data_class是User，而您的objecttype的类是Interest：

class UserInterestsType extends AbstractType
{
    /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
    $builder
        ->add('interests', EntityType::class, [
            'class' => Interest::class,
            'choice_label' => 'name',
            'expanded' => true,
            'multiple' => true,
            'query_builder' => function (EntityRepository $er) {
    return $er->createQueryBuilder('u')
        ->orderBy('u.name', 'ASC');
}, 
        ])
        ->add('update', SubmitType::class);
    }
 }

然后假设您要检索共享特定兴趣或特定用户兴趣的用户，您将在现有的用户和兴趣存储库中进行以下查询：

// UserRepository: All users sharing one (or more) interests
$em = $this->getEntityManager();
$repository = $em->getRepository('YourNamespace:User');
$query = $repository->createQueryBuilder('u')
    ->innerJoin('u.interests', 'i')
    ->where('i.id = :interest_id')
    ->setParameter('interest_id', 5) // just for the example
    ->getQuery()->getResult();

// InterestRepository: All interests of one user 
$em = $this->getEntityManager();
$repository = $em->getRepository('YourNamespaceYourBundle:Interest');
$query = $repository->createQueryBuilder('i')
    ->innerJoin('i.users', 'u')
    ->where('u.id = :user_id')
    ->setParameter('user_id', 1)
    ->getQuery()->getResult();

Symfony表单，用于将用户兴趣的选择存储到UserInterests

问题

1 个答案: