在python中的指定html块之间提取文本

Question

我有下面一段 html，只需要从中提取文本

<p>Current</p> and <p>Archive</p>

Html 块看起来像：

<p>Current</p>
<a href="some link to somewhere 1">File1</a>
<br>
<a href="some link to somewhere 2">File2</a>
<br>
<a href="some link to somewhere 3">File3</a>
<br>
<p>Archive</p>
<a href="Some another link to another file">Some another file</a>

因此所需的输出应该类似于 File1、File2、File3。

这是我迄今为止尝试过的

import re
m = re.compile('<p>Current</p>(.*?)<p>Archive</p>').search(text)

但没有按预期工作。

有没有简单的解决方案如何在python中指定的html标签块之间提取文本？

Answer 1

如果您坚持使用正则表达式，您可以将它与 list comp 结合使用，如下所示：

chunk="""<p>Current</p>
<a href="some link to somewhere 1">File1</a>
<br>
<a href="some link to somewhere 2">File2</a>
<br>
<a href="some link to somewhere 3">File3</a>
<br>
<p>Archive</p>
<a href="Some another link to another file">Some another file</a>"""

import re 

# find all things between > and < the shorter the better  
found = re.findall(r">(.+?)<",chunk) 

# only use the stuff after "Current" before "Archive"
found[:] = found[ found.index("Current")+1:found.index("Archive")]

print(found) # python 3 syntax, remove () for python2.7 

输出：

['File1', 'File2', 'File3']

Answer 2

from bs4 import BeautifulSoup as bs


html_text = """
<p>Current</p>
<a href="some link to somewhere 1">File1</a>
<br>
<a href="some link to somewhere 2">File2</a>
<br>
<a href="some link to somewhere 3">File3</a>
<br>
<p>Archive</p>
<a href="Some another link to another file">Some another file</a>"""

a_tag = soup.find_all("a")

text = []
for i in a_tag:
   text.append(get_text())

print (text)

输出：

['File1', 'File2', 'File3', 'Some another file']

BeautifulSoup 库对于解析 html 文件并从中获取文本非常有用。