Question

我正在使用Symfony 2.7

我收到了这个表格

<form name="Document" method="post" action enctype="multipart/form-data">
    <div class="col-md-6">
        <input type="file" id="documento" name="Document[Ruta]" required="required">
    </div>
    <div class="col-md-3">
        <button type="button" onclick="saveDoc()" class="btn btn-success" data-dismiss="modal">Subir</button>
    </div>
</form>

和这个ajax：

function saveDoc() {
    var inputFile = document.getElementById('documento');
    var file = inputFile.files[0];
    var data = new FormData();
    data.append('doc', file, file.name);
    $.ajax({
        url: "{{ path('fileupload') }}",
        type: "POST",
        data: {'data': data},
        processData: false,
        contentType: false,
        cache: false
    });

}

使用这个symfony控制器：

public function fileUpload(Request $request)
{
    $file = $request->files->get('doc')->getClientOriginalName;
    $user = $this->getUser();

    $filename = md5(uniqid()).'.'.$file->guessExtension();

    $request->files->get('doc')->move(
        $this->getParameter('upload_directory'),
        $filename
    );
    $document = new Document();
    $document->setRuta($filename);
    $document->setIdUser($user);
    $document->setFechaCreacion(new \DateTime("now"));
    $em = $this->getDoctrine()->getManager();
    $em->persist($document);
    $em->flush();

    return new JsonResponse(true);

}

但是我收到了这个错误：

注意：尝试获取非对象的属性 500内部服务器错误 - ContextErrorException

我看到235897个页面上传了symfony和ajax的文件，我无法理解我的代码有什么问题，请问各位帮帮我吧？

Answer 1

更改

$file = $request->files->get('doc')->getClientOriginalName;

到

$file = $request->files->get('doc');

Answer 2

Change this

function saveDoc() {
    var inputFile = document.getElementById('documento');
    var file = inputFile.files[0];
    var data = new FormData();
    data.append('doc', file, file.name);
    $.ajax({
        url: "{{ path('fileupload') }}",
        type: "POST",
        data: {'data': data},
        processData: false,
        contentType: false,
        cache: false
    });

}


To

function saveDoc() {
    var inputFile = document.getElementById('documento');
    var file = inputFile.files[0];
    var data = new FormData();
    data.append('doc', file, file.name);
    $.ajax({
        url: "{{ path('fileupload') }}",
        type: "POST",
        **data: data**,
        processData: false,
        contentType: false,
        cache: false
    });

}

并在控制器中

更改 $ file = $ request-＆gt; files-＆gt; get（'doc'） - ＆gt; getClientOriginalName; 至 $ file = $ request-＆gt; files-＆gt; get（'doc'）;

我试图使用ajax和symfony上传文件

2 个答案: