how to "openurl"

>>> import urllib
>>> page = urllib.urlopen('http://www.python.org/').read()
>>> print page
...

y = ['http://whatever.com/directory/', dir1]
path = "".join(y)

if os.path.isdir(newpath):
    k = urllib.urlopen ('http://www.python.org')
    open_url = k.read ()
    print "Content-Type: text/plain \n\n"
    print open_url
else:
    print "<b><blink>DOES NOT EXIST: </blink></b>"

#!/usr/bin/env python

import cgi, os, sys, urllib, urlparse

sys.stderr = sys.stdout

def view():
	
	#Load form values and make them available though the 'form' variable. Put
	#the form value 'url' into 'path'.
	
	form = cgi.FieldStorage()
	path = form['url'].value
	
	#Split path using urlparse.urlsplit().
	
	split = urlparse.urlsplit(path)
	
	if os.path.isdir(split[2][1:]):
		
		#If the directory exits then open it using urllib.urlopen().
		
		print urllib.urlopen(path).read()
	else:
		print 'Directory does not exist..'

if __name__ == '__main__':

	print 'Content-Type: text/html\n'
	view()