Question

我正在处理带有子标记的HTML元素，我想“忽略”或删除它们，以便文本仍然存在。刚才，如果我尝试.string任何带标签的元素，我得到的只是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.

我希望第二行为This is a paragraph with a tag.。我该怎么做？

Answer 1

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

而且，你会得到：

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

Answer 2

请改用.strings iterable。使用''.join()拉入所有字符串并将它们连接在一起：

print ''.join(main.strings)

迭代.strings会直接或在子标记中生成每个包含字符串。

演示：

>>> print ''.join(main.strings)

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

获取Beautiful Soup元素的.string时如何忽略标记？

2 个答案: