Question

我想将utidy的结果传递给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：

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想要一个字符串。有没有办法来cast_html？或者我做错了，应采取不同的方法吗？

Answer 1

只需将str()包裹在cleaned_html左右当它传递给BeautifulSoup时。

Answer 2

将传递给BeautifulSoup的值转换为字符串。在您的情况下，请执行以下编辑到最后一行：

soup = BeautifulSoup(str(cleaned_html))

美丽的汤和uTidy

2 个答案: