Python ile Brute Force Yapılır mı?
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- 15 Ağu 2023
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| """ | |
| Password brute-force algorithm. | |
| List of most probable passwords and english names can be found, respectively, at: | |
| - SecLists/Passwords/probable-v2-top12000.txt at master · danielmiessler/SecLists | |
| - random-name/middle-names.txt at master · dominictarr/random-name | |
| Author: Raphael Vallat | |
| Date: May 2018 | |
| Python 3 | |
| """ | |
| import string | |
| from itertools import product | |
| from time import time | |
| from numpy import loadtxt | |
| def product_loop(password, generator): | |
| for p in generator: | |
| if ''.join(p) == password: | |
| print('\nPassword:', ''.join(p)) | |
| return ''.join(p) | |
| return False | |
| def bruteforce(password, max_nchar=8): | |
| """Password brute-force algorithm. | |
| Parameters | |
| ---------- | |
| password : string | |
| To-be-found password. | |
| max_nchar : int | |
| Maximum number of characters of password. | |
| Return | |
| ------ | |
| bruteforce_password : string | |
| Brute-forced password | |
| """ | |
| print('1) Comparing with most common passwords / first names') | |
| common_pass = loadtxt('probable-v2-top12000.txt', dtype=str) | |
| common_names = loadtxt('middle-names.txt', dtype=str) | |
| cp = [c for c in common_pass if c == password] | |
| cn = [c for c in common_names if c == password] | |
| cnl = [c.lower() for c in common_names if c.lower() == password] | |
| if len(cp) == 1: | |
| print('\nPassword:', cp) | |
| return cp | |
| if len(cn) == 1: | |
| print('\nPassword:', cn) | |
| return cn | |
| if len(cnl) == 1: | |
| print('\nPassword:', cnl) | |
| return cnl | |
| print('2) Digits cartesian product') | |
| for l in range(1, 9): | |
| generator = product(string.digits, repeat=int(l)) | |
| print("\t..%d digit" % l) | |
| p = product_loop(password, generator) | |
| if p is not False: | |
| return p | |
| print('3) Digits + ASCII lowercase') | |
| for l in range(1, max_nchar + 1): | |
| print("\t..%d char" % l) | |
| generator = product(string.digits + string.ascii_lowercase, | |
| repeat=int(l)) | |
| p = product_loop(password, generator) | |
| if p is not False: | |
| return p | |
| print('4) Digits + ASCII lower / upper + punctuation') | |
| # If it fails, we start brute-forcing the 'hard' way | |
| # Same as possible_char = string.printable[:-5] | |
| all_char = string.digits + string.ascii_letters + string.punctuation | |
| for l in range(1, max_nchar + 1): | |
| print("\t..%d char" % l) | |
| generator = product(all_char, repeat=int(l)) | |
| p = product_loop(password, generator) | |
| if p is not False: | |
| return p | |
| # EXAMPLE | |
| start = time() | |
| bruteforce('sunshine') # Try with '123456' or '751345' or 'test2018' | |
| end = time() | |
| print('Total time: %.2f seconds' % (end - start)) |
github'dan üsttekini buldum hazır tool olarakta hydra var
Moderatör tarafında düzenlendi:
Merhabalar.Elbette yapılabilir.Resim olarak örnek bırakıyorum.
GitHub - Antu7/python-bruteForce: Brute Force Attack Tools Using Python
Brute Force Attack Tools Using Python. Contribute to Antu7/python-bruteForce development by creating an account on GitHub.
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