python http status code

Question

I\'m writing my own directory buster in python, and I\'m testing it against a web server of mine in a safe and secure environment. This script basically tries to retrieve common

Answer 1

I would be adviced you to use http://docs.python-requests.org/en/latest/# for http.

Answer 2

I did not found any problems with your code, except it is almost unreadable. I have rewritten it into this working snippet:

import httplib

host = 'www.google.com'
directories = ['aosicdjqwe0cd9qwe0d9q2we', 'reader', 'news']

for directory in directories:
    conn = httplib.HTTPConnection(host)
    conn.request('HEAD', '/' + directory)

    url = 'http://{0}/{1}'.format(host, directory)
    print '    Trying: {0}'.format(url)

    response = conn.getresponse()
    print '    Got: ', response.status, response.reason

    conn.close()

    if response.status == 200:
        print ("[!] The subdirectory '{0}' "
               "could be interesting.").format(directory)

Outputs:

$ python snippet.py
    Trying: http://www.google.com/aosicdjqwe0cd9qwe0d9q2we
    Got:  404 Not Found
    Trying: http://www.google.com/reader
    Got:  302 Moved Temporarily
    Trying: http://www.google.com/news
    Got:  200 OK
[!] The subdirectory 'news' could be interesting.

Also, I did use HEAD HTTP request instead of GET, as it is more efficient if you do not need the contents and you are interested only in the status code.