这是另一种观点(主要源自上述解决方案)，但是

uses null for padding does not use lambda (never been a fan) tested with python 2.7 and 3.6.5 #!/usr/bin/python2.7 # you'll have to adjust for your setup, e.g., #!/usr/bin/python3 import base64, re from Crypto.Cipher import AES from Crypto import Random from django.conf import settings class AESCipher: """ Usage: aes = AESCipher( settings.SECRET_KEY[:16], 32) encryp_msg = aes.encrypt( 'ppppppppppppppppppppppppppppppppppppppppppppppppppppppp' ) msg = aes.decrypt( encryp_msg ) print("'{}'".format(msg)) """ def __init__(self, key, blk_sz): self.key = key self.blk_sz = blk_sz def encrypt( self, raw ): 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('utf-8') iv = Random.new().read( AES.block_size ) cipher = AES.new( self.key.encode('utf-8'), AES.MODE_CBC, iv ) return base64.b64encode( iv + cipher.encrypt( raw ) ).decode('utf-8') def decrypt( self, enc ): if enc is None or len(enc) == 0: raise NameError("No value given to decrypt") enc = base64.b64decode(enc) iv = enc[:16] cipher = AES.new(self.key.encode('utf-8'), AES.MODE_CBC, iv ) return re.sub(b'\x00*$', b'', cipher.decrypt( enc[16:])).decode('utf-8')

你可以使用新的django-mirage-field包。

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:])]

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())

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:])

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。