用python抓取diggrss提要

Question

有没有办法通过digg的rss提要获取链接？或者我必须获得网站，然后用正则表达式手动刮取它？在

我想从rss获得digg指向的真正链接，而不是评论提要。在

示例- http://feeds.digg.com/~r/digg/popular/~3/Hx0VATaafSw/Apple_Scaling_Final_Cut_Studio_Apps_to_Fit_Prosumers_2

去

http://www.appleinsider.com/articles/10/05/18/apple_scaling_final_cut_studio_apps_to_fit_prosumers.html

Answer 1

您可以使用diggapi的story.getInfo方法。其中一个可能的参数是clean_title，您可以从RSS提要中的链接解析它。下面是一个示例实现：

import feedparser
import urllib2
from xml.etree import ElementTree

rss_link = 'http://feeds.digg.com/digg/popular.rss'
api_link = 'http://services.digg.com/1.0/endpoint?method=story.getInfo&clean_title=%s'

data = feedparser.parse(rss_link)

for i, e in enumerate(data.entries, 1):
  print '%d. Digg link: %s' % (i, e.link)
  title = e.link[e.link.rfind('/') + 1 :]
  xml = urllib2.urlopen(api_link % title).read()
  tree = ElementTree.fromstring(xml)
  print '%d. Real link: %s' % (i, tree.find('story').get('link'))

。。。哪些输出：

^{pr2}$

Answer 2

看一下feedparser模块。在

>>> import feedparser
>>> d = feedparser.parse('http://feeds.digg.com/digg/popular.rss')
>>> for entry in d.entries:
...     print entry.link
...
http://feeds.digg.com/~r/digg/popular/~3/Hx0VATaafSw/Apple_Scaling_Final_Cut_Studio_Apps_to_Fit_Prosumers_2
http://feeds.digg.com/~r/digg/popular/~3/mXb8b0QH3Rc/Skateboarder_Lives_Any_Man_s_Worst_Nightmare_video
http://feeds.digg.com/~r/digg/popular/~3/61N9gFUth1k/CBS_A_bloodbath_of_cancellations
http://feeds.digg.com/~r/digg/popular/~3/vZ3_6F1RAcI/Red_Dead_Redemption_Free_Roam_Done_Right
(snip)