Question

我想与Symfony3 / doctrine创建一个ManyToMany关系（实体是'Categorie'和'Service'）

所以我有两种形式来创造这种吸引力。

第一个表单（Categorie）正常工作但不是第二个表单（服务）：新服务与类别无关，我不明白：

Categorie.php

/**
 * Categorie
 *
 * @ORM\Table(name="categorie")
 * @ORM\Entity(repositoryClass="GestionBundle\Repository\CategorieRepository")
 */
class Categorie 
{
    /**
     * @var int
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;

    /**
     * @ORM\ManyToMany(targetEntity="Service", inversedBy="categories")
     */
    private $services;

    /**
     * Categorie constructor.
     */
    public function __construct()
    {
        $this->services = new ArrayCollection();
    }

    [...]
}

Service.php

/**
 * Service
 *
 * @ORM\Table(name="service")
 * @ORM\Entity(repositoryClass="GestionBundle\Repository\ServiceRepository")
 */
class Service
{
    /**
     * @var int
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;

    /**
     * @ORM\ManyToMany(targetEntity="Categorie", mappedBy="services")
     */
    private $categories;

    /**
     * @var string
     *
     * @ORM\Column(name="nom", type="string", length=255, unique=true)
     */
    private $nom;

    /**
     * Categorie constructor.
     */
    public function __construct()
    {
        $this->categories = new ArrayCollection();
    }

    [...]
}

CategorieType.php

class CategorieType extends AbstractType
{
    /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('nom')
            ->add('services', EntityType::class, array(
                'class'        => 'GestionBundle:Service',
                'choice_label' => 'nom',
                'multiple'     => true,
                'required'     => false))
            ->add('Ajouter', SubmitType::class)
        ;
    }

    [...]
}

ServiceType.php

class ServiceType extends AbstractType
{
    /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('nom')
            ->add('categories', EntityType::class, array(
                'class'        => 'GestionBundle:Categorie',
                'choice_label' => 'nom',
                'multiple'     => true,
                'required'     => false))
            ->add('Ajouter', SubmitType::class)
        ;
    }

    [...]
}

控制器：

/**
     * @Route("/Categories/Creation", name="route_gestion_eltcoord_categories_creation")
     * @Template()
     */
    public function CreationCategorieAction(Request $request)
    {
        $Categorie = new Categorie();

        $form = $this->createForm(CategorieType::class, $Categorie);
        $form->handleRequest($request);

        if ($form->isSubmitted() && $form->isValid()) {
            $em = $this->getDoctrine()->getManager();
            $em->persist($Categorie);
            $em->flush();
            return $this->redirectToRoute('route_gestion_eltcoord_categories');
        }

        return array('form' => $form->createView());
    }

/**
     * @Route("/Services/Creation", name="route_gestion_eltcoord_services_creation")
     * @Template()
     */
    public function CreationServiceAction(Request $request)
    {
        $Service = new Service();

        $form = $this->createForm(ServiceType::class, $Service);
        $form->handleRequest($request);

        if ($form->isSubmitted() && $form->isValid()) {
            $em = $this->getDoctrine()->getManager();
            $em->persist($Service);
            $em->flush();
            return $this->redirectToRoute('route_gestion_eltcoord_services');
        }

        return array('form' => $form->createView());
    }

谢谢。

Answer 1

据我所知，你不会从类别根表单中嵌套表单？

在服务表单构建器中尝试CollectionType::class而不是EntityType::class，配置选项（在较少的entry_type处映射实体类）。

配置here的相关文档。

并且here有来自Symfony食谱的例子。

Answer 2

在ManyToMany关系中，您必须创建一个类似于此post的连接表，因此以这种方式，doctrine可以创建第三个实体来描述关系。

希望这对你有所帮助。

Answer 3

我已经有了联合表。

以下是一个例子：

新服务＆＃39; S1（表格服务）：

     Service          Categorie        categorie_service
+------+-------+   +------+-------+   +--------+--------+
|  id  |  Name |   |  id  |  Name |   |  id_c  |  id_s  |
+------+-------+   +------+-------+   +--------+--------+
|   1  |   S1  | 
+------+-------+

新＆＃39;分类＆＃39; C1与S1（表格分类）相关：

     Service          Categorie        categorie_service
+------+-------+   +------+-------+   +--------+--------+
|  id  |  Name |   |  id  |  Name |   |  id_c  |  id_s  |
+------+-------+   +------+-------+   +--------+--------+
|   1  |   S1  |   |   1  |   C1  |   |    1   |    1   |
+------+-------+   +------+-------+   +--------+--------+

新服务＆＃39; S2与C1（表格服务）有关：

     Service          Categorie        categorie_service
+------+-------+   +------+-------+   +--------+--------+
|  id  |  Name |   |  id  |  Name |   |  id_c  |  id_s  |
+------+-------+   +------+-------+   +--------+--------+
|   1  |   S1  |   |   1  |   C1  |   |    1   |    1   |
+------+-------+   +------+-------+   +--------+--------+
|   2  |   S2  |
+------+-------+

不起作用......

我应该这样：

 categorie_service
+--------+--------+
|  id_c  |  id_s  |
+--------+--------+
|    1   |    1   |
+--------+--------+
|    1   |    2   |
+--------+--------+

P

Answer 4

我补充说：

$categories = $form->getData()->getCategories(); 
foreach ($categories as $categorie) {
    $categorie->addService($service);
}

进入＆＃39; CreationServiceAction＆＃39;并且在致电$ em-＆gt;持续...

之前

现在没问题。

关系ManyToMany与doctrine / symfony3

4 个答案: