How to replace URL encoding with nothing or space using .replace or regex

Question

My UI is sending %20 for space
I need to replace %20 with space
The incoming data may be string or list

Below are the condition need to satisfied

My input are input Lets say my search bar is searching below pattern

ABC,%20CDE Expected out ABC,CDE

ABC%20CDE Expected out ABC CDE

ABC%20%20%20CDE Expected out ABC CDE

[ABC,%20CDE] Expected out [ABC,CDE]

ABC,%20%20%20 Expected out ABC

[ABC,%20%20%20] Expected out [ABC]

[,%20%20%20CDE] Expected out [CDE]

,%20%20%20CDE Expected out CDE

My Code only satisfied one condition

string.replace('%20',' ')

I am searching with replace. If not regex also would help

@WiktorStribiżew `s = ',%20%20%20CDE'`; `s.replace('%20', ' ').strip() ` o/p `', CDE'` E0 `'CDE'` — , Aug 17 '20 at 09:52
`text.replace('%20', ' ').strip(", ")` - https://rextester.com/CZKU92290 — Wiktor Stribiżew, Aug 17 '20 at 10:05
Can you describe the algorithm of expected results ? There is a URL decoding. But what about the extra steps of removing spaces and `,` ? — forzagreen, Aug 17 '20 at 10:13

score 0 · Answer 1 · answered Aug 17 '20 at 10:02

0

Use `unquote` from `urlib.parse`

from urllib.parse import unquote


unquote("ABC,%20CDE")       # 'ABC, CDE'
unquote("[ABC,%20CDE]")     # '[ABC, CDE]'
unquote("ABC%20CDE")        # 'ABC CDE'
unquote("ABC%20%20%20CDE")  # 'ABC   CDE'
unquote(",%20%20%20CDE")    # ',   CDE'

Then you can use replace to replace extra spaces.

Docs: https://docs.python.org/3/library/urllib.parse.html

answered Aug 17 '20 at 10:02

forzagreen

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unquote(",%20%20%20CDE") expected is 'CDE' – Aug 17 '20 at 10:08

How to replace URL encoding with nothing or space using .replace or regex

1 Answers1

Use unquote from urlib.parse

Use `unquote` from `urlib.parse`