Django ListView中的返回地址

如何使用ListView返回经销商列表中的地址

我在views.py中尝试

class DealershipList(ListView):
    template_name = 'dealership_list.html'
    model = Dealership

    def get_queryset(self):
        pass

    def get_context_data(self, **kwargs):
        context = super(DealershipList, self).get_context_data(**kwargs)
        address = Dealership.objects.get(pk=self.kwargs['address'])
        context['address'] = self.address
        return context

我需要在模板

{% for dealership in dealership_list %}
    <div class="col-lg-6">
        <h4>{{ dealership.dealership }}</h4>
        <p>{{ address.address }}</p>
        <p>Site: <a href="{{ dealership.site }}" target="_blank">{{ dealership.site }}</a></p>
    </div>
{% endfor %}

现在我试着：

class DealershipList(ListView):
    template_name = 'core/dealership/dealership_list.html'
    model = Dealership

    def get_context_data(self, **kwargs):
        id_address = Address.objects.get(pk=self.kwargs['pk'])
        address = Dealership.objects.filter(address=id_address)
        context = super(DealershipList, self).get_context_data(**kwargs)
        context['address'] = address
        return context

但不工作

url.py

url(r'^dealerships/$', DealershipList.as_view(), name='dealership_list'),

决心

{{ dealership.address.address }}

---

你能更好地解释一下你想完成什么，你采取了哪些步骤来完成它吗？你也可以显示你的网址。py？我需要返回经销商和地址。

{{ dealership.address.address }}