How to obtain field names of RSS feed(xml file) in python dynamically using feedparser?

First of all, the link you're going through has a 404 error. So you're not going to get any rss from that link.

Secondly, an RSS link ends with a .rss file most of the times.

ex: http://timesofindia.feedsportal.com/c/33039/f/533916/index.rss

Once you get an actual working RSS link, all you have to do is this:

fee = feedparser.parse('http://timesofindia.feedsportal.com/c/33039/f/533916/index.rss')
for feed in fee.entries:
    print feed.title
    print feed.link

What I wrote above was for the getting the item elements.

Let me provide you with a better example.

import feedparser
rss_document = """
<?xml version="1.0" encoding="utf-8"?>
<rss version="2.0">
<channel>
<title>Sample Feed</title>
<description>For documentation &lt;em&gt;only&lt;/em&gt;</description>
<link>http://example.org/</link>
<pubDate>Sat, 07 Sep 2002 00:00:01 GMT</pubDate>
<!-- other elements omitted from this example -->
<item>
<title>First entry title</title>
<link>http://example.org/entry/3</link>
<description>Watch out for &lt;span style="background-image:
url(javascript:window.location='http://example.org/')"&gt;nasty
tricks&lt;/span&gt;</description>
<pubDate>Thu, 05 Sep 2002 00:00:01 GMT</pubDate>
<guid>http://example.org/entry/3</guid>
<!-- other elements omitted from this example -->
</item>
</channel>
</rss>
"""
rss = feedparser.parse(rss_document)

# Channel Details

print "-----Channel Details-----"

print rss.feed.title
print rss.feed.description
print rss.feed.link

# Item Details

print "-----Item Details-----"
for fee in rss.entries:
    print fee.title
    print fee.summary,
    print fee.link

Use below code it will give you all keys name,

import feedparser
feeds_all = feedparser.parse(URL)
feed_all_keys = feeds_all.keys()
feed_keys = feeds_all.feed.keys()
entries_keys = feeds_all.entries.keys()

1. feed_all_keys holds all keys
2. feed_keys holds keys related to feed
3. entries_keys holds keys related to entries(items)