I want to have SEO-friendly URL,my current url in urls.py :
(ur'^company/news/(?P<news_title>.*)/(?P<news_id>\d+)/$','CompanyHub.views.getNews')
I use it in template:
{% for n in news %}
     <a href="{% url CompanyHub.views.getNews n.title,n.pk %}" >{{n.description}}</a>
{% endfor %}
I use news_id to get news object with that PK .
I want to convert this url:
../company/news/tile of news,with comma/11
to:
../company/news/tile-of-news-with-comma/11
by doing some thing like this in template:
{% for n in news %}
      <a href="{% url CompanyHub.views.getNews slugify(n.title),n.pk %}" >{{n.description}}</a>
{% endfor %}
I checked out these questions:
question1
question2
question3 and this article but they save an slugify field in database while I wanna generate it on demand.in addition I want to run a query by news_id.
I think this question is good,but I don't know how to use news_id to fetch my news object
 
     
     
    