COnverting从函数到类继承密码错误

Question

我正在尝试将我的3个类似的函数视图转换为基于继承类的视图。

视图看上去几乎完全相同，只有表单和它们的名称不同。其中一项如下：

@login_required
def add_video(request):
    if request.method == 'POST':
        movie_form = MovieForm(request.POST)
        if movie_form.is_valid():
            user = request.user
            new_post = movie_form.save(commit=False)
            new_post.author = user
            new_post.save()
            movie_form.save_m2m()
            messages.success(request, 'Post successfully added')
            return redirect(new_post)
    else:
        movie_form = MovieForm()
    return render(request, 'ak47/movie.html', {'movie_form': movie_form})

作为CBV，它看起来如下：

@login_required
class AddingView(CreateView):
        def form_valid(self, form):
            user = self.request.user
            new_post = form.save(commit=False)
            new_post.author = user
            new_post.save()
            form.save_m2m()
            messages.success(request, 'Post successfully added')

class AddMovieView(AddingView):
    form_class = MovieForm
class AddUrlView(AddingView):
    form_class = UrlImgForm

不幸的是，错误日志显示了一些奇怪的东西：

  File "/home/cruel/PycharmProjects/silverak/ak47/views.py", line 249, in <module>
    class AddMovieView(AddingView):
TypeError: function() argument 1 must be code, not str

Answer 1

不能在类上使用函数装饰符。必须编写一个装饰器，特别是在类上工作；而login_required不这样做。

相反，使用 mixin

class AddingView(LoginRequired, CreateView):
    pass