Symfony2将Doctrine\Common\Collections\ArrayCollection保存在数据库中，而不是名称 [英] Symfony2 saves Doctrine\Common\Collections\ArrayCollection in database not the name

问题描述

我面临一个我找不到解决方案的问题，因此需要帮助。

Hi I am facing an that i cant not find the solution of it, so for a help.

我有两个实体：演员和艺术家。从我有演员的女演员演员表中，我使用以下代码：

I have two entities: Cast and Artists. In cast from i have actor, actress which will be field by Artist table, I used this code:

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。因此在db中保存如下：

there can be multiple actor or actress. so in db it saves like:

Doctrine\Common\Collections\ArrayCollection@000000006f69bd7b000000001772666a    

。我不知道为什么，它应该保存ID或名称。

in the actor field. i dont why, it should save the id or name.

这是演员表：

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

Artist ORM

Artist 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

感谢帮助。.

推荐答案

您的方案，我建议您在 Cast 和 Artist ManyToMany 关联c $ c>实体每个演员都有很多艺术家，每个艺术家可以出现在一个以上的演员中

According to your scenario i can suggest you have a ManyToMany association between your Cast and Artist entity each cast have many artists and each artist can appear in morethan one cast

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();
    }
} 

和 Cast 实体将具有

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;
    }
}

添加所有艺术家记录后，您可以创建通过选择多位演员（无论演员是演员）来投射唱片

Once you have added all artists record you can create a cast record by selecting multiple artists whether its actor/actress

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