On 10Jul2010 09:11, pavan kumar maddali wrote:
| Thank You Steve,
| I am not using the urllib.
Your example did:
from urllib.request import urlopen
response = urlopen('http://python.org/') html =
response.read()
| I am using the xmlrpc and http modules from the
| python library.
[.
est of the code is just to get the result from the api
regards
Pavan
From: Steven D'Aprano
To: python-list@python.org
Sent: Sat, July 10, 2010 11:35:06 AM
Subject: Re: About new urllib.request in python 3.1.2
On Sat, 10 Jul 2010 07:06:47 -0700, pcchen wr
On Sat, 10 Jul 2010 07:06:47 -0700, pcchen wrote:
> And for the following three simple lines of code, borrowed from official
> python-doc 3.1.2:
>
from urllib.request import urlopen
response = urlopen('http://python.org/') html = response.read()
>
> They could cause this error:
>
> Fi
Sorry I have searched related topic but none could point out this
problem.
I am currently using python 3.1.2:
>>>Python 3.1.2 (r312:79147, Jun 30 2010, 11:58:11)
>>>[GCC 4.2.1 20070719 [FreeBSD]] on freebsd8
And for the following three simple lines of code, borrowed from
official python-doc 3.1
