好得很程序员自学网

<tfoot draggable='sEl'></tfoot>

全面了解Python爬虫之xlml解析库

下面小编就为大家带来一篇Python爬虫之xlml解析库(全面了解)。小编觉得挺不错的,现在就分享给大家,也给大家做个参考。一起跟随小编过来看看吧

#!/usr/bin/python
#_*_coding:utf-8_*_

from lxml import etree

text='''
<p>
 <ul>
  <li class="item-0"><a href="link1.html" rel="external nofollow" rel="external nofollow" >first item</a></li>
  <li class="item-1"><a href="link2.html" rel="external nofollow" >second item</a></li>
  <li class="item-inactive"><a href="link3.html" rel="external nofollow" >third item</a></li>
  <li class="item-1"><a href="link4.html" rel="external nofollow" >fourth item</a></li>
  <li class="item-0"><a href="link5.html" rel="external nofollow" >fifth item</a>
 </ul>
</p>
  '''

# html=etree.HTML(text) #html对象,存储在地址中,有自动修正功能
# result=etree.tostring(html) #将html对象转化为字符串

html=etree.parse('hello.html')
# result=etree.tostring(html,pretty_print=True)
# print result
print type(html)
result= html.xpath('//li')
print result
print len(result)
print type(result)
print type(result[0])
print html.xpath('//li/@class') # 获取li标签下的所有的class
print html.xpath('//li/a[@href="link1.html" rel="external nofollow" rel="external nofollow" ]') #获取li标签下href为link1的<a>标签
print html.xpath('//li//span') #获取li标签下所有的span标签
print html.xpath('//li[last()-1]/a')[0].text #获取倒数第二个元素的内容 

以上就是全面了解Python爬虫之xlml解析库的详细内容,更多请关注Gxl网其它相关文章!

查看更多关于全面了解Python爬虫之xlml解析库的详细内容...

  阅读:40次