Python 在DetailView中获取Slug

url.py

要访问的URL：

因此，我的应用程序必须做的是，根据URL检索slug并使用Intrinio的API获取响应

响应部分都可以工作，但目前它始终是同一家公司（？identifier=SHOP）。我想根据url使其动态化

---

但是我对Django非常陌生，我不确定如何将slug传递给DetailView。希望您能提供帮助。

您可以在

self.kwargs

中访问slug

class DetailView(generic.DetailView):
    model = Company
    template_name = 'news/detail.html'

    def get_context_data(self, **kwargs):
         # Add in a QuerySet of all the books
         context = super(DetailView, self).get_context_data(**kwargs)
         response = requests.get('https://api.intrinio.com/news?identifier=SHOP', auth=requests.auth.HTTPBasicAuth(
        'xxxx',
        'xxxx'))
         context['articleList'] = response.json()
         return context

---

您可以在

self.kwargs

中访问slug

class DetailView(generic.DetailView):
    model = Company
    template_name = 'news/detail.html'

    def get_context_data(self, **kwargs):
         # Add in a QuerySet of all the books
         context = super(DetailView, self).get_context_data(**kwargs)
         response = requests.get('https://api.intrinio.com/news?identifier=SHOP', auth=requests.auth.HTTPBasicAuth(
        'xxxx',
        'xxxx'))
         context['articleList'] = response.json()
         return context

---

太好了，这很有效！我需要更深入地了解夸尔格家族，看起来很棒，很管用！我需要多看看kwargs，看起来不错

def get_context_data(self, **kwargs):
     # Add in a QuerySet of all the books
     context = super(DetailView, self).get_context_data(**kwargs)
     slug = self.kwargs['slug']
     response = requests.get('https://api.intrinio.com/news?identifier=%s' % slug,
     ...
     )