django逆向动态id

Question

在我运行没有模板的 View 后，我想返回到之前的 View 。然而，这个观点有一个动态的部分，现在我不知道如何用我的反面来传递这个论点:

我的网址:

url(r'^building/(?P<building_id>\d+)/$', views.building, name='building'),
url(r'^building/upgrade/(?P<building_id>\d+)/$', views.building_upgrade, name='building_upgrade'),

我的看法:

@login_required
def building_upgrade(request, building_id):
    building = get_object_or_404(Building, pk=building_id)
    if building.town.user != request.user:
        return HttpResponseRedirect(reverse('index'))
    create_upgrade(building)
    return HttpResponseRedirect(reverse('building'))

我怎样才能将building_id反向发送回来，以便它知道要加载什么网址？

Answer 1

reverse('building', kwargs={'building_id': building_id})