在获取Beautiful Soup元素的.string时如何忽略标签？

一尘不染

在获取Beautiful Soup元素的.string时如何忽略标签？

python

我正在使用带有子标记的HTML元素，这些子标记我想“忽略”或删除，以便文本仍然存在。刚才，如果我尝试.string使用带有标签的任何元素，那么我得到的仅仅是None。

import bs4

soup = bs4.BeautifulSoup("""
    <div id="main">
      <p>This is a paragraph.</p>
      <p>This is a paragraph <span class="test">with a tag</span>.</p>
      <p>This is another paragraph.</p>
    </div>
""")

main = soup.find(id='main')
for child in main.children:
    print child.string

输出：

This is a paragraph.
None
This is another paragraph.

我要第二行This is a paragraph with a tag.。我该怎么做呢？

阅读 247

2021-01-20

共1个答案

一尘不染

for child in soup.find(id='main'):
    if isinstance(child, bs4.Tag):
        print child.text

并且，您将获得：

This is a paragraph.
This is a paragraph with a tag.
This is another paragraph.

2021-01-20