使用AES-256和utf8mb4加密和解密拉丁字符和特殊字符(中文):

对于那些需要加密和解密拉丁文和特殊值(如中文)的人，这里有一个@MIkee代码的修改来完成这项任务。

记住，UTF-8本身不能处理这种类型的编码。

import base64, re
from Crypto.Cipher import AES
from Crypto import Random
from django.conf import settings

import codecs

# Make utf8mb4 recognizable.
codecs.register(lambda name: codecs.lookup('utf8') if name == 'utf8mb4' else None)


class AESCipher:

    def __init__(self, key, blk_sz):
        self.key = key
        self.blk_sz = blk_sz

    def encrypt( self, raw ):
        # raw is the main value
        if raw is None or len(raw) == 0:
            raise NameError("No value given to encrypt")
        raw = raw + '\0' * (self.blk_sz - len(raw) % self.blk_sz)
        raw = raw.encode('utf8mb4')
        # Initialization vector to avoid same encrypt for same strings.
        iv = Random.new().read( AES.block_size )
        cipher = AES.new( self.key.encode('utf8mb4'), AES.MODE_CFB, iv )
        return base64.b64encode( iv + cipher.encrypt( raw ) ).decode('utf8mb4')

    def decrypt( self, enc ):
        # enc is the encrypted value
        if enc is None or len(enc) == 0:
            raise NameError("No value given to decrypt")
        enc = base64.b64decode(enc)
        iv = enc[:16]
        # AES.MODE_CFB that allows bigger length or Latin values
        cipher = AES.new(self.key.encode('utf8mb4'), AES.MODE_CFB, iv )
        return re.sub(b'\x00*$', b'', cipher.decrypt( enc[16:])).decode('utf8mb4')

用法:

>>> from django.conf import settings
>>> from aesencryption import AESCipher
>>>
>>> aes = AESCipher(settings.SECRET_KEY[:16], 32)
>>>
>>> value = aes.encrypt('漢字')
>>>
>>> value
'hnuRwBjwAHDp5X0DmMF3lWzbjR0r81WlW9MRrWukgQwTL0ZI88oQaWvMfBM+W87w9JtSTw=='
>>> dec_value = aes.decrypt(value)
>>> dec_value
'漢字'
>>>

拉丁字母也是如此，例如ã， á， à， â， ã， ç等。

注意点

请记住，如果要在数据库中存储拉丁值，则需要将其设置为允许这种类型的数据。因此，如果您的数据库设置为utf-8，它将不接受这种类型的数据。你也需要在那里换衣服。

你可以通过使用像SHA-1或SHA-256这样的加密哈希函数(不是Python的内置哈希)从任意密码中获得密码短语。Python在其标准库中包含了对这两者的支持:

import hashlib

hashlib.sha1("this is my awesome password").digest() # => a 20 byte string
hashlib.sha256("another awesome password").digest() # => a 32 byte string

您可以使用[:16]或[:24]截断加密哈希值，它将保留其安全性，直到您指定的长度。

您可以使用类似PKCS#7填充的方案。您可以使用它来代替前面的函数来填充(进行加密时)和解封(进行解密时)。我将在下面提供完整的源代码。

import base64
import hashlib
from Crypto import Random
from Crypto.Cipher import AES
import pkcs7

class Encryption:

    def __init__(self):
        pass

    def Encrypt(self, PlainText, SecurePassword):
        pw_encode = SecurePassword.encode('utf-8')
        text_encode = PlainText.encode('utf-8')

        key = hashlib.sha256(pw_encode).digest()
        iv = Random.new().read(AES.block_size)

        cipher = AES.new(key, AES.MODE_CBC, iv)
        pad_text = pkcs7.encode(text_encode)
        msg = iv + cipher.encrypt(pad_text)

        EncodeMsg = base64.b64encode(msg)
        return EncodeMsg

    def Decrypt(self, Encrypted, SecurePassword):
        decodbase64 = base64.b64decode(Encrypted.decode("utf-8"))
        pw_encode = SecurePassword.decode('utf-8')

        iv = decodbase64[:AES.block_size]
        key = hashlib.sha256(pw_encode).digest()

        cipher = AES.new(key, AES.MODE_CBC, iv)
        msg = cipher.decrypt(decodbase64[AES.block_size:])
        pad_text = pkcs7.decode(msg)

        decryptedString = pad_text.decode('utf-8')
        return decryptedString

import StringIO
import binascii


def decode(text, k=16):
    nl = len(text)
    val = int(binascii.hexlify(text[-1]), 16)
    if val > k:
        raise ValueError('Input is not padded or padding is corrupt')

    l = nl - val
    return text[:l]


def encode(text, k=16):
    l = len(text)
    output = StringIO.StringIO()
    val = k - (l % k)
    for _ in xrange(val):
        output.write('%02x' % val)
    return text + binascii.unhexlify(output.getvalue())

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。

请看mnothic的回答。

兼容UTF-8编码:

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

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:] ))