image - Symfony 3 中的 ContextErrorException

问题描述

我正在尝试在 SF3 中上传图片，上传时出现此错误：

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

这是我的实体中错误所在的部分（此处为第 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()
        ));
    }
}

标签： imagesymfonyupload

似乎您正在做不必要的工作，并使这比它可能需要的复杂一点。您是否遵循了如何上传文件的 Symfony 指南？

同时，看起来图像名称就是其中的内容，$this->image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可能只来自表单提交，并且在您的实体中，您可能希望使用File代替 - 如下所示：

use Symfony\Component\HttpFoundation\File\File;

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

image - Symfony 3 中的 ContextErrorException

问题描述

解决方案

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