查找<li>下href属性值为link1. html的名称为a的子节点
时间: 2024-05-02 17:23:33 浏览: 6
我可以帮您查找<li>下href属性值为link1.html的名称为a的子节点。您可以使用以下代码:
```python
from bs4 import BeautifulSoup
html = """
<ul>
<li><a href="link1.html">Link 1</a></li>
<li><a href="link2.html">Link 2</a></li>
<li><a href="link3.html">Link 3</a></li>
</ul>
"""
soup = BeautifulSoup(html, 'html.parser')
li = soup.find('li')
a = li.find('a', href='link1.html')
print(a.text) # 输出 'Link 1'
```
这段代码会输出`Link 1`,表示名称为a且href属性值为link1.html的子节点被成功找到。希望对您有所帮助!
相关问题
使用lxml技术,查找<li>标签中href属性值为link1.html的<a>标签并打印
可以使用lxml库的XPath语法来查找所有的`<li>`标签中`href`属性值为`link1.html`的`<a>`标签。以下是使用lxml和Python代码示例:
```python
from lxml import etree
# 假设HTML代码存储在变量html中
html = '<ul><li><a href="link1.html">Link 1</a></li><li><a href="link2.html">Link 2</a></li><li><a href="link3.html">Link 3</a></li></ul>'
# 使用lxml解析HTML代码
tree = etree.HTML(html)
# 使用XPath查找所有的li标签中href属性值为link1.html的a标签
a_tags = tree.xpath('//li[a/@href="link1.html"]/a')
# 打印查找到的a标签的文本内容
for a in a_tags:
print(a.text)
```
输出结果为:
```
Link 1
```
吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 吴昊 20200722041 题量: 27 满分:100.0 截止日期:2023-05-23 12:00 吴昊 39' 57'' 数据采集与网络爬虫第一次阶段测试 返回 26 交卷 已知hello.html文件的内容如下。 <html> <body> <div> <ul> <li class="item-0"><a href="link1.html">first item </a></li> <li class="item-1"><a href="link2.html">second item </a></li> <li class="item-inactive"><a href="link3.html">third item </a></li> <li class="item-0"><a href="link4.html">fourth item </a></li> <li class="item-0"><a href="link5.html">fifth item </a></li> </ul></div></body></html> 请使用lxml库分别按照如下要求查找hello.html文件中的指定节点 (1)编写程序,查找所有名称为li的节点,并输出查找的结果 (2)编写程序,查找class属性值为item-0的所有节点,并输出查找的结果。 (3)编写程序,查找<li>下href属性值为link1.html的名为a的子节点,并输出查找的结果
以下是三个问题的答案:
(1)查找所有名称为li的节点,并输出查找的结果:
```python
from lxml import etree
html = '''
<html>
<body>
<div>
<ul>
<li class="item-0"><a href="link1.html">first item </a></li>
<li class="item-1"><a href="link2.html">second item </a></li>
<li class="item-inactive"><a href="link3.html">third item </a></li>
<li class="item-0"><a href="link4.html">fourth item </a></li>
<li class="item-0"><a href="link5.html">fifth item </a></li>
</ul>
</div>
</body>
</html>
'''
# 使用lxml解析html字符串
html_tree = etree.HTML(html)
# 查找所有名称为li的节点
li_nodes = html_tree.xpath('//li')
# 输出查找结果
for li_node in li_nodes:
print(etree.tostring(li_node, encoding='utf-8').decode('utf-8'))
```
输出结果如下:
```
<li class="item-0"><a href="link1.html">first item </a></li>
<li class="item-1"><a href="link2.html">second item </a></li>
<li class="item-inactive"><a href="link3.html">third item </a></li>
<li class="item-0"><a href="link4.html">fourth item </a></li>
<li class="item-0"><a href="link5.html">fifth item </a></li>
```
(2)查找class属性值为item-0的所有节点,并输出查找的结果:
```python
from lxml import etree
html = '''
<html>
<body>
<div>
<ul>
<li class="item-0"><a href="link1.html">first item </a></li>
<li class="item-1"><a href="link2.html">second item </a></li>
<li class="item-inactive"><a href="link3.html">third item </a></li>
<li class="item-0"><a href="link4.html">fourth item </a></li>
<li class="item-0"><a href="link5.html">fifth item </a></li>
</ul>
</div>
</body>
</html>
'''
# 使用lxml解析html字符串
html_tree = etree.HTML(html)
# 查找class属性值为item-0的所有节点
item_0_nodes = html_tree.xpath('//li[@class="item-0"]')
# 输出查找结果
for item_0_node in item_0_nodes:
print(etree.tostring(item_0_node, encoding='utf-8').decode('utf-8'))
```
输出结果如下:
```
<li class="item-0"><a href="link1.html">first item </a></li>
<li class="item-0"><a href="link4.html">fourth item </a></li>
<li class="item-0"><a href="link5.html">fifth item </a></li>
```
(3)查找<li>下href属性值为link1.html的名为a的子节点,并输出查找的结果:
```python
from lxml import etree
html = '''
<html>
<body>
<div>
<ul>
<li class="item-0"><a href="link1.html">first item </a></li>
<li class="item-1"><a href="link2.html">second item </a></li>
<li class="item-inactive"><a href="link3.html">third item </a></li>
<li class="item-0"><a href="link4.html">fourth item </a></li>
<li class="item-0"><a href="link5.html">fifth item </a></li>
</ul>
</div>
</body>
</html>
'''
# 使用lxml解析html字符串
html_tree = etree.HTML(html)
# 查找<li>下href属性值为link1.html的名为a的子节点
a_node = html_tree.xpath('//li/a[@href="link1.html"]')
# 输出查找结果
print(etree.tostring(a_node[0], encoding='utf-8').decode('utf-8'))
```
输出结果如下:
```
<a href="link1.html">first item </a>
```