Python BeautifulSoup:如何找到所有最低级别的div(不包含嵌套div的div)?

[d for d in soup.findAll('div') if not d.find('div')]
#[<div id="one"><p>1. Get this div!</p></div>, 
# <div id="three"><p>2. Get this div!</p></div>, 
# <div id="four"><p>3. Get this div!</p></div>]

最简单的方法是用所需的id创建一个列表,然后使用 re.compile :

from bs4 import BeautifulSoup as soup
import re
ids = ['one', 'three', 'four']
results = soup(content, 'html.parser').find_all('div', {'id':re.compile('|'.join(ids))})
for i in results:
 print(i)
 print('-'*20)

输出:

<div id="one">
<p>1. Get this div!</p>
</div>
--------------------
<div id="three">
<p>2. Get this div!</p>
</div>
--------------------
<div id="four">
<p>3. Get this div!</p>
</div>
--------------------

def get_ids(_d):
  if not any(getattr(i, '__dict__', {}).get('name') == 'div' for i in _d.__dict__['contents']):
     return _d 
  _r = [get_ids(i) for i in _d.__dict__['contents'] if getattr(i, '__dict__', {}).get('name') == 'div']
  return None if not _r else _r[0]

final_results = []
for i in [get_ids(i) for i in soup(content, 'html.parser').find_all('div')]:
  if i not in s:
     s.append(i)

print(final_results)

输出:

[<div id="one"><p>1. Get this div!</p></div>, <div id="three"><p>2. Get this div!</p></div>, <div id="four"><p>3. Get this div!</p></div>]