Question

我正在尝试在SF3中上传图像，并且在上传时出现此错误：

Symfony \ Component \ HttpFoundation \ File \ UploadedFile :: __ construct（）缺少参数2。

这是错误所在的我实体的一部分（此处是第9行）：

public function preUpload()
{
    // if there is no file (optional field)
    if (null === $this->image) {
        return;
    }

    // $file = new File($this->getUploadRootDir() . '/' . $this->image);
    $file = new File($this->getUploadRootDir() .'/' . $this->image);
    $uploadedfile = new UploadedFile($this->getUploadRootDir() .'/' . $this->image);

    // the name of the file is its id, one should just store also its extension
    // to make clean, we should rename this attribute to "extension" rather than "url" 
    $this->url = $file->guessExtension();

    // and we generate the alt attribute of the <img> tag,
    // the value of the file name on the user's PC
    $this->alt = $uploadedfile->getClientOriginalName();
}

然后是我的控制器：

public function mediaEditAction(Request $request)
{
    $media = new Media();
    $form = $this->createForm(MediaType::class, $media);
    $form->handleRequest($request);

    if ($form->isSubmitted() && $form->isValid()) {
        $file = $media->getImage();
        $fileName = md5(uniqid()).'.'.$file->guessExtension();
        $file->move(
            $this->getParameter('images_directory'),
            $fileName
        );
        $media->setImage($fileName);

        $em = $this->getDoctrine()->getManager();
        $em->persist($media);
        $em->flush();

        $request->getSession()->getFlashBag()->add('Notice', 'Photo added with success');

        // redirection
        $url = $this->generateUrl('medecin_parametre');

        // permanent redirection with the status http 301
        return $this->redirect($url, 301);
    } else {
        return $this->render('DoctixMedecinBundle:Medecin:mediaedit.html.twig', array(
            'form' => $form->createView()
        ));
    }
}

Answer 1

似乎您正在做不必要的工作，并使此工作比可能需要的复杂一些。您遵循了这个Symfony guide for How to Upload Files吗？

与此同时，图像名称似乎是$this->image中的名称，因此您可以将其作为第二个构造函数参数传递。

$uploadedfile = new UploadedFile($this->getUploadRootDir().'/'.$this->image, $this->image);

但是， UploadedFile 可能应该仅来自表单提交，并且在您的实体中，您想使用 File -像这样：

use Symfony\Component\HttpFoundation\File\File;

$uploadedfile = new File($this->getUploadRootDir() .'/' . $this->image);

Symfony 3中的ContextErrorException

1 个答案: