如何让 user_url 零件在现场。com/user_url/gallery/slug 在 DetailView 的情况下？

Question

我想在模板中将 link 从 site.com/user_url/gallery/slug 返回到 site.com/user_url/gallery/ 使用类似的东西：

<a href="{% url 'profiles_user:profiles_gallery' -->>>????<<<--- %}" 
class="btn btn-default">"Come back to all galleries and photos"</a>

我需要提供 user_url 参数而不是 -->>>????<<<--- 以获得这样的 url 作为 site.com/user_url/gallery.

# site.com/user_url/gallery/slug - gallery details
class ProfileGalleryDetailView(DetailView):
    template_name = 'profiles/gallery_detail.html'

    def get_queryset(self):
        print(self.__dict__)
        user = get_object_or_404(UserProfile, user_url=self.kwargs['user_url'])
        return Gallery.objects.filter(galleryextended__user=user, slug=self.kwargs['slug']).on_site().is_public()

print(self.__dict__) 告诉我：

{'args': (), 'kwargs': {'slug': 'time-sleep', 'user_url': '1-plus-1'}, 
'request': <WSGIRequest: GET '/1-plus-1/gallery/time-sleep/'>, 
'head': <bound method BaseDetailView.get of <profiles.views.ProfileGalleryDetailView object at 0x7fe912b41860>>}

如何从模板中的 kwargs 获取 'user_url': '1-plus-1'？我是否需要使用 get_context_data 才能将 user_url 添加到上下文？

# Core urls.py
urlpatterns = [
    url(r'^(?P<user_url>[\w.-]+)/', include('profiles.urls', namespace='profiles_user')),
]

# profiles.urls
urlpatterns = [
    url(r'^$', views.ProfileDetailView.as_view(), name='profiles_home'),
    url(r'^gallery/$', views.ProfileGalleryArchiveIndexView.as_view(), name='profiles_gallery'),
    url(r'^gallery/(?P<slug>[\-\w]+)/$', views.ProfileGalleryDetailView.as_view(), name='profiles_gallery-details'),
]

Answer 1

在基于 class 的视图中，get_context_data 方法包括 view 视图。因此，您可以使用 view.kwargs.user_url.

访问 user_url kwarg

<a href="{% url 'profiles_user:profiles_gallery' view.kwargs.user_url %}">

如何让 user_url 零件在现场。com/user_url/gallery/slug 在 DetailView 的情况下？

How to get user_url part being at site.com/user_url/gallery/slug in case of DetailView?

django

django-templates

django-urls