美丽的汤和整洁 [英] Beautiful Soup and uTidy

问题描述

我想将 utidy 的结果传递给Beautiful Soup，ala:

I want to pass the results of utidy to Beautiful Soup, ala:

page = urllib2.urlopen(url)
options = dict(output_xhtml=1,add_xml_decl=0,indent=1,tidy_mark=0)
cleaned_html = tidy.parseString(page.read(), **options)
soup = BeautifulSoup(cleaned_html)

运行时，会出现以下错误:

When run, the following error results:

Traceback (most recent call last):
  File "soup.py", line 34, in <module>
    soup = BeautifulSoup(cleaned_html)
  File "/var/lib/python-support/python2.6/BeautifulSoup.py", line 1499, in __init__
    BeautifulStoneSoup.__init__(self, *args, **kwargs)
  File "/var/lib/python-support/python2.6/BeautifulSoup.py", line 1230, in __init__
    self._feed(isHTML=isHTML)
  File "/var/lib/python-support/python2.6/BeautifulSoup.py", line 1245, in _feed
    smartQuotesTo=self.smartQuotesTo, isHTML=isHTML)
  File "/var/lib/python-support/python2.6/BeautifulSoup.py", line 1751, in __init__
    self._detectEncoding(markup, isHTML)
  File "/var/lib/python-support/python2.6/BeautifulSoup.py", line 1899, in _detectEncoding
    xml_encoding_match = re.compile(xml_encoding_re).match(xml_data)
TypeError: expected string or buffer

我收集utidy返回一个XML文档，而BeautifulSoup需要一个字符串.有没有办法投射cleaned_html?还是我做错了，应该采取其他方法?

I gather utidy returns an XML document while BeautifulSoup wants a string. Is there a way to cast cleaned_html? Or am I doing it wrong and should take a different approach?

推荐答案

只需包装> 在cleaned_html附近 将其传递给BeautifulSoup时.

Just wrap str() around cleaned_html when passing it to BeautifulSoup.

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