假设“之前”只是一个 URL 列表：
- 查找 > 替换...
- 单击
.*以启用正则表达式
- 进入
(.+)“查找内容”
- 输入
<em><a href="http://\1" target="blank">\1</a></em>“替换为”
- 点击“全部替换”
如果“之前”不是所有 URL，那么“查找内容”将更加棘手。
根据评论，这是一种（hacky）Python 方法。
文件.html
<html>
<body>
<p>
Dummy text. website.dk/info
Dummy text (website.com) Dummy text.
Dummy text. website.dk
Dummy text. www.website.com
</p>
<p>
Dummy text. <em><a href="http://website.dk/info" target="blank">website.dk/info</a></em>
Dummy text (<em><a href="http://website.com" target="blank">website.com</a></em>) dummy text.
Dummy text. <em><a href="http://website.dk" target="blank">website.dk</a></em>
Dummy text. <em><a href="http://website.com" target="blank">www.website.com</a></em>
</p>
</body>
</html>
链接链接.py
import re;
def link_links(m):
# Link all links.
return re.sub(
# Experiment with this pattern; e.g., search for "URL regex".
r'(?<=\W)((?:www\.)?\w+\.\w+(?:\/\S+)*)',
'<em><a href="http://\\1" target="blank">\1</a></em>',
m.group(0)
)
with open("file.html", "r") as html:
match_non_html_re = re.compile(r'''
(?<=>) # After a closing HTML tag
[^<]+ # Match all non-HTML
(?=<) # Ensure it is followed by an opening HTML tag (since we cannot use atomic grouping)
(?!<\/a>) # Ensure we were not within a link tag already
''', re.VERBOSE)
print re.sub(match_non_html_re, link_links, html.read())
