make url from umlaut words in python

Question

I have this sample code. I read about how superior the Requests is. But I wrote this long time ago with urllib2.

My question is:

When I run this code in Spyder it works nice. But when I run it from command line (windows 8 64bit) using python mycode.py it throws error. May anybody please guide me how can I get through this problem? I need to run it from command line. Thanks a lot.

# -*- coding: utf-8 -*-

import urllib2 

city = u'Köln'


def make_url(word):
    Word = unicode(word)
    print type(Word)    
    url_Word = urllib2.quote(Word, "utf-8")
    print "\ntype = %s \n" %type(Word)
    print "url_word = %s \n" %url_Word


make_url(city)

---

result from Spyder:

<type 'unicode'>

type = <type 'unicode'> 

url_word = K%F6ln 

---

result from command line:

<type 'unicode'>
C:\Python27\lib\urllib.py:1285: UnicodeWarning: Unicode equal comparison failed
to convert both arguments to Unicode - interpreting them as being unequal
  return ''.join(map(quoter, s))
Traceback (most recent call last):
  File "to_Ask_question.py", line 21, in <module>
    make_url(city)
  File "to_Ask_question.py", line 16, in make_url
    url_Word = urllib2.quote(Word, "utf-8")
  File "C:\Python27\lib\urllib.py", line 1285, in quote
    return ''.join(map(quoter, s))
KeyError: u'\xf6'

Answer 1

I got idea from this question.

By just converting to string before feeding to urllib2.quote()

Here is the code:

import urllib2 
import codecs

city = u'Köln'

def make_url(word):
    word = codecs.encode(word,'utf-8')
    print type(word)    
    url_Word = urllib2.quote(word)
    print "\ntype = %s \n" %type(word)
    print "url_word = %s \n" %url_Word


make_url(city)