python: encode a url with percentage signs?

Question

I am trying to convert the following url

http://www.website.com/search/si/1/doctors/Vancouver, BC

to

http://www.website.com/search/si/1/doctors/Vancouver%2C%20BC

I tried

urllib.quote('http://www.website.com/search/si/1/doctors/Vancouver, BC', '')

and it resulted in replacing everything with a percentage sign.

What's the proper way to do this?

score 1 · Accepted Answer · edited May 23 '17 at 12:27

1

Use urllib.quote() for the url path leaving everything else as is:

from urllib import quote
from urlparse import urlparse, urlunparse

url = "http://www.website.com/search/si/1/doctors/Vancouver, BC"

scheme, netloc, path, params, query, fragment = urlparse(url)
path = quote(path)
print urlunparse((scheme, netloc, path, params, query, fragment))

prints:

http://www.website.com/search/si/1/doctors/Vancouver%2C%20BC

See also:

How can I normalize a URL in python

edited May 23 '17 at 12:27

Community

1
1

answered Jul 26 '14 at 03:23

alecxe

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omg....finally a real answer – user299709 Jul 26 '14 at 03:27
No, not a real answer: imagine you have a `:` in the query string: `'http://www.website.com/Vancouver:BC'`. – Alexander Gelbukh Jul 26 '14 at 03:36
@AlexanderGelbukh good point, what about now? – alecxe Jul 26 '14 at 03:42
@alecxe Now perfect. I wonder if this can't be done with one function. – Alexander Gelbukh Jul 26 '14 at 03:51
@AlexanderGelbukh yup, `werkzeug` helps not to reinvent the wheel here , see [`url_fix()`](https://github.com/mitsuhiko/werkzeug/blob/master/werkzeug/urls.py#L481-L497). – alecxe Jul 26 '14 at 03:54

Alexander Gelbukh · Answer 2 · 2014-07-26T03:34:34.300

0

urllib.quote('www.website.com/search/si/1/doctors/Vancouver, BC')

without the second parameter AND without the protocol part http://. The second parameter is a list of safe characters NOT to be replaced, default is '/', which is OK in your case.

edited Jul 26 '14 at 03:34

answered Jul 26 '14 at 03:14

Alexander Gelbukh

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this gives me `http%3A//www.website.com/search/si/1/doctors/Vancouver%2C%20BC` – user299709 Jul 26 '14 at 03:17
You should use the path part without the protocol part. Otherwise it thinks that what you want to fetch is http://http%3A//www.website.com/search/si/1/doctors/Vancouver%2C%20BC. – Alexander Gelbukh Jul 26 '14 at 03:24
You can use [urlparse](https://docs.python.org/2/library/urlparse.html) to separate the path from the URL. – Alexander Gelbukh Jul 26 '14 at 03:27
This can work for you: `urllib.quote(url, '/:'])` – Alexander Gelbukh Jul 26 '14 at 03:31

python: encode a url with percentage signs?

2 Answers2