我正在处理具有子标记的HTML元素,我想"忽略"或删除这些子标记,以便文本仍然存在。刚才,如果我尝试.string
任何带有标记的元素,我得到的都是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.
我希望第二行是This is a paragraph with a tag.
。我该怎么做?
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.
请改用.strings
可迭代函数。使用''.join()
拉入所有字符串并将它们连接在一起:
print ''.join(main.strings)
在.strings
上迭代会直接或在子标记中生成每个包含的字符串。
演示:
>>> print ''.join(main.strings)
This is a paragraph.
This is a paragraph with a tag.
This is another paragraph.