如何在Python中解密来自JavaScript CryptoJS.AES.encrypt(密码,密码)的密码

Question

我有一个密码,可以通过JavaScript加密

  var password = 'sample'
  var passphrase ='sample_passphrase'
  CryptoJS.AES.encrypt(password, passphrase)

然后我尝试解密密码来自Python中的JavaScript:

  from Crypto.Cipher import AES
  import base64

  PADDING = '\0'

  pad_it = lambda s: s+(16 - len(s)%16)*PADDING
  key = 'sample_passphrase'
  iv='11.0.0.101'        #------> here is my question, how can I get this iv to restore password, what should I put here?
  key=pad_it(key)        #------> should I add padding to keys and iv?
  iv=pad_it(iv)          ##
  source = 'sample'
  generator = AES.new(key, AES.MODE_CFB,iv)
  crypt = generator.encrypt(pad_it(source))
  cryptedStr = base64.b64encode(crypt)
  print cryptedStr
  generator = AES.new(key, AES.MODE_CBC,iv)
  recovery = generator.decrypt(crypt)
  print recovery.rstrip(PADDING)

我从浏览器控制台检查了JS,它显示IV in CryptoJS.AES.encrypt(password, passphrase)是一个具有某些属性的对象(如sigBytes:16, words: [-44073646, -1300128421, 1939444916, 881316061]).它似乎是随机生成的.

从一个网页,它告诉我JS有两种加密密码的方法(参考链接):

• 一个. crypto.createCipher(algorithm, password)
• 湾 crypto.createCipheriv(algorithm, key, iv)

我在JavaScript中看到的应该是选项a.但是,只有选项b相当于python中的AES.new().

问题是:

1. 如何在不更改JavaScript代码的情况下在Python中恢复此密码？

2. 如果我需要Python中的IV,我怎样才能从JavaScript中使用的密码中获取它？

Answer 1

您必须实现OpenSSL EVP_BytesToKey,因为这是CryptoJS用于从提供的密码派生密钥和IV的内容,但pyCrypto仅支持密钥+ IV类型加密.CryptoJS还生成一个随机盐,也必须发送到服务器.如果密文对象被转换为字符串,则它会自动使用包含随机盐的OpenSSL兼容格式.

var data = "Some semi-long text for testing";
var password = "some password";
var ctObj = CryptoJS.AES.encrypt(data, password);
var ctStr = ctObj.toString();

out.innerHTML = ctStr;

<script src="https://cdn.rawgit.com/CryptoStore/crypto-js/3.1.2/build/rollups/aes.js"></script>
<div id="out"></div>

可能的输出:

U2FsdGVkX1 + ATH716DgsfPGjzmvhr + 7 + + pzYfUzR + 25u0D7Z5Lw04IJ LmvPXJMpz

对于AES,PKCS#7填充和CBC模式,CryptoJS默认为256位密钥大小.AES具有128位块大小,也是IV大小.这意味着我们必须从EVP_BytesToKey请求32 + 16 = 48字节.我找到了一个半功能的实现在这里和进一步扩展它.

这是完整的Python(使用2.7和3.4测试)代码,它与CryptoJS兼容:

from Crypto import Random
from Crypto.Cipher import AES
import base64
from hashlib import md5

BLOCK_SIZE = 16

def pad(data):
    length = BLOCK_SIZE - (len(data) % BLOCK_SIZE)
    return data + (chr(length)*length).encode()

def unpad(data):
    return data[:-(data[-1] if type(data[-1]) == int else ord(data[-1]))]

def bytes_to_key(data, salt, output=48):
    # extended from https://gist.github.com/gsakkis/4546068
    assert len(salt) == 8, len(salt)
    data += salt
    key = md5(data).digest()
    final_key = key
    while len(final_key) < output:
        key = md5(key + data).digest()
        final_key += key
    return final_key[:output]

def encrypt(message, passphrase):
    salt = Random.new().read(8)
    key_iv = bytes_to_key(passphrase, salt, 32+16)
    key = key_iv[:32]
    iv = key_iv[32:]
    aes = AES.new(key, AES.MODE_CBC, iv)
    return base64.b64encode(b"Salted__" + salt + aes.encrypt(pad(message)))

def decrypt(encrypted, passphrase):
    encrypted = base64.b64decode(encrypted)
    assert encrypted[0:8] == b"Salted__"
    salt = encrypted[8:16]
    key_iv = bytes_to_key(passphrase, salt, 32+16)
    key = key_iv[:32]
    iv = key_iv[32:]
    aes = AES.new(key, AES.MODE_CBC, iv)
    return unpad(aes.decrypt(encrypted[16:]))


password = "some password".encode()
ct_b64 = "U2FsdGVkX1+ATH716DgsfPGjzmvhr+7+pzYfUzR+25u0D7Z5Lw04IJ+LmvPXJMpz"

pt = decrypt(ct_b64, password)
print("pt", pt)

print("pt", decrypt(encrypt(pt, password), password))

类似的代码可以在我的Java和PHP答案中找到.

没有HTTPS的浏览器中的JavaScript AES加密是简单的混淆,并且不提供任何真正的安全性,因为密钥必须与密文一起传输.