python - Python和BeautifulSoup，没有找到'a'

Question

这是一段HTML代码（来自delicious）：

<h4>
<a rel="nofollow" class="taggedlink " href="http://imfy.us/" >Generate Secure Links with Anonymous Referers &amp; Anti-Bot Protection</a>
<span class="saverem">
  <em class="bookmark-actions">
    <strong><a class="inlinesave action" href="/save?url=http%3A%2F%2Fimfy.us%2F&amp;title=Generate%20Secure%20Links%20with%20Anonymous%20Referers%20%26%20Anti-Bot%20Protection&amp;jump=%2Fdux&amp;key=fFS4QzJW2lBf4gAtcrbuekRQfTY-&amp;original_user=dux&amp;copyuser=dux&amp;copytags=web+apps+url+security+generator+shortener+anonymous+links">SAVE</a></strong>
  </em>
</span>
</h4>

我正在尝试查找 class="inlinesave action" 的所有链接。这是代码：

sock = urllib2.urlopen('http://delicious.com/theuser')
html = sock.read()
soup = BeautifulSoup(html)
tags = soup.findAll('a', attrs={'class':'inlinesave action'})
print len(tags)

但它什么也没找到！

有什么想法吗？

谢谢

Answer 1

如果您想寻找具有您想要的这两个类的锚点，则必须使用正则表达式，我认为：

tags = soup.findAll('a', attrs={'class': re.compile(r'\binlinesave\b.*\baction\b')})

class="action inlinesave"请记住，如果类名的顺序颠倒（ ），则此正则表达式将不起作用。

以下语句应该适用于所有情况（即使它看起来很丑 imo。）：

soup.findAll('a', 
    attrs={'class': 
        re.compile(r'\baction\b.*\binlinesave\b|\binlinesave\b.*\baction\b')
    })

Answer 2

可能这个问题在版本 3.1.0中已修复，我可以解析你的，

>>> html="""<h4>
... <a rel="nofollow" class="taggedlink " href="http://imfy.us/" >Generate Secure Links with Anony
... <span class="saverem">
...   <em class="bookmark-actions">
...     <strong><a class="inlinesave action" href="/save?url=http%3A%2F%2Fimfy.us%2F&amp;title=Gen
...   </em>
... </span>
... </h4>"""
>>>
>>> from BeautifulSoup import BeautifulSoup
>>> soup = BeautifulSoup(html)
>>> tags = soup.findAll('a', attrs={'class':'inlinesave action'})
>>> print len(tags)
1
>>> tags
[<a class="inlinesave action" href="/save?url=http%3A%2F%2Fimfy.us%2F&amp;title=Generate%20Secure%
>>>

我也尝试过 BeautifulSoup 2.1.1，它根本不起作用。

Answer 3

Python 字符串方法

html=open("file").read()
for item in html.split("<strong>"):
    if "class" in item and "inlinesave action" in item:
        url_with_junk = item.split('href="')[1]
        m = url_with_junk.index('">') 
        print url_with_junk[:m]

Answer 4

You might make some forward progress using pyparsing:

from pyparsing import makeHTMLTags, withAttribute

htmlsrc="""<h4>... etc."""

atag = makeHTMLTags("a")[0]
atag.setParseAction(withAttribute(("class","inlinesave action")))

for result in atag.searchString(htmlsrc):
    print result.href

Gives (long result output snipped at '...'):

/save?url=http%3A%2F%2Fimfy.us%2F&title=Genera...+anonymous+links