Python Regex：将字符串中的所有网址替换为<img>和<a>标签</a>

Question

I have a string with many urls to some pages and images:

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

And I need to convert it to:

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

Image formats are unpredictable, they can be .png .JPEG etc., and any links can be found multiple times per string

I understand, that there are some strange javascript examples here, but I can not get how to convert them to python.

But I found this as a starting point:

url_regex = /(\\b(https?|ftp|file):\\/\\/[-A-Z0-9+&@#\\/%?=~_|!:,.;]*[-A-Z0-9+&@#\\/%=~_|])/ig img_regex = /^ftp|http|https?:\\/\\/(?:[az\\-]+\\.)+[az]{2,6}(?:\\/[^\\/#?]+)+\\.(?:jpe?g|gif|png)$/ig

Big thx for help

Answer 1

You can do this without regex , if you want.

stng = 'La-la-la https://example.com/ la-la-la https://example.com/example.PNG'

sentance = '{f_txt} <a href="{f_url}">{f_url}</a> {s_txt} <img src="{s_url}">'

f_txt, f_url, s_txt, s_url = stng.split()

print(sentance.format(f_txt=f_txt, f_url=f_url, s_txt=s_txt, s_url=s_url))

Output

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

Answer 2

You may use the following regular expression:

(https?.*?\\.com\\/)(\\s+[\\w-]*\\s+)(https?.*?\\.com\\/[\\w\\.]+)

• (https?.*?\\.com\\/) First capture group. Capture http or https , anything up to .com and forward slash / .
• (\\s+[\\w-]*\\s+) Second capture group. Capture whitespace, alphanumerical characters and hypens, and whitespace. You can add more characters to the character set if needed.
• (https?.*?\\.com\\/[\\w\\.]+) Third capture group. Capture http or https , anything up to .com , forward slash / , alphanumerical characters and full stop . for the extension. Again you can add more characters to the character set in this capture group if you are expecting other characters.

You can test the regex live here .

Alternatively, if you are expecting variable urls and domains you may use:

(\\w*\\:.*?\\.\\w*\\/)(\\s+[\\w-]*\\s+)(\\w*\\:?.*?\\.\\w*\\/[\\w\\.]+)

Where first and third capture groups now do match any alphanumerical characters followed by colon : , and anything up to a . , alphanumerical characters \\w and forward slash. You can test this here .

You may replace captured groups with:

<a href="\\1">\\1</a>\\2<img src="\\3">

Where \\1 , \\2 , and \\3 are backreferences to captured groups one, two and three respectively.

---

Python snippet:

>>import re
>>str = "La-la-la https://example.com/ la-la-la https://example.com/example.PNG"

>>out = re.sub(r'(https?.*?\.com\/)(\s+[\w-]*\s+)(https?.*?\.com\/[\w\.]+)',
       r'<a href="\1">\1</a>\2<img src="\3">',
       str)
>>print(out)
La-la-la <a href="https://example.com/">https://example.com/</a> la-la-la <img src="https://example.com/example.PNG">