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symfony - 无法加载类型 "sonata_media_type": class does not exist

转载 作者:行者123 更新时间:2023-12-02 03:42:11 25 4
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您好,我第一次使用 SonataMedia Bundle,但我无法访问我的实体帖子的链接;它给了我#“无法加载类型“sonata_media_type”:类不存在。”#这是我的代码:Config.php

    sonata_media:
db_driver: doctrine_orm
default_context: default
contexts:
Post:
providers:
- sonata.media.provider.image

formats:
small: { width: 100 , quality: 70}
big: { width: 1680 , quality: 100}



**PostType:**


public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder->add('media', 'sonata_media_type', array(
'provider' => 'sonata.media.provider.image',
'context' => 'post'
));
}
**My Class Post**



class Post
{
/**
* @var int
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;


/**
* @var \Application\Sonata\MediaBundle\Entity\Media
* @ORM\ManyToOne(targetEntity="Application\Sonata\MediaBundle\Entity\Media", cascade={"persist"})
*/
protected $media;


/**
* Get id
*
* @return int
*/
public function getId()
{
return $this->id;
}

/**
* @param MediaInterface $media
*/
public function setMedia(MediaInterface $media)
{
$this->media = $media;
}

/**
* @return MediaInterface
*/
public function getMedia()
{
return $this->media;
}

}

我按照这个文档来做到这一点: [https://sonata-project.org/blog/2013/10/11/mediabundle-mediatype-improved][1]请问如何才能将我的实体帖子正确链接到实体媒体???

最佳答案

您应该使用完全限定的类名MediaType::class,就像您在 v.2.8 以来的任何 Symfony 表单中所做的那样:

**PostType:**


public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder->add('media', MediaType::class, array(
'provider' => 'sonata.media.provider.image',
'context' => 'post'
));
}

关于symfony - 无法加载类型 "sonata_media_type": class does not exist,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48211323/

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