I have spent plenty of time as far as I am newbie in Python.
How could I ever decode such a URL:
example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0to this one in python 2.7: example.com?title==правовая+защита
url=urllib.unquote(url.encode("utf8")) is returning something very ugly.
Still no solution, any help is appreciated.
14 Answers
The data is UTF-8 encoded bytes escaped with URL quoting, so you want to decode, with urllib.parse.unquote(), which handles decoding from percent-encoded data to UTF-8 bytes and then to text, transparently:
from urllib.parse import unquote
url = unquote(url)Demo:
>>> from urllib.parse import unquote
>>> url = 'example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0'
>>> unquote(url)
'example.com?title=правовая+защита'The Python 2 equivalent is urllib.unquote(), but this returns a bytestring, so you'd have to decode manually:
from urllib import unquote
url = unquote(url).decode('utf8') 2 If you are using Python 3, you can use urllib.parse
url = """example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0"""
import urllib.parse
urllib.parse.unquote(url)gives:
'example.com?title=правовая+защита' 1 You can achieve an expected result with requests library as well:
import requests
url = ""
print(f"Before: {url}")
print(f"After: {requests.utils.unquote(url)}")Output:
$ python3 test_url_unquote.py
Before:
After: Set.zipMight be handy if you are already using requests, without using another library for this job.
In HTML the URLs can contain html entities. This replaces them, too.
#from urllib import unquote #earlier python version
from urllib.request import unquote
from html import unescape
unescape(unquote(')) 3