django - 干燥 Django View request.user 对象

Question

我正在构建一个 Web 应用程序，我发现在检查 request.user 是否与创建电影的用户匹配时有很多重复。电影对用户有 FK，因此只有创建特定电影的用户才能对其执行适当的操作。

@login_required
def edit_movie(request, slug, template_name="movies/edit_movie.html"):
    movie = get_object_or_404(Movie, slug=slug)
    if movie.user != request.user:
        raise HttpResponseForbidden
    # Rest of code omitted for brevity.

@login_required
def edit_screener(request, slug, template_name="movies/edit_screener.html"):
    movie = get_object_or_404(Movie, slug=slug)
    movie_media = movie.moviemedia_set.get(movie_type='screener')
    if movie.user != request.user:
        raise HttpResponseForbidden()
    # Rest of code omitted for brevity.


@login_required
def dashboard(request, template_name='movies/dashboard.html'):
    movies = Movie.objects.active().filter(
        user=request.user).order_by('-created_at')
    # Rest of code omitted for brevity.

我在 Ruby on Rails 方面有很强的背景，我们只需在 Controller 上使用 before_filter :find_user 就可以避免重复。 Django 中处理此类重复的最佳方法是什么？

J

Answer 1

如果可以显示 404 而不是 403，您可以这样做:

movie = get_object_or_404(Movie, slug=slug, user=request.user)