Question

我正在构建一个Web应用程序，我发现在检查request.user与创建影片的用户匹配时有很多复制。电影对用户具有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中有很强的背景，我们只需在控制器上使用before_filter：find_user，这样就可以避免重复。 Django处理这种重复的最佳方法是什么？

Ĵ

Answer 1

如果可以显示404而不是403，则可以执行此操作：

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

Answer 2

为什么不创建简单的函数来代替复制相同的代码？

def get_movie(slug, user):
    movie = get_object_or_404(Movie, slug=slug)
    if movie.user != user:
        raise Http403
    return movie

DRYing Django查看request.user对象

2 个答案: