Django表单上传request.files为空

Question

我以前没有在这里发布过问题，主要阅读。

我正在学习 Django 并在 .但现在我以某种方式打破了它。

我上传时 request.FILES 为空，但我可以在 request.raw_post_data 中看到文件名。

这是 html 的代码

<form enctype="multipart/form-data" method="post" action="">{% csrf_token %}
{{ form.as_p }}
<input type="submit" name="submit" value="Upload Photo" />
</form

表格

class PhotoUploadForm(forms.Form):
    title = forms.CharField(max_length=50)
    description = forms.CharField(required=False,max_length="254")
    photo = forms.ImageField()

View

class PhotoUploadView(FormView):

    template_name ="album/photo_upload.html"
    form_class = PhotoUploadForm

    def get_context_data(self,**kwargs):
        context = super(PhotoUploadView,self).get_context_data(**kwargs)
        context['user_info'] = self.request.user
        if 'upload_form' in kwargs:
            context['upload_form'] = kwargs['upload_form']
        else:
            context['upload_form'] = PhotoUploadForm()
        album = get_object_or_404(Album,id=self.kwargs['album_id'])
        context['album'] = album
        context['form'] = self.form_class
        return context

    def post(self,*args,**kwargs):
        print self.request.FILES
        print self.request.raw_post_data
        if self.request.method == "POST":
            form = PhotoUploadForm(self.request.POST,self.request.FILES)
            if form.is_valid():
                photo = Photo()
                photo.title = form.cleaned_data['title']
                photo.summary = form.cleaned_data['description']
                photo.album = get_object_or_404(Album,id = kwargs['album_id'])
                photo.is_cover_photo = True                 
                path = self.generate_filename(self.request.FILES['photo'].name,self.request.user,kwargs['album_id'])
                destination = open(path,"wb+")
                for chunk in self.request.FILES['photo'].chunks():
                    destination.write(chunk)
                destination.close()
                photo.imagePath = path
                photo.save()
        return self.render_to_response(self.get_context_data(upload_form=form)

Answer 1

我也使用 FormView 做到了(我主要是学习如何使用它，最新通用类 View 的 django doc 非常有限)；虽然@bezidejni 的回答可能是一个更好的解决方案，但这就是您解决问题的方法:

class PhotoUploadView(FormView):

    template_name ="album/photo_upload.html"
    form_class = PhotoUploadForm

    def form_valid(self, form):
        """
        This is what's called when the form is valid.
        """
        photo = form.cleaned_data['photo'] # <= this is your uploaded file in memory
        # read your photo object by chunks, save it to disk, or whatever else
        # ....
        # then keep on going! you can make the 'get_success_url' more complex should you need to.
        return HttpResponseRedirect(self.get_success_url())

附注。
如果您使用 FormView，则不应使用 render_to_response；过载 get_context_data如果您需要将上下文传递给已指定为 template_name 的模板