表单的视图数据应该是 class 的一个实例...但是是一个 (n) 字符串

Question

我目前收到以下错误：

"The form's view data is expected to be an instance of class Symfony\Component\HttpFoundation\File\File, but is a(n) string. You can avoid this error by setting the "data_class" option to null or by adding a view transformer that transforms a(n) string to an instance of Symfony\Component\HttpFoundation\File\File."

SoundController - 上传功能

 /**
 * @Security("is_granted('IS_AUTHENTICATED_FULLY')")
 * @Route("/song/upload", name="upload_song")
 * @param Request $request
 * @return \Symfony\Component\HttpFoundation\Response
 */
public function uploadSong(Request $request)
{
    $song = new Sound();

    $form = $this->createForm(SoundType::class, $song);
    $form->handleRequest($request);

    if ($form->isSubmitted() && $form->isValid())
    {
        $file = $song->getFile();
        $user = $this->getUser();

        $fileName = $this
            ->get('app.file_uploader')
            ->setDir($this->get('kernel')->getRootDir()."/../web".$this->getParameter('songs_directory'))
            ->upload($file);

        $song->setFile($fileName);

        $file = $song->getCoverFile();
        if ($file === null)
        {
            $song->setCoverFile($this->getParameter('default_cover'));
        }
        else
        {
            $fileName = $this
                ->get('app.file_uploader')
                ->setDir($this->get('kernel')->getRootDir()."/../web".$this->getParameter('covers_directory'))
                ->upload($file);

            $song->setCoverFile($fileName);
        }

        $song->setUploader($user);
        $song->setUploaderID($user->getId());
        $user->addSong($song);

        $entityManager = $this->getDoctrine()->getManager();
        $entityManager->persist($song);
        $entityManager->flush();

        return $this->redirectToRoute('song_view', [
            'id' => $song->getId()
        ]);
    }

    return $this->render('song/upload.html.twig', [
        'form' => $form->createView()
    ]);
}

声音类型 - 形式

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('file', FileType::class)
        ->add('coverFile', FileType::class, [
            'required' => false
        ])
        ->add('songName', TextType::class)
        ->add('songAuthor', TextType::class);
}

Answer 1

/**
 * @ORM\Column(type="string")
 *
 * @Assert\NotBlank(message="Please, upload the song as a MP3 file.")
 * @Assert\File(mimeTypes={ "audio/mpeg", "audio/wav", "audio/x-wav", "application/octet-stream" })
 */
private $file;

你告诉 doctrine 你想要存储一个字符串，但是你在表单中呈现了一个上传按钮，它向你发送了一个你根本不想存储在数据库中的物理文件。相反，您想将文件从临时目录移动到上传目录，并且您想记住文件的名称到数据库中，因此您需要这个属性这是一个字符串。

最好的方法是遵循这个 page

Answer 2

答案如下：

{
      $builder
          ->add('file', FileType::class, array('data_class' => null))
          ->add('coverFile', FileType::class, array('data_class' => null))
          ->add('coverFile', FileType::class, array('data_class' => null,'required' => false))
          ->add('songName', TextType::class)
          ->add('songAuthor', TextType::class);
  }

表单的视图数据应该是 class 的一个实例...但是是一个 (n) 字符串

The form's view data is expected to be an instance of class ... but is a(n) string

php

symfony

data-class