Python Django基于函数的视图到基于类的视图

我知道这个url是

http://127.0.0.1:8000/upload/picturelist/1

，使用户id=1

在myurl.py中

url(r'^picturelist/(?P<user_id>\d+)$', views.pictureList),

如何使此基于函数的视图使用createview

class pictureList(CreateView):

---

我从未使用过

CreateView

，但以下是我从阅读文档中获得的信息：

您可以通过定义：

查看：

class pictureList(CreateView):
  model = YourModelHere
  fields = ['whatever','fields','you','want','edited']

  def form_valid(self, form):
    record = form.save(commit = False)

    # assuming the user id is associated 
    # to the model with fieldname user_id

    if (self.request.user == record.user_id):
      record.save()
      return HttpResponseRedirect(self.get_success_url())
    # not sure if this works:
    return self.form_invalid()

然后模板将位于

“yourappname/yourmodelhere\u form.html”

---

请参见示例。

您可以执行以下操作：

在url.py中：

url（r'^picturelist/（？P\d+）$），views.MakeItView.as_view（），

在views.py中：

class MakeItView(CreateView):
    model = myModel
    template_name = 'whatever.html'

    def get_context_data(self, **kwargs):
        context = super(MakeItView, self).get_context_data(**kwargs)
        if int(self.kwargs['user_id']) != self.request.user.id:
            raise PermissionDenied
        return context

---

嗨，我在/author/add/4上找到了NoReverseMatch。

class MakeItView(CreateView):
    model = myModel
    template_name = 'whatever.html'

    def get_context_data(self, **kwargs):
        context = super(MakeItView, self).get_context_data(**kwargs)
        if int(self.kwargs['user_id']) != self.request.user.id:
            raise PermissionDenied
        return context