How to percent-encode URL parameters in Python?

浪子不回头ぞ 提交于 2019-11-26 11:03:44
Nadia Alramli

From the docs:

urllib.quote(string[, safe])

Replace special characters in string using the %xx escape. Letters, digits, and the characters '_.-' are never quoted. By default, this function is intended for quoting the path section of the URL.The optional safe parameter specifies additional characters that should not be quoted — its default value is '/'

That means passing '' for safe will solve your first issue:

>>> urllib.quote('/test')
'/test'
>>> urllib.quote('/test', safe='')
'%2Ftest'

About the second issue, there is a bug report about it here. Apparently it was fixed in python 3. You can workaround it by encoding as utf8 like this:

>>> query = urllib.quote(u"Müller".encode('utf8'))
>>> print urllib.unquote(query).decode('utf8')
Müller

By the way have a look at urlencode

Note that urllib.quote moved to urllib.parse.quote in Python3

In Python 3, urllib.quote has been moved to urllib.parse.quote and it does handle unicode by default.

>>> from urllib.parse import quote
>>> quote('/test')
'/test'
>>> quote('/test', safe='')
'%2Ftest'
>>> quote('/El Niño/')
'/El%20Ni%C3%B1o/'

My answer is similar to Paolo's answer.

I think module requests is much better. It's based on urllib3. You can try this:

>>> from requests.utils import quote
>>> quote('/test')
'/test'
>>> quote('/test', safe='')
'%2Ftest'

If you're using django, you can use urlquote:

>>> from django.utils.http import urlquote
>>> urlquote(u"Müller")
u'M%C3%BCller'

Note that changes to Python since this answer was published mean that this is now a legacy wrapper. From the Django 2.1 source code for django.utils.http:

A legacy compatibility wrapper to Python's urllib.parse.quote() function.
(was used for unicode handling on Python 2)

It is better to use urlencode here. Not much difference for single parameter but IMHO makes the code clearer. (It looks confusing to see a function quote_plus! especially those coming from other languates)

In [21]: query='lskdfj/sdfkjdf/ksdfj skfj'

In [22]: val=34

In [23]: from urllib.parse import urlencode

In [24]: encoded = urlencode(dict(p=query,val=val))

In [25]: print(f"http://example.com?{encoded}")
http://example.com?p=lskdfj%2Fsdfkjdf%2Fksdfj+skfj&val=34

Docs

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

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

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