title: Python, Django 1.7 : Redirect all URLs to a single controller
body: <p>let's base URL is <code>so.com</code> 
  So, if URL start with abc e.g. </p>
  
  <pre><code>so.com/abc/
  so.com/abc/123
  so.com/abc?newtab=123
  so.com/abc#123
  so.com/abc/123?tab=new
  ...
  </code></pre>
  
  <p>Then all this URL patterns should go to a Class Abc</p>
  
  <pre><code>myapp/urls.py 
  ...
  url(r'^abc[a-zA-Z0-9=#_\?\-/]+$',views.Abc.as_view(),name='abc')
  </code></pre>
  
  <p>myapp/myviews/abc.py</p>
  
  <pre><code>class Abc(View):
     def get(self,request):
      ...
     def foo(user_id):
     ...
     def bar(post_id):
     ...
  </code></pre>
  
  <p>In function <code>get(self,request):</code> how to get everything after abc  that  was requested .
  e.g.</p>
  
  <pre><code>so.com/abc/xyz =&gt; /xyz
  so.com/abc#123 =&gt; 123
  so.com/abc?tab=new =&gt; ?tab=new 
   so.com/abc/123?tab=new =&gt; tab = new and 123 
  </code></pre>
  
  <p>When <code>#123</code> is added after abc then it automatic converts to <code>abc/#123</code> </p>
  
  <p>How to get this work ?</p>
  
  <p>I have seen many question but they are not helpful.</p>
  
  <p><a href="https://stackoverflow.com/questions/17866114/django-get-absolute-url">Django Get Absolute URL</a></p>
  
  <p><a href="https://stackoverflow.com/questions/427102/what-is-a-slug-in-django">What is a &quot;slug&quot; in Django?</a></p>
  
  <p><a href="https://stackoverflow.com/questions/2882490/get-the-current-url-within-a-django-template">How to get the current URL within a Django template?</a></p>
  
  <p>...</p>