In previous articles, we found out how to develop a sitemap for the Django website A legitimate sitemap increases your site’s online search engine ranking. Hence great for seo.

Likewise, including a robot.txt file benefits your website. It tells crawlers, which page to crawl and which page not to crawl for indexing.

In this article, we will see how to create RSS eat your Django website.

The RSS feed help to keep up readers with their favorite blogs, news sites, and other websites. RSS enables the content and brand-new updates to come to the reader. Generally, you use RSS to syndicate or sign up for the feed of a website, blog site or almost any media content that is updated online

Create a file in your app directory, parallel to urls.py file and name it feeds.py Paste the listed below code in it.

In the below sample example, we are bring posts/articles from the database of pythoncircle.com. We have actually executed four techniques, item, item_title, item_description and item_link The code has been updated with comments.

 class  LatestEntriesFeed( Feed): 

 title = " PythonCircle.com: New post for Python developers each week"

 link = "/ feed/"

 description = " Updates on changes and additions to python short articles on pythoncircle.com."


# return 10 recently created/updated posts

 def  items( self): 

 return  get_recent_updated_posts( number_of_posts =-LRB- ***********)10)
 def  item_title( self, item): 

 return  item.title


# return a short description of short article

 def  item_description( self, item): 

 return  item.description


# develop and return the post URL

 def  item_link( self, product): 

 return  reverse(' appname: index',  args=( item.post _ id,))

Now in your task’s urls.py file (not in any app’s urls.py file) include listed below code.

 from appname.feeds import LatestEntriesFeed().

# add feeds course
urlpatterns  = [

    path(r'feed/', LatestEntriesFeed()),

]