How to parse a xml feed using feed parser python?

Question

Im trying to parse a feed in python using feedparser. But all I get is None returned. Im not sure what im missing. Here is my code:

import feedparser

def rss(self):
    rss = 'https://news.google.com/news?q=fashion&output=rss'
    feed = feedparser.parse(rss)
    for key in feed.entries: 
        return key.title

If you think there is a better rss/xml feed parse. Please let me know. (Im new to python)

print(key) displays none and print(len(feed.entries)) also displays none

print(feed)
{'feed': {}, 'entries': [], 'bozo': 1, 'bozo_exception': URLError(SSLError(1, '[SSL: CERTIFICATE_VERIFY_FAILED] certificate verify failed (_ssl.c:749)'),)}

print(feedparser)
<module 'feedparser' from '/Users/User_name/python-projects/my_env/lib/python3.6/site-packages/feedparser.py'>

Answer 1

Figured out the issue was actually with the SSL handshake fixed it by adding ssl._create_default_https_context = ssl._create_unverified_context.

For anyone else facing the issue. Full code is:

import feedparser
import ssl
if hasattr(ssl, '_create_unverified_context'):
    ssl._create_default_https_context = ssl._create_unverified_context
rss = 'https://news.google.com/news?q=fashion&output=rss'
feed = feedparser.parse(rss)

print(feed)

Answer 2

Try the following basic code, which works fine for me and gave me 10 items in the feed when I ran it.

1. Install feedparser from pip

pip install feedparser

2. Usage

import urllib2
import feedparser

url = "https://news.google.com/news?q=fashion&output=rss"
response = urllib2.urlopen(url).read()

print response

d = feedparser.parse(response)
print len(d.entries)
for item in d.entries:
    print "------"
    print item.title
    if 'subtitle' in item:
        print item.subtitle
    print item.link
    print item.description
    print item.published
    print item.id
    print item.updated
    if 'content' in item:
        print item.content

Or, paste the FULL code you're running, and I'll take a look.