Python3 【解析库XPath】

一.XPath简介

  对网页的层级关系进行解析,XPath的选择功能十分强大,它提供了非常简洁明了的路径选择表达式。

另外,它还提供了超过100个内建函数,用于字符串、数值、时间的匹配以及节点、序列的处理等,

几乎所有的定位节点,都可以用XPath进行选择。

官网: https://www.w3.org/TR/xpath

1.XPath常用规则:

                             

二.基本的使用

from lxml import etree

text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two"><a href="link2">2</a></li>
        <li class="three"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#将网页整体补为网页结构,打开文件路径
#html = etree.parse('demo.html',etree.HTMLParser())

print(html)

#将网页转换为文本类型,为bytes
result = etree.tostring(html)

#转化为str类型
result = result.decode("utf-8")

print(result)

1.匹配选择(所有节点)

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two"><a href="link2">2</a></li>
        <li class="three"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配
result = html.xpath('//*')

print(result)

2.子节点

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two"><a href="link2">2</a></li>
        <li class="three"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配
result = html.xpath('//li/a')

print(result)

这里"/"代表的是直接的子节点,"//"代表是所有的子孙节点

 3.父节点

父节点:使用"..",也可以使用parent::代表父级

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two"><a href="link2">2</a></li>
        <li class="three"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配
#属性为link4的a标签的父级的class属性
result = html.xpath('//a[@href="link4"]/../@class')

#@表示属性
result1 = html.xpath('//a[@href="link4"]/parent::*/@class')

print(result)
print(result1)

4.文本获取

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two"><a href="link2">2</a></li>
        <li class="three"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配
#属性为link4的a标签的父级的class属性
result = html.xpath('//a[@href="link4"]/text()')

print(result)

5.属性多值匹配

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配
#contains(@属性,值)
result = html.xpath('//li[contains(@class,"three")]/a/text()')

print(result)

6.多属性匹配

 多个属性确定一个节点,这时就需要匹配多个属性

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two three" name="item"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配
#contains(@属性,值)
result = html.xpath('//li[contains(@class,"three") and @name="item"]/a/text()')

print(result)

7.按序选择

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two three" name="item"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配

#匹配第一个li
result1 = html.xpath('//li[1]/a/text()')

#最后一个倒数2
result2 = html.xpath('//li[last()-2]/a/text()')

#最后一个
result3 = html.xpath('//li[last()]/a/text()')

#小于3
result4 = html.xpath('//li[position()<3]/a/text()')


#内置函数100,http://www.w3school.com.cn/xpath/xpath_functions.asp
print(result1)
print(result2)
print(result3)
print(result4)

8.节点轴选择

   1.ancestor

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two three" name="item"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配

result = html.xpath('//li[1]/ancestor::*')


#内置函数100,http://www.w3school.com.cn/xpath/xpath_functions.asp
print(result)

     我们调用了ancestor轴,可以获取所有祖先节点。其后需要跟两个冒号,然后是节点的选择器,这里直接使用*,表示匹配所有的节点,因此返回结果是第一个li节点的所有祖先节点,包括html,body,div和ul.

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two three" name="item"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配

result = html.xpath('//li[1]/ancestor::div')


#内置函数100,http://www.w3school.com.cn/xpath/xpath_functions.asp
print(result)
View Code

我们又加了限制条件,这次在冒号后面加了div,这样得到的结果就只有div这个祖先节点了。

   2.atrribute

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two three" name="item"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配

result = html.xpath('//li[1]/attribute::*')


#内置函数100,http://www.w3school.com.cn/xpath/xpath_functions.asp
print(result)
View Code

我们调用了attribute轴,可以获取所有属性值,其后跟的选择器还是*,这代表获取节点的所有属性,返回值就是li节点的所有属性值

   3.child

我们调用了child轴,可以获取所有直接子节点。这里我们又加了限定条件,选取href
属性为link1的a节点

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1">1</a></li>
        <li class="two three" name="item"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配

result = html.xpath('//li[1]/child::a[@href="link1"]')


#内置函数100,http://www.w3school.com.cn/xpath/xpath_functions.asp
print(result)
View Code

   4.descendant

我们调用了descendant轴,可以获取所有子孙节点。这里我们又加了限制条件获取
span节点,所以返回的结果只包含span节点而不包含a节点

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1"><span>1</span></a></li>
        <li class="two three" name="item"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配

result = html.xpath('//li[1]/descendant::span')


#内置函数100,http://www.w3school.com.cn/xpath/xpath_functions.asp
print(result)
View Code

   5.following

我们调用了following轴,可以获取当前节点之后的所有节点。这里我们虽然使用的是
*匹配,但又加了索引选择,所以只获取了第二个后续节点。

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1"><span>1</span></a></li>
        <li class="two three" name="item"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配

result = html.xpath('//li[1]/following::*[2]')


#内置函数100,http://www.w3school.com.cn/xpath/xpath_functions.asp
print(result)
View Code

   6.following-siblings

from lxml import etree


text = '''
<div>
    <ul>
        <li class="one"><a href="link1"><span>1</span></a></li>
        <li class="two three" name="item"><a href="link2">2</a></li>
        <li class="three two"><a href="link3">3</a></li>
        <li class="four"><a href="link4">4</a></li>
        <li class="five"><a href="link5">5</a>
    </ul>
</div>

'''
#将文本转换为网页类型,并修复补全
html = etree.HTML(text)

#选择内容匹配

result = html.xpath('//li[1]/following-sibling::*')


#内置函数100,http://www.w3school.com.cn/xpath/xpath_functions.asp
print(result)
View Code

我们调用了following-sibling轴,可以获取当前节点之后的所有同级节点。这里我们
使用*匹配,所以获取了所有后续同级节点。

 
原文地址:https://www.cnblogs.com/Crown-V/p/12725652.html