Tell me more ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.

I am trying to parse this xml (http://www.reddit.com/r/videos/top/.rss) and am having troubles doing so. I am trying to save the youtube links in each of the items, but am having trouble because of the "channel" child node. How do I get to this level so I can then iterate through the items?

#reddit parse
reddit_file = urllib2.urlopen('http://www.reddit.com/r/videos/top/.rss')
#convert to string:
reddit_data = reddit_file.read()
#close file because we dont need it anymore:
reddit_file.close()

#entire feed
reddit_root = etree.fromstring(reddit_data)
channel = reddit_root.findall('{http://purl.org/dc/elements/1.1/}channel')
print channel

reddit_feed=[]
for entry in channel:   
    #get description, url, and thumbnail
    desc = #not sure how to get this

    reddit_feed.append([desc])
share|improve this question

2 Answers

up vote 1 down vote accepted

You can try findall('channel/item')

import urllib2
from xml.etree import ElementTree as etree
#reddit parse
reddit_file = urllib2.urlopen('http://www.reddit.com/r/videos/top/.rss')
#convert to string:
reddit_data = reddit_file.read()
print reddit_data
#close file because we dont need it anymore:
reddit_file.close()

#entire feed
reddit_root = etree.fromstring(reddit_data)
item = reddit_root.findall('channel/item')
print item

reddit_feed=[]
for entry in item:   
    #get description, url, and thumbnail
    desc = entry.findtext('description')  
    reddit_feed.append([desc])
share|improve this answer

I wrote that for you using Xpath expressions (tested successfully ):

from lxml import etree
import urllib2

headers = { 'User-Agent' : 'Mozilla/5.0' }
req = urllib2.Request('http://www.reddit.com/r/videos/top/.rss', None, headers)
reddit_file = urllib2.urlopen(req).read()

reddit = etree.fromstring(reddit_file)

for item in reddit.xpath('/rss/channel/item'):
    print "title =", item.xpath("./title/text()")[0]
    print "description =", item.xpath("./description/text()")[0]
    print "thumbnail =", item.xpath("./*[local-name()='thumbnail']/@url")[0]
    print "link =", item.xpath("./link/text()")[0]
    print "-" * 100
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.