为什么BeautifulSoup4中标记的同级可以是字符串？

Question

乍一看，我认为.next_sibling和previous_sibling应该是兄弟标记是很自然的。但我今天玩它的时候，它就产生了类似于"\n"的NavigableString

在仔细检查its documentation之后，注意到：

In real documents, the .next_sibling or .previous_sibling of a tag will usually be a string containing whitespace. Going back to the “three sisters” document:

<a href="http://example.com/elsie" class="sister" id="link1">Elsie</a>
<a href="http://example.com/lacie" class="sister" id="link2">Lacie</a>
<a href="http://example.com/tillie" class="sister" id="link3">Tillie</a>
You might think that the .next_sibling of the first <a> tag would be the second <a> tag. But actually, it’s a string: the comma and newline that separate the first <a> tag from the second:

link = soup.a
link
# <a class="sister" href="http://example.com/elsie" id="link1">Elsie</a>

link.next_sibling
# u',\n'
The second <a> tag is actually the .next_sibling of the comma:

link.next_sibling.next_sibling
# <a class="sister" href="http://example.com/lacie" id="link2">Lacie</a>

为什么

Answer 1

文件第16页 """ “”“

我希望我回答了你的问题

Answer 2

.find_next_sibling属性用于对HTML文档进行细粒度搜索。CSS选择器不能做的事情（它们可以选择标记，而不是标记之间的字符串，例如，不能用CSS选择器选择字符串SELECT THIS：<p>some text</p>SELECT THIS<p>some text</p>）

如果要搜索同级标记，请使用find_next_sibling()方法。您还可以通过将text=True参数传递给find_next_sibling()来模拟.find_next_sibling行为：

data = '''
<a href="http://example.com/elsie" class="sister" id="link1">Elsie</a>
<a href="http://example.com/lacie" class="sister" id="link2">Lacie</a>
<a href="http://example.com/tillie" class="sister" id="link3">Tillie</a>'''


from bs4 import BeautifulSoup

soup = BeautifulSoup(data, 'html.parser')

link = soup.a
print(link)                                     # <a class="sister" href="http://example.com/elsie" id="link1">Elsie</a>
print(type(link.next_sibling))                  # <class 'bs4.element.NavigableString'>
print(link.find_next_sibling())                 # <a class="sister" href="http://example.com/lacie" id="link2">Lacie</a>
print(type(link.find_next_sibling(text=True)))  # <class 'bs4.element.NavigableString'>