Python正则表达式：用HTML链接替换字符串中的所有url，但不包括.png，.gif，.jpg，.jpeg

Question

I have a big multiline string that might contain lots of very different urls in any places of the line, like:

La-la-la https://example.com/ https://example.com/ https://example.com/ la-la-la https://example.com/ la-la-la https://example.com/ la-la-la

And I need to replace them all with <a href="https://example.com/">https://example.com/</a>

Conditions:

• URL - is everything that starts with https?:// and goes till whitespace/linebreak/end of the entire string.

• But. URLs, that end strictly with .png, .gif, .jpg, .jpeg in any cases should not be matched (Later they'll be replaced them with tag)

Answer 1

You can use re.sub :

import re
def href(d, skip = ['.png', '.gif', '.jpg', '.jpeg']):
  return f'<a href="{d}">{d}</a>' if not re.findall('|'.join(skip), d) else d

s = """
La-la-la https://example.com/ https://example.com/
https://example.com/ la-la-la https://example.com/
la-la-la https://example.com/ la-la-la
"""
new_s = re.sub('https*://.*?(?=[\s$])', lambda x:href(x.group()), s)

Output:

La-la-la <a href="https://example.com/">https://example.com/</a> <a href="https://example.com/">https://example.com/</a> <a href="https://example.com/">https://example.com/</a> la-la-la <a href="https://example.com/">https://example.com/</a> la-la-la <a href="https://example.com/">https://example.com/</a> la-la-la

Answer 2

`badtags = ['.png', '.gif', 'jpg', 'jpeg']
 goodurls = ['https://', 'http://']
 for line in string:
     for word in line.strip().split():
         if(word[0:7] == 'https://' or word[0:6] == 'http://'):
             if(not word[-4:] in badtags):
                 // replace logic
`

This is a fairly simply way to do it, you may have to use a regular iterating for loop to be able to index your original array.