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')

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。

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。

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。

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)