处理与BeautifulSoup的不良链接的最佳方法是什么?

最后发布: 2009-01-17 06:10:41


问题

我正在研究从美味中获取网址,然后使用这些网址来发现关联的供稿的东西。

但是,美味的某些书签不是html链接,并且会导致BS发出响声。 基本上,如果BS提取链接并且它看起来不像html,我想扔掉链接。

现在,这就是我要得到的。

trillian:Documents jauderho$ ./d2o.py "green data center" 
processing http://www.greenm3.com/
processing http://www.eweek.com/c/a/Green-IT/How-to-Create-an-EnergyEfficient-Green-Data-Center/?kc=rss
Traceback (most recent call last):
  File "./d2o.py", line 53, in <module>
    get_feed_links(d_links)
  File "./d2o.py", line 43, in get_feed_links
    soup = BeautifulSoup(html)
  File "/Library/Python/2.5/site-packages/BeautifulSoup.py", line 1499, in __init__
    BeautifulStoneSoup.__init__(self, *args, **kwargs)
  File "/Library/Python/2.5/site-packages/BeautifulSoup.py", line 1230, in __init__
    self._feed(isHTML=isHTML)
  File "/Library/Python/2.5/site-packages/BeautifulSoup.py", line 1263, in _feed
    self.builder.feed(markup)
  File "/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/HTMLParser.py", line 108, in feed
    self.goahead(0)
  File "/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/HTMLParser.py", line 150, in goahead
    k = self.parse_endtag(i)
  File "/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/HTMLParser.py", line 314, in parse_endtag
    self.error("bad end tag: %r" % (rawdata[i:j],))
  File "/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/HTMLParser.py", line 115, in error
    raise HTMLParseError(message, self.getpos())
HTMLParser.HTMLParseError: bad end tag: u'</b  />', at line 739, column 1

更新:

耶希亚的答案就解决了。 作为参考,下面是获取内容类型的一些代码:

def check_for_html(link):
    out = urllib.urlopen(link)
    return out.info().getheader('Content-Type')
python parsing beautifulsoup
回答

我只是包装了BeautifulSoup处理并查找HTMLParser.HTMLParseError异常

import HTMLParser,BeautifulSoup
try:
    soup = BeautifulSoup.BeautifulSoup(raw_html)
    for a in soup.findAll('a'):
        href = a.['href']
        ....
except HTMLParser.HTMLParseError:
    print "failed to parse",url

但除此之外,您还可以在抓取页面时检查响应的内容类型,并在尝试解析之前确保它类似于text/htmlapplication/xml+xhtml类。 那应该避免大多数错误。