Python error when using urllib.open

Question

When I run this:

import urllib

feed = urllib.urlopen(\"http://www.yahoo.com\")

print feed

I get this output in the interactive window (Py

Answer 1

In python 3.0:

import urllib
import urllib.request

fh = urllib.request.urlopen(url)
html = fh.read().decode("iso-8859-1")
fh.close()

print (html)