我试图使用PyCrypto构建两个函数,它们接受两个参数:消息和密钥,然后加密/解密消息。

我在网上找到了几个链接来帮助我,但每一个都有缺陷:

codekoala的这个使用os。这是PyCrypto不鼓励的。

此外,我给函数的键不能保证具有预期的确切长度。我要怎么做才能做到呢?

还有,有几种模式,推荐哪种模式?我不知道用什么:/

最后,静脉注射到底是什么?我是否可以提供不同的IV来加密和解密,或者这将返回不同的结果?


当前回答

让我来回答你关于“模式”的问题。AES-256是一种分组密码。它以一个32字节的键和一个16字节的字符串(称为块)作为输入,并输出一个块。为了加密,我们在操作模式中使用AES。上面的解决方案建议使用CBC,这是一个例子。另一种叫做CTR,它更容易使用:

from Crypto.Cipher import AES
from Crypto.Util import Counter
from Crypto import Random

# AES supports multiple key sizes: 16 (AES128), 24 (AES192), or 32 (AES256).
key_bytes = 32

# Takes as input a 32-byte key and an arbitrary-length plaintext and returns a
# pair (iv, ciphtertext). "iv" stands for initialization vector.
def encrypt(key, plaintext):
    assert len(key) == key_bytes

    # Choose a random, 16-byte IV.
    iv = Random.new().read(AES.block_size)

    # Convert the IV to a Python integer.
    iv_int = int(binascii.hexlify(iv), 16)

    # Create a new Counter object with IV = iv_int.
    ctr = Counter.new(AES.block_size * 8, initial_value=iv_int)

    # Create AES-CTR cipher.
    aes = AES.new(key, AES.MODE_CTR, counter=ctr)

    # Encrypt and return IV and ciphertext.
    ciphertext = aes.encrypt(plaintext)
    return (iv, ciphertext)

# Takes as input a 32-byte key, a 16-byte IV, and a ciphertext, and outputs the
# corresponding plaintext.
def decrypt(key, iv, ciphertext):
    assert len(key) == key_bytes

    # Initialize counter for decryption. iv should be the same as the output of
    # encrypt().
    iv_int = int(iv.encode('hex'), 16)
    ctr = Counter.new(AES.block_size * 8, initial_value=iv_int)

    # Create AES-CTR cipher.
    aes = AES.new(key, AES.MODE_CTR, counter=ctr)

    # Decrypt and return the plaintext.
    plaintext = aes.decrypt(ciphertext)
    return plaintext

(iv, ciphertext) = encrypt(key, 'hella')
print decrypt(key, iv, ciphertext)

这通常被称为AES-CTR。我建议谨慎使用AES-CBC与PyCrypto。原因是它要求您指定填充方案,正如给出的其他解决方案所示。一般来说,如果不小心填充,就会出现完全破坏加密的攻击!

现在,重要的是要注意键必须是一个随机的32字节字符串;密码是不够的。通常,键的生成是这样的:

# Nominal way to generate a fresh key. This calls the system's random number
# generator (RNG).
key1 = Random.new().read(key_bytes)

密钥也可以由密码派生:

# It's also possible to derive a key from a password, but it's important that
# the password have high entropy, meaning difficult to predict.
password = "This is a rather weak password."

# For added # security, we add a "salt", which increases the entropy.
#
# In this example, we use the same RNG to produce the salt that we used to
# produce key1.
salt_bytes = 8
salt = Random.new().read(salt_bytes)

# Stands for "Password-based key derivation function 2"
key2 = PBKDF2(password, salt, key_bytes)

上面的一些解决方案建议使用SHA-256来获得密钥,但这通常被认为是糟糕的加密实践。 查看维基百科了解更多操作模式。

其他回答

为了其他人的利益,这里是我的解密实现,我通过组合@Cyril和@Marcus的答案得到的。这假设它是通过HTTP请求传入的,加密文本加引号,base64编码。

import base64
import urllib2
from Crypto.Cipher import AES


def decrypt(quotedEncodedEncrypted):
    key = 'SecretKey'

    encodedEncrypted = urllib2.unquote(quotedEncodedEncrypted)

    cipher = AES.new(key)
    decrypted = cipher.decrypt(base64.b64decode(encodedEncrypted))[:16]

    for i in range(1, len(base64.b64decode(encodedEncrypted))/16):
        cipher = AES.new(key, AES.MODE_CBC, base64.b64decode(encodedEncrypted)[(i-1)*16:i*16])
        decrypted += cipher.decrypt(base64.b64decode(encodedEncrypted)[i*16:])[:16]

    return decrypted.strip()

下面是我的实现,它通过一些修复为我工作。它将密钥和秘密短语的对齐增强为32字节,IV为16字节:

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

class AESCipher(object):

    def __init__(self, key):
        self.bs = AES.block_size
        self.key = hashlib.sha256(key.encode()).digest()

    def encrypt(self, raw):
        raw = self._pad(raw)
        iv = Random.new().read(AES.block_size)
        cipher = AES.new(self.key, AES.MODE_CBC, iv)
        return base64.b64encode(iv + cipher.encrypt(raw.encode()))

    def decrypt(self, enc):
        enc = base64.b64decode(enc)
        iv = enc[:AES.block_size]
        cipher = AES.new(self.key, AES.MODE_CBC, iv)
        return self._unpad(cipher.decrypt(enc[AES.block_size:])).decode('utf-8')

    def _pad(self, s):
        return s + (self.bs - len(s) % self.bs) * chr(self.bs - len(s) % self.bs)

    @staticmethod
    def _unpad(s):
        return s[:-ord(s[len(s)-1:])]

PyCrypto已经过时了。

如今,密码学有了更好的支持。

这是另一个实现。注意,这将返回字节,您需要使用base64将它们转换为用于传输的字符串。

import os
import hashlib
from cryptography.hazmat.primitives.ciphers import Cipher, algorithms, modes
from cryptography.hazmat.backends import default_backend

_BLOCK_SIZE = 16

class AesStringCipher:
    def __init__(self, key):
        self._key = hashlib.sha256(key.encode()).digest()

    def encrypt_str(self, raw:str) -> bytes:
        iv = os.urandom(_BLOCK_SIZE)
        cipher = Cipher(algorithms.AES(self._key), modes.CBC(iv), default_backend())
        encryptor = cipher.encryptor()
        raw = _pad(raw)
        return iv + encryptor.update(raw.encode('utf-8')) + encryptor.finalize()

    def decrypt_str(self, enc:bytes) -> str:
        iv = enc[:_BLOCK_SIZE]
        enc = enc[_BLOCK_SIZE:]
        cipher = Cipher(algorithms.AES(self._key), modes.CBC(iv), default_backend())
        decryptor = cipher.decryptor()
        raw = decryptor.update(enc) + decryptor.finalize()
        raw = raw.decode('utf-8')
        return _unpad(raw)

def _pad(s:str) -> str:
    padding = (_BLOCK_SIZE - (len(s) % _BLOCK_SIZE))
    return s + padding * chr(padding)

def _unpad(s:str) -> str:
    return s[:-ord(s[len(s)-1:])]


if __name__ == '__main__':
    cipher = AesStringCipher('my secret password')

    secret_msg = 'this is a super secret msg ...'
    enc_msg = cipher.encrypt_str(secret_msg)
    dec_msg = cipher.decrypt_str(enc_msg)

    assert secret_msg == dec_msg

您可能需要以下两个函数:当输入长度不是BLOCK_SIZE的倍数时,pad- to pad(加密时)和unpad- to unpad(解密时)。

BS = 16
pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS)
unpad = lambda s : s[:-ord(s[len(s)-1:])]

你问的是键的长度?您可以使用该密钥的MD5哈希,而不是直接使用它。

而且,根据我使用PyCrypto的一点经验,当输入相同时,IV用于混合加密的输出,因此IV被选择为随机字符串,并将其用作加密输出的一部分,然后使用它来解密消息。

这是我的实现:

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

class AESCipher:
    def __init__( self, key ):
        self.key = key

    def encrypt( self, raw ):
        raw = pad(raw)
        iv = Random.new().read( AES.block_size )
        cipher = AES.new( self.key, AES.MODE_CBC, iv )
        return base64.b64encode( iv + cipher.encrypt( raw ) )

    def decrypt( self, enc ):
        enc = base64.b64decode(enc)
        iv = enc[:16]
        cipher = AES.new(self.key, AES.MODE_CBC, iv )
        return unpad(cipher.decrypt( enc[16:] ))
from Crypto import Random
from Crypto.Cipher import AES
import base64

BLOCK_SIZE=16
def trans(key):
     return md5.new(key).digest()

def encrypt(message, passphrase):
    passphrase = trans(passphrase)
    IV = Random.new().read(BLOCK_SIZE)
    aes = AES.new(passphrase, AES.MODE_CFB, IV)
    return base64.b64encode(IV + aes.encrypt(message))

def decrypt(encrypted, passphrase):
    passphrase = trans(passphrase)
    encrypted = base64.b64decode(encrypted)
    IV = encrypted[:BLOCK_SIZE]
    aes = AES.new(passphrase, AES.MODE_CFB, IV)
    return aes.decrypt(encrypted[BLOCK_SIZE:])