How to URL encode in Python 3?

PythonUrlencode

Python Problem Overview


I have tried to follow the documentation but was not able to use urlparse.parse.quote_plus() in Python 3:

from urllib.parse import urlparse

params = urlparse.parse.quote_plus({'username': 'administrator', 'password': 'xyz'})

I get

>AttributeError: 'function' object has no attribute 'parse'

Python Solutions


Solution 1 - Python

You misread the documentation. You need to do two things:

  1. Quote each key and value from your dictionary, and
  2. Encode those into a URL

Luckily urllib.parse.urlencode does both those things in a single step, and that's the function you should be using.

from urllib.parse import urlencode, quote_plus

payload = {'username':'administrator', 'password':'xyz'}
result = urlencode(payload, quote_via=quote_plus)
# 'password=xyz&username=administrator'

Solution 2 - Python

For Python 3 you could try using quote instead of quote_plus:

import urllib.parse

print(urllib.parse.quote("http://www.sample.com/", safe=""))

Result:

http%3A%2F%2Fwww.sample.com%2F

Or:

from requests.utils import requote_uri
requote_uri("http://www.sample.com/?id=123 abc")

Result:

'https://www.sample.com/?id=123%20abc'

Solution 3 - Python

You’re looking for urllib.parse.urlencode

import urllib.parse

params = {'username': 'administrator', 'password': 'xyz'}
encoded = urllib.parse.urlencode(params)
# Returns: 'username=administrator&password=xyz'

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionamphibientView Question on Stackoverflow
Solution 1 - PythonAdam SmithView Answer on Stackoverflow
Solution 2 - PythonRich RajahView Answer on Stackoverflow
Solution 3 - PythonrumpelView Answer on Stackoverflow