Question

我目前收到以下错误：

＆＃34;表单的视图数据应该是Symfony \ Component \ HttpFoundation \ File \ File类的实例，但是是（n）字符串。您可以通过设置＆＃34; data_class＆＃34;来避免此错误。 null的选项或添加视图转换器，将（n）字符串转换为Symfony \ Component \ HttpFoundation \ File \ File的实例。＆＃34;

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()
    ]);
}

SoundType - 表格

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

以下是答案：

{
      $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);
  }

Answer 2

/**
 * @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;

您告诉学说您要存储字符串，但是您在表单中呈现一个上传按钮，该按钮会向您发送一个您根本不想存储在数据库中的物理文件。相反，您希望将文件从临时目录移动到上传目录，并且您想要记住数据库中的文件名称，因此您需要此属性为字符串。

最好的方法是遵循此page

表单的视图数据应该是类的实例...但是是（n）字符串

2 个答案: