From the Python 3 documentation:
urllib.parse.quote(string, safe='/', encoding=None, errors=None)
Replace special characters in string using the
%xxescape. Letters, digits, and the characters'_.-~'are never quoted. By default, this function is intended for quoting the path section of a URL. The optional safe parameter specifies additional ASCII characters that should not be quoted — its default value is'/'.
That means passing '' for safe will solve your first issue:
>>> import urllib.parse
>>> urllib.parse.quote('/test')
'/test'
>>> urllib.parse.quote('/test', safe='')
'%2Ftest'
(The function quote was moved from urllib to urllib.parse in Python 3.)
By the way, have a look at urlencode.
About the second issue, there was a bug report about it and it was fixed in Python 3.
For Python 2, you can work around it by encoding as UTF-8 like this:
>>> query = urllib.quote(u"Müller".encode('utf8'))
>>> print urllib.unquote(query).decode('utf8')
Müller
From the Python 3 documentation:
urllib.parse.quote(string, safe='/', encoding=None, errors=None)
Replace special characters in string using the
%xxescape. Letters, digits, and the characters'_.-~'are never quoted. By default, this function is intended for quoting the path section of a URL. The optional safe parameter specifies additional ASCII characters that should not be quoted — its default value is'/'.
That means passing '' for safe will solve your first issue:
>>> import urllib.parse
>>> urllib.parse.quote('/test')
'/test'
>>> urllib.parse.quote('/test', safe='')
'%2Ftest'
(The function quote was moved from urllib to urllib.parse in Python 3.)
By the way, have a look at urlencode.
About the second issue, there was a bug report about it and it was fixed in Python 3.
For Python 2, you can work around it by encoding as UTF-8 like this:
>>> query = urllib.quote(u"Müller".encode('utf8'))
>>> print urllib.unquote(query).decode('utf8')
Müller
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/'
urllib - How to urlencode a querystring in Python? - Stack Overflow
How to url encode User Search Query Syntax for Management API in Python
Encode url including slashes?
URL encoding python
Python 3
In Python 3, the urllib package has been broken into smaller components. You'll use urllib.parse.quote_plus (note the parse child module)
import urllib.parse
safe_string = urllib.parse.quote_plus(...)
Python 2
What you're looking for is urllib.quote_plus:
safe_string = urllib.quote_plus('string_of_characters_like_these:$#@=?%^Q^$')
#Value: 'string_of_characters_like_these%3A%24%23%40%3D%3F%25%5EQ%5E%24'
Python 3
Use urllib.parse.urlencode:
>>> import urllib.parse
>>> f = { 'eventName' : 'myEvent', 'eventDescription' : 'cool event'}
>>> urllib.parse.urlencode(f)
eventName=myEvent&eventDescription=cool+event
Note that this does not do url encoding in the commonly used sense (look at the output). For that use urllib.parse.quote_plus.
Python 2
You need to pass your parameters into urllib.urlencode() as either a mapping (dict), or a sequence of 2-tuples, like:
>>> import urllib
>>> f = { 'eventName' : 'myEvent', 'eventDescription' : 'cool event'}
>>> urllib.urlencode(f)
'eventName=myEvent&eventDescription=cool+event'
Hello i would like to encode an url like this:
http://www.foodserv.co.za
I tried it with the following code
import urllib link = urllib.parse.quote(link) print(link)
But with that i only get: http%3A//www.foodserv.co.za
But i would like to have also the // encoded and in the end i should look like this: http%3A%2F%2Fwww.foodserv.co.za
Is there some other way or parameter to encode the url like i need it?
hello I am trying to encode the special characters in a URL so that for example:
https://www.google.com/ would equal https%3A%2F%2Fwww.google.com%2F
I tried
url = "https://www.google.com/"
url = url.encode("ascii")but I got:
b'https://www.google.com/'
and I also tried urllib:
url = "https://www.google.com/"
url = urllib.parse.urlencode(url)
but got:
line 906, in urlencode
raise TypeErrorany help would be much appreciated thank you in advance!!!!!!!!!