Python xpath表达式 Python xpath表达式怎样实现数据处理

<bookstore>
<book>
 <title lang="eng">Harry Potter</title>
 <price>999</price>
</book>
<book>
 <title lang="eng">Learning XML</title>
 <price>888</price>
</book>
</bookstore>

from lxml import etree
text = ''' <div> <ul> 
    <li class="item-1"><a href="link1.html" 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> </div> '''

html = etree.HTML(text)

#获取href的列表和title的列表
href_list = html.xpath("//li[@class='item-1']/a/@href")
title_list = html.xpath("//li[@class='item-1']/a/text()")


#组装成字典
for href in href_list:
  item = {}
  item["href"] = href
  item["title"] = title_list[href_list.index(href)]
  print(item)

# 如果取到的是一个节点，返回的是element对象，可以继续使用xpath方法，对此我们可以在后面的数据提取过程中：先根据某个标签进行分组，分组之后再进行数据的提取
li_list = html.xpath("//li[@class='item-1']")

#在每一组中继续进行数据的提取
for li in li_list:
  item = {}
  item["href"] = li.xpath("./a/@href")[0] if len(li.xpath("./a/@href"))>0 else None
  item["title"] = li.xpath("./a/text()")[0] if len(li.xpath("./a/text()"))>0 else None
  print(item)