我想用urllib2来获取我打开的网页的标题。有什么好的方法可以做到这一点呢？我需要解析网页的HTML，找到我想要的内容（现在只需要
标签，以后可能还需要其他的）。</p>
<p>有没有什么好的解析库可以用来实现这个目的？</p>
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<a target="_blank" href="/tags/%E6%A0%87%E9%A2%98%E6%8F%90%E5%8F%96"
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<p>使用 <a href="http://www.crummy.com/software/BeautifulSoup/" rel="nofollow noreferrer">Beautiful Soup</a> 这个工具。</p>
<pre><code>html = urllib2.urlopen("...").read()
from BeautifulSoup import BeautifulSoup
soup = BeautifulSoup(html)
print soup.title.string
</code></pre>
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回答于 2025-04-15 由 <a href="#"
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<p>试试这个叫做 <a href="http://www.crummy.com/software/BeautifulSoup/" rel="noreferrer">Beautiful Soup</a> 的工具：</p>
<pre><code>url = 'http://www.example.com'
response = urllib2.urlopen(url)
html = response.read()
soup = BeautifulSoup(html)
title = soup.html.head.title
print title.contents
</code></pre>
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回答于 2025-04-15 由 <a href="#"
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<p>是的，我推荐使用 <a href="http://www.crummy.com/software/BeautifulSoup/" rel="noreferrer">BeautifulSoup</a></p>
<p>如果你想获取网页的标题，可以简单地这样做：</p>
<pre><code>soup = BeautifulSoup(html)
myTitle = soup.html.head.title
</code></pre>
<p>或者</p>
<pre><code>myTitle = soup('title')
</code></pre>
<p>这个内容来自于 <a href="http://www.crummy.com/software/BeautifulSoup/documentation.html" rel="noreferrer">官方文档</a></p>
<p>它非常强大，可以处理各种杂乱的HTML代码。</p>
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