Question

我正在创建一个从任何网页获取图片网址的代码，代码在python中并使用BeutifulSoup和httplib2。当我运行代码时，我得到下一个错误：

Look me http://movies.nytimes.com          (this line is printed by the code)
Traceback (most recent call last):
File "main.py", line 103, in <module>
visit(initialList,profundidad)
File "main.py", line 98, in visit
visit(dodo[indice], bottom -1)
File "main.py", line 94, in visit
getImages(w)
File "main.py", line 34, in getImages
iSoupList = BeautifulSoup(response, parseOnlyThese=SoupStrainer('img'))
File "/usr/local/lib/python2.6/dist-packages/BeautifulSoup.py", line 1499, in __init__
BeautifulStoneSoup.__init__(self, *args, **kwargs)
File "/usr/local/lib/python2.6/dist-packages/BeautifulSoup.py", line 1230, in __init__
self._feed(isHTML=isHTML)
File "/usr/local/lib/python2.6/dist-packages/BeautifulSoup.py", line 1263, in _feed
self.builder.feed(markup)
File "/usr/lib/python2.6/HTMLParser.py", line 108, in feed
self.goahead(0)
File "/usr/lib/python2.6/HTMLParser.py", line 148, in goahead
k = self.parse_starttag(i)
File "/usr/lib/python2.6/HTMLParser.py", line 226, in parse_starttag
endpos = self.check_for_whole_start_tag(i)
File "/usr/lib/python2.6/HTMLParser.py", line 301, in check_for_whole_start_tag
self.error("malformed start tag")
File "/usr/lib/python2.6/HTMLParser.py", line 115, in error
raise HTMLParseError(message, self.getpos())
HTMLParser.HTMLParseError: malformed start tag, at line 942, column 118

有人可以解释我如何修复或制作错误的例外

Answer 1

您使用的是BeautifulSoup的最新版本吗？这似乎是版本3.1.x的一个已知问题，因为它开始使用一个新的解析器（HTMLParser，而不是SGMLParser），它在处理格式错误的HTML时更糟糕。您可以在BeautifulSoup website上找到有关此内容的更多信息作为一种快速解决方案，您只需使用旧版本（3.0.7a）。

Answer 2

要专门捕获该错误，请将代码更改为：

try:
    iSoupList = BeautifulSoup(response, parseOnlyThese=SoupStrainer('img'))

except HTMLParseError:
    #Do something intelligent here

除了块之外，还有一些关于Python的尝试的更多内容： http://docs.python.org/tutorial/errors.html

Answer 3

我在HTML文档中输入字符串 =＆amp; 时出错了。当我替换该字符串时（在我的情况下使用 =和），我不再收到该解析错误。

如何修复或使此错误成为例外

3 个答案: