替换文本而不在BeautifulSoup中转义 [英] Replace text without escaping in BeautifulSoup
问题描述
我想在BeautifulSoup中包装一些尚未与锚链接链接的单词.我用它来实现它:
I would like to wrap some words that are not already links with anchor links in BeautifulSoup. I use this to achieve it:
from bs4 import BeautifulSoup
import re
text = ''' replace this string '''
soup = BeautifulSoup(text)
pattern = 'replace'
for txt in soup.findAll(text=True):
if re.search(pattern,txt,re.I) and txt.parent.name != 'a':
newtext = re.sub(r'(%s)' % pattern,
r'<a href="#\1">\1</a>',
txt)
txt.replaceWith(newtext)
print(soup)
不幸的是返回
<html><body><p><a href="#replace">replace</a> this string </p></body></html>
我要寻找的地方:
<html><body><p><a href="#replace">replace</a> this string </p></body></html>
有没有一种方法可以告诉BeautifulSoup不要逃脱链接元素?
Is there a way in which I can tell BeautifulSoup not to escape the link elements?
这里要替换的简单正则表达式将不起作用，因为我最终不仅将拥有一个我想替换的模式，而且拥有多个模式.这就是为什么我决定使用BeautifulSoup排除已经是链接的所有内容的原因.
A simple regex to replace will not do here because I will eventually not only have one pattern that I want to replace but multiple. This is why I decided to use BeautifulSoup to exclude everything that already is a link.
推荐答案
You need to create new tag using new_tag use insert_after to insert part of your text after your newly created a tag.
for txt in soup.find_all(text=True):
if re.search(pattern, txt, re.I) and txt.parent.name != 'a':
newtag = soup.new_tag('a')
newtag.attrs['href'] = "#{}".format(pattern)
newtag.string = pattern
txt.replace_with(newtag)
newtag.insert_after(txt.replace(pattern, ""))
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