(Python) socket.gaierror on every addres...except http://www.reddit.com?

I'm just playing around and I'm trying to grab information from websites. Unfortunately, with the following code:

import sys
import socket
import re
from urlparse import urlsplit

url = urlsplit(sys.argv[1])


sock = socket.socket()
sock.connect((url[0] + '://' + url[1],80))
path = url[2]
if not path:
    path = '/'

print path
sock.send('GET ' + path + ' HTTP/1.1\r\n'
    + 'User-Agent: Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US) AppleWebKit/525.19 (KHTML, like Gecko) Chrome/0.3.154.9 Safari/525.19\r\n'
    + 'Accept: */*\r\n'
    + 'Accept-Language: en-US,en\r\n'
    + 'Accept-Charset: ISO-8859-1,*,utf-8\r\n'
    + 'Host: 68.33.143.182\r\n'
    + 'Connection: Keep-alive\r\n'
    + '\r\n')

I get the following error:

Traceback (most recent call last):
File "D:\Development\Python\PyCrawler\PyCrawler.py", line 10, in sock.connect((url[0] + '://' + url[1],80)) File "", line 1, in connect socket.gaierror: (11001, 'getaddrinfo failed')

The only time I do not get an error is if the url passed is http://www.reddit.com. Every other url I have tried comes up with the socket.gaierror. Can anyone explain this? And possibly give a solution?

From stackoverflow

• you forgot to resolve the hostname:

  addr = socket.gethostbyname(url[1])
  ...
  sock.connect((addr,80))
  

• sock.connect((url[0] + '://' + url[1],80))
  

  Do not do that, instead do this:

  sock.connect((url[1], 80))
  

  connect expects a hostname, not a URL.

  Actually, you should probably use something higher-level than sockets to do HTTP. Maybe httplib.

  The.Anti.9 : I've tried that too. It gives me Access Denied errors everywhere.

• Please please please please please please please don't do this.

  urllib and urllib2 are your friends.

  Read the "missing" urllib2 manual if you are having trouble with it.

• Have you ever altered your Hosts file? If it has an entry for Reddit but not much else, that might explain that site's unique result.

• Use urllib2. Or BeautifulSoup.