Hex to plain ASCII? Python 2.7.13

Question

Im trying to turn this: %73%6c%61%70%72%69%73%65%40%6c%69%65%6e%6d%75%6c%74%69%6d%65%64%69%61%2e%63%6f%6d

into this: slaprise@lienmultimedia.com

and my brain is exploding.. Any help would be appreciated.

Thank you

It is not hex encoded string. It is url encoded string. – narendra-choudhary Apr 29 '20 at 06:30 — narendra-choudhary, Apr 29 '20 at 06:30

score 1 · Accepted Answer · answered Apr 29 '20 at 06:28

1

Python 2.7.17 (should work for Python 2.7.13)

import urllib2

url = urllib2.unquote("%73%6c%61%70%72%69%73%65%40%6c%69%65%6e%6d%75%6c%74%69%6d%65%64%69%61%2e%63%6f%6d")
print(url)
# slaprise@lienmultimedia.com

answered Apr 29 '20 at 06:28

narendra-choudhary

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If the answer solved the problem you were having, [please accept it](https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work). – narendra-choudhary Apr 30 '20 at 01:39

CodeCop · Answer 2 · 2020-04-30T00:20:29.850

0

You are trying to URL-decode that string, use urllib:

from urllib import unquote

url = unquote("%73%6c%61%70%72%69%73%65%40%6c%69%65%6e%6d%75%6c%74%69%6d%65%64%69%61%2e%63%6f%6d")
# url = slaprise@lienmultimedia.com

edited Apr 30 '20 at 00:20

answered Apr 29 '20 at 06:24

CodeCop

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Your solution will not work for Python2.7 – narendra-choudhary Apr 29 '20 at 06:31
@narendra-choudhary I've updated my code, btw I think he can still use urllib – CodeCop Apr 29 '20 at 06:32

Hex to plain ASCII? Python 2.7.13

2 Answers2