使用 BeautifulSoup 删除标签但保留其内容 [英] Remove a tag using BeautifulSoup but keep its contents

问题描述

目前我有这样的代码:

soup = BeautifulSoup(value)

for tag in soup.findAll(True):
    if tag.name not in VALID_TAGS:
        tag.extract()
soup.renderContents()

除了我不想扔掉无效标签内的内容.调用soup.renderContents()时如何去掉标签但保留内容?

Except I don't want to throw away the contents inside the invalid tag. How do I get rid of the tag but keep the contents inside when calling soup.renderContents()?

推荐答案

我使用的策略是，如果标签属于 NavigableString 类型，则用其内容替换标签，如果不是，则递归到它们并用 NavigableString 等替换它们的内容.试试这个:

The strategy I used is to replace a tag with its contents if they are of type NavigableString and if they aren't, then recurse into them and replace their contents with NavigableString, etc. Try this:

from BeautifulSoup import BeautifulSoup, NavigableString

def strip_tags(html, invalid_tags):
    soup = BeautifulSoup(html)

    for tag in soup.findAll(True):
        if tag.name in invalid_tags:
            s = ""

            for c in tag.contents:
                if not isinstance(c, NavigableString):
                    c = strip_tags(unicode(c), invalid_tags)
                s += unicode(c)

            tag.replaceWith(s)

    return soup

html = "<p>Good, <b>bad</b>, and <i>ug<b>l</b><u>y</u></i></p>"
invalid_tags = ['b', 'i', 'u']
print strip_tags(html, invalid_tags)

结果是:

<p>Good, bad, and ugly</p>

我在另一个问题上给出了同样的答案.似乎出现了很多.

I gave this same answer on another question. It seems to come up a lot.

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