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:] ))

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())

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:] ))

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。

我用过Crypto和PyCryptodomex库，它非常快…

import base64
import hashlib
from Cryptodome.Cipher import AES as domeAES
from Cryptodome.Random import get_random_bytes
from Crypto import Random
from Crypto.Cipher import AES as cryptoAES

BLOCK_SIZE = AES.block_size

key = "my_secret_key".encode()
__key__ = hashlib.sha256(key).digest()
print(__key__)

def encrypt(raw):
    BS = cryptoAES.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(cryptoAES.block_size)
    cipher = cryptoAES.new(key= __key__, mode= cryptoAES.MODE_CFB,iv= iv)
    a= base64.b64encode(iv + cipher.encrypt(raw))
    IV = Random.new().read(BLOCK_SIZE)
    aes = domeAES.new(__key__, domeAES.MODE_CFB, IV)
    b = base64.b64encode(IV + aes.encrypt(a))
    return b

def decrypt(enc):
    passphrase = __key__
    encrypted = base64.b64decode(enc)
    IV = encrypted[:BLOCK_SIZE]
    aes = domeAES.new(passphrase, domeAES.MODE_CFB, IV)
    enc = aes.decrypt(encrypted[BLOCK_SIZE:])
    unpad = lambda s: s[:-ord(s[-1:])]
    enc = base64.b64decode(enc)
    iv = enc[:cryptoAES.block_size]
    cipher = cryptoAES.new(__key__, cryptoAES.MODE_CFB, iv)
    b=  unpad(base64.b64decode(cipher.decrypt(enc[cryptoAES.block_size:])).decode('utf8'))
    return b

encrypted_data =encrypt("Hi Steven!!!!!")
print(encrypted_data)
print("=======")
decrypted_data = decrypt(encrypted_data)
print(decrypted_data)