Question

嗨，我正面临一个我无法找到解决方案，所以求助。

我有两个实体：演员和艺术家。在演员阵容中我有演员，女演员，这将是艺术家表的领域，我使用这个代码：

为此：

namespace Bbd\MyAppBundle\Form;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;

class CastType extends AbstractType
{
/**
 * @param FormBuilderInterface $builder
 * @param array $options
 */
public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('actor', 'entity', array(
            'class'    => 'BbdMyAppBundle:Artist',
            'property' => 'name',
            'multiple' => true,
            'label'    => 'Artist',
            'required' => false,

        ))
        ->add('actress')
        ->add('content')
    ;
}

可以有多个演员或女演员。所以在db中它保存如下：

Doctrine\Common\Collections\ArrayCollection@000000006f69bd7b000000001772666a

在演员字段中。我不知道为什么，它应该保存id或名称。

这是施法者：

Bbd\MyAppBundle\Entity\Cast:
type: entity
repositoryClass: Bbd\MyAppBundle\Repository\CastRepository
table: cast
id:
    id:
        type: integer
        generator: { strategy: AUTO }
fields:
    actor:
        type: text
        nullable: true
    actress:
         type: text
         nullable: true
oneToOne:
    content:
        targetEntity: Content
        inversedBy: cast
        joinColumn:
             name: content_id
             referencedColumnName: id
             onDelete: CASCADE

艺术家ORM

Bbd\MyAppBundle\Entity\Artist:
type: entity
repositoryClass: Bbd\MyAppBundle\Repository\ArtistRepository
table: artist
id:
    id:
        type: integer
        generator: { strategy: AUTO }
fields:
    name:
        type: string
        length: 255
        unique: true
    bangla_name:
        type: string
        length: 255
        unique: true
    priority:
        type: integer
    birth:
        type: date
    sex:
        type: string
        length: 6
    bio_english:
        type: text
    bio_bangla:
        type: text

感谢您的帮助..

Answer 1

根据您的情况，我可以建议您在ManyToMany和Cast实体之间建立Artist关联，每个演员都有很多艺术家，每个艺术家都可以出现超过一个演员

您的Artist实体看起来像

use Doctrine\ORM\Mapping as ORM;
/** @Entity **/
class Artist
{
    /**
     * @ORM\ManyToMany(targetEntity="Cast", inversedBy="artists")
     * @JORM\oinTable(name="cast_artists")
     **/
    private $cast;
    public function __construct() {
        $this->cast = new \Doctrine\Common\Collections\ArrayCollection();
    }
}

并且Cast实体将具有类似

的映射

use Doctrine\ORM\Mapping as ORM;
/** @Entity **/
class Cast
{
    /**
     * @ORM\ManyToMany(targetEntity="Artist", mappedBy="cast")
     **/
    private $artists;

    public function __construct() {
        $this->artists = new \Doctrine\Common\Collections\ArrayCollection();
    }
    public function addArtist($artists) {
        $this->artists[] = $artists;
        return $this;
    }
    public function removeArtist($artists) {
        $this->artists->removeElement($artists);
    }
    public function getArtists() {
        return $this->artists;
    }
}

添加完所有艺术家记录后，您可以通过选择多位艺术家来创建演员记录，无论其演员是谁/

Symfony2在数据库中保存Doctrine \ Common \ Collections \ ArrayCollection而不是名称

1 个答案: