Trying to get HTTP code. Can someone try this code for me in their Python interpretor and see why it doesn't work?

Question

import httplib
def httpCode(theurl):
    if theurl.startswith("http://"): theurl = theurl[7:]
    head = theurl[:theurl.find('/')]
    tail = theurl[theurl.find('/'):]
    response_code = 0
    conn = httplib.HTTPConnection(head)
    conn.request("HEAD",tail)
    res = conn.getresponse()
    response_code = int(res.status)
    return response_code

Basically, this function will take a URL and return its HTTP code (200, 404, etc) The error I got was:

Exception Value:  (-2, 'Name or service not known')

I must do it with this method. That is, I am usually passing in large video files. I need to get the "header" and get the HTTP code. I cannot download the file and then get the HTTP code, because it would take too long.

Python 2.6.2 (release26-maint, Apr 19 2009, 01:58:18)
[GCC 4.3.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import httplib
>>> def httpCode(theurl):
...     if theurl.startswith("http://"): theurl = theurl[7:]
...     head = theurl[:theurl.find('/')]
...     tail = theurl[theurl.find('/'):]
...     response_code = 0
...     conn = httplib.HTTPConnection(head)
...     conn.request("HEAD",tail)
...     res = conn.getresponse()
...     response_code = int(res.status)
...     print response_code
...
>>> httpCode('http://youtube.com')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 7, in httpCode
  File "/usr/lib/python2.6/httplib.py", line 874, in request
    self._send_request(method, url, body, headers)
  File "/usr/lib/python2.6/httplib.py", line 911, in _send_request
    self.endheaders()
  File "/usr/lib/python2.6/httplib.py", line 868, in endheaders
    self._send_output()
  File "/usr/lib/python2.6/httplib.py", line 740, in _send_output
    self.send(msg)
  File "/usr/lib/python2.6/httplib.py", line 699, in send
    self.connect()
  File "/usr/lib/python2.6/httplib.py", line 683, in connect
    self.timeout)
  File "/usr/lib/python2.6/socket.py", line 498, in create_connection
    for res in getaddrinfo(host, port, 0, SOCK_STREAM):
socket.gaierror: [Errno -2] Name or service not known
>>>

Answer 1

Your code worked for me, and for one other person who commented. This implies that the URL you're using is causing a problem with your parsing somehow. head and tail should both be examined in order to determine what it thinks the host is. For example:

head = theurl[:theurl.find('/')]
print head
tail = theurl[theurl.find('/'):]
print tail

Once you can see what head and tail are, you can determine if it really should be able to resolve head. For example, what if the url was:

http://myhost.com:8080/blah/blah

It would fail because of the port number.

Answer 2

As suggested in comment by Adam Crossland, you should be checking your head and tail values. In your case, without a trailing slash you end up with

head = "youtube.co"
tail = "m"

string.find returns -1 if it is not found, hence you are grabbing all but last character for head and only the last character for tail.