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。

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()

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())

我很感激其他启发我的答案，但它们对我不起作用。

在花了几个小时试图弄清楚它是如何工作的之后，我用最新的PyCryptodomex库提出了下面的实现(这是另一个故事，我如何设法在代理后面设置它，在Windows上，在virtualenv中…唷)

它正在处理您的实现。记得写下填充、编码和加密步骤(反之亦然)。你必须打包和拆包，记住顺序。

import base64
import hashlib
from Cryptodome.Cipher import AES
from Cryptodome.Random import get_random_bytes

__key__ = hashlib.sha256(b'16-character key').digest()

def encrypt(raw):
    BS = AES.block_size
    pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS)

    raw = base64.b64encode(pad(raw).encode('utf8'))
    iv = get_random_bytes(AES.block_size)
    cipher = AES.new(key= __key__, mode= AES.MODE_CFB,iv= iv)
    return base64.b64encode(iv + cipher.encrypt(raw))

def decrypt(enc):
    unpad = lambda s: s[:-ord(s[-1:])]

    enc = base64.b64decode(enc)
    iv = enc[:AES.block_size]
    cipher = AES.new(__key__, AES.MODE_CFB, iv)
    return unpad(base64.b64decode(cipher.decrypt(enc[AES.block_size:])).decode('utf8'))

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:] ))