def parse_feed(feed_url):
return feedparser.parse(feed_url)<EOL>
Parses a given feed and returns the result in a given context
variable.

It is **highly** recommended that you use `Django's template
fragment caching`_ to cache the output of this tag for a
reasonable amount of time (e.g., one hour); polling a feed too
often is impolite, wastes bandwidth and may lead to the feed
provider banning your IP address.

.. _Django's template fragment caching: http://www.djangoproject.com/documentation/cache/#template-fragment-caching

Arguments should be:

1. The URL of the feed to parse.

Requires the Universal Feed Parser, which can be obtained at
http://feedparser.org/. See `its documentation`_ for details of the
parsed feed object.

.. _its documentation: http://feedparser.org/docs/

Syntax::

    {% parse_feed [feed_url] as [varname] %}

Example::

    {% parse_feed "http://www2.ljworld.com/rss/headlines/" as ljworld_feed %}