python - 从 URL 获取参数 slugs

Question

我正在尝试从 URL 中获取两个参数以添加到我的上下文中。

这是网址:

  url(r'^company/(?P<company>[\w\-\_]+)/?/(?P<program>[\w\-\_]+)/?$', RegistrationView.as_view(),
                       name='test'), 

View :

class RegistrationView(RegistrationMixin, BaseCreateView):
    form_class = AppUserIntroducerCreateForm
    template_name = "registration/register_introducer.html"
    slug_field = 'company'



    def get_context_data(self, *args, **kwargs):
        context = super(RegistrationIntroducerView, self).get_context_data(**kwargs)
        print(self.get_slug_field())
        context['company'] = ??????
        context['program'] = ??????
        return context

我已经尝试了所有方法来获取值 self.company、kwargs['company'] 等，我在这里做错了什么？

Answer 1

Here对你来说是引用。

context = super(RegistrationView, self).get_context_data(**kwargs)
print(self.get_slug_field())
context['company'] = self.kwargs['company']
context['program'] = self.kwargs['program']