如何处理媒体文件而不将其保存在Django中

Question

在我的django项目中，我收到了客户端发布的媒体文件。 像这样

def upload(request):
    if request.method == 'POST':
        form = UploadFileForm(request.POST, request.FILES)
        if form.is_valid():
            handle_uploaded_file(request.FILES['file'], request.POST['fid'])
            return JsonResponse({"result":"ok"})
    return JsonResponse({"result":"failed","msg":"unkown"})


def handle_uploaded_file(f, fid):
    with open(STEP_DIR + '/' + fid, 'wb+') as destination:
        for chunk in f.chunks():
            destination.write(chunk)

另一方面，我想通过另一个模块来处理此文件。 然后此模块将打开一个文件并按以下方式处理它：

Import thirdModule
thirdModule.open('path_to_url').exporter(...)

由于thirdModule将通过给定路径打开文件。 所以我必须保存刚刚从django收到的文件？
有什么办法可以直接处理文件而不保存它？

def handle_uploaded_file(f, fid):
    thirdModule.open(convert_media_to_stream(f))
    ...

Answer 1

if form.is_valid():
    data = form.cleaned_data
    inmemory_uploaded_file = data['file_field_name']
    file_name = str(inmemory_uploaded_file)
    process_file_without_saving_into_disk(inmemory_uploaded_file)
    # rest of your code

更新资料

def handle_upload(file, file_name, upload_path):
    if not os.path.exists(upload_path):
        os.mkdir(upload_path)

    file_path = os.path.join(upload_path, file_name)
    with open(file_path, 'wb+') as destination:
        for chunk in file.chunks():
            destination.write(chunk)
    return file_path