lxml，是python中用来处理xml和html的功能最丰富和易用的库
from lxml import etree
from lxml import html
h = '''
<html>
<head>
<meta name="content-type" content="text/html; charset=utf-8" />
<title>友情链接查询 - 站长工具</title>
<!-- uRj0Ak8VLEPhjWhg3m9z4EjXJwc -->
<meta name="Keywords" content="友情链接查询" />
<meta name="Description" content="友情链接查询" />
</head>
<body>
<h1 class="heading">Top News</h1>
<p style="font-size: 200%">World News only on this page</p>
Ah, and here's some more text, by the way.
<p>... and this is a parsed fragment ...</p>
<a href="http://www.cydf.org.cn/" rel="nofollow" target="_blank">青少年发展基金会</a>
<a href="http://www.4399.com/flash/32979.htm" target="_blank">洛克王国</a>
<a href="http://www.4399.com/flash/35538.htm" target="_blank">奥拉星</a>
<a href="http://game.3533.com/game/" target="_blank">手机游戏</a>
<a href="http://game.3533.com/tupian/" target="_blank">手机壁纸</a>
<a href="http://www.4399.com/" target="_blank">4399小游戏</a>
<a href="http://www.91wan.com/" target="_blank">91wan游戏</a>
</body>
</html>
'''
# 第一种使用方法
page = etree.HTML(h)
#hrefs = page.xpath('//a')
href = page.cssselect('a')
for href in hrefs:
print(href.attrib)
第二种使用方法
def parse_from():
tree = html.fromstring(h)
for href in tree.cssselect('a'):
#for hfre in tree.xpath('//a'):
a = href
print(a.text)
print(a.attrib)
paese_from()
parse_from()