美丽的汤和 uTidy

发布于 2024-07-22 04:32:20 字数 1303 浏览 14 评论 0原文

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,出现以下错误结果:

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

I Gather 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? 或者我做错了,应该采取不同的方法?

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)

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

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?

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评论(2

薔薇婲 2024-07-29 04:32:20

只需将 str() 包裹在 cleaned_html 上
将其传递给 BeautifulSoup 时。

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

秋风の叶未落 2024-07-29 04:32:20

将传递给 BeautifulSoup 的值转换为字符串。
对于您的情况,请对最后一行进行以下编辑:

soup = BeautifulSoup(str(cleaned_html))

Convert the value passed to BeautifulSoup into a string.
In your case, do the following edit to the last line:

soup = BeautifulSoup(str(cleaned_html))
~没有更多了~
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