title: correct way to obtain the rest of the parameters of URL in Django rest framework
body: <p>Hi everyone I am using  Django rest framework to create an API</p>
  
  <p>In my URLs.py file I have this</p>
  
  <pre><code>  url(r'^cpuProjects/$', cpuProjectsViewSet.as_view({'get': 'list'})),
    url(r'^cpuProjects/(?P&lt;project_name&gt;[a-zA-Z0-9]+)$', cpuProjectsViewSet.as_view({'get': 'retrieve'})),
  </code></pre>
  
  <p>This work perfectly, and I have this to url</p>
  
  <pre><code>http://127.0.0.1:8000/cpuProjects/
  http://127.0.0.1:8000/cpuProjects/ad
  </code></pre>
  
  <p>in my retrieve function, I have this to get the parameter</p>
  
  <pre><code>  def retrieve(self, request, project_name=None):
      try:
        opc = self.kwargs.get(self.lookup_url_kwarg)
        print(opc)
        ...
  </code></pre>
  
  <p>Now, I add this to my URLs.py files</p>
  
  <pre><code>url(r'^cpuProjects/(?P&lt;project_name&gt;[a-zA-Z0-9]+/[a-zA-Z0-9]+)$', cpuProjectsViewSet.as_view({'get': 'retrieve'})),
  
  
  http://127.0.0.1:8000/cpuProjects/name_project/whatever_string_here
  </code></pre>
  
  <p>My <code>print(opc)</code> in this last case return this <code>ad/pending</code>.</p>
  
  <p>Is this the correct way to obtain the rest of the parameters of URL in Django rest framework?</p>