用Python抓取RSS feed

Question

我是Python和编程的新手,所以如果问题非常愚蠢,请原谅.

我一直在逐步学习RSS抓取这个教程,但是当我试图收集相应链接到正在收集的文章的标题时,我得到了Python的"列表索引超出范围"错误.

这是我的代码:

from urllib import urlopen
from BeautifulSoup import BeautifulSoup
import re

source  = urlopen('http://feeds.huffingtonpost.com/huffingtonpost/raw_feed').read()

title = re.compile('<title>(.*)</title>')
link = re.compile('<link>(.*)</link>')

find_title = re.findall(title, source)
find_link = re.findall(link, source)

literate = []
literate[:] = range(1, 16)

for i in literate:
    print find_title[i]
    print find_link[i]

这时候我只告诉它来获取冠军,但立即引发时,我想检索标题索引错误执行罚款和其对应的链接.

任何帮助将不胜感激.

Answer 1

您可以使用feedparser模块解析来自给定URL的RSS源:

#!/usr/bin/env python
import feedparser # pip install feedparser

d = feedparser.parse('http://feeds.huffingtonpost.com/huffingtonpost/latestnews')
# .. skipped handling http errors, cacheing ..

for e in d.entries:
    print(e.title)
    print(e.link)
    print(e.description)
    print("\n") # 2 newlines

产量

Even Critics Of Safety Net Increasingly Depend On It
http://www.huffingtonpost.com/2012/02/12/safety-net-benefits_n_1271867.html
<p>Ki Gulbranson owns a logo apparel shop, deals in 
<!-- ... snip ... -->

Christopher Cain, Atlanta Anti-Gay Attack Suspect, Arrested And
Charged With Aggravated Assault And Robbery
http://www.huffingtonpost.com/2012/02/12/atlanta-anti-gay-suspect-christopher-cain-arrested_n_1271811.html
<p>ATLANTA -- Atlanta police have arrested a suspect 
<!-- ... snip ... -->

使用正则表达式解析rss(xml)可能不是一个好主意.