我正在尝试使用PyCrypto构建两个接受两个参数的函数:消息和密钥,然后对消息进行加密/解密。
我在网络上找到了几个链接可以帮助我,但是每个链接都有缺陷:
[This one at codekoala使用os.urandom,PyCrypto不建议这样做。
此外,不能保证我提供给函数的键具有期望的确切长度。我该怎么做才能做到这一点?
另外,有几种模式,推荐哪种?我不知道该用什么:/
最后,IV到底是什么?我可以提供不同的IV进行加密和解密,还是会返回不同的结果?
这是我到目前为止所做的:
from Crypto import Random
from Crypto.Cipher import AES
import base64
BLOCK_SIZE=32
def encrypt(message, passphrase):
# passphrase MUST be 16, 24 or 32 bytes long, how can I do that ?
IV = Random.new().read(BLOCK_SIZE)
aes = AES.new(passphrase, AES.MODE_CFB, IV)
return base64.b64encode(aes.encrypt(message))
def decrypt(encrypted, passphrase):
IV = Random.new().read(BLOCK_SIZE)
aes = AES.new(passphrase, AES.MODE_CFB, IV)
return aes.decrypt(base64.b64decode(encrypted))
这是我的实现,并通过一些修复为我工作,并以32字节和iv到16字节增强了密钥和秘密短语的对齐方式:
import base64
import hashlib
from Crypto import Random
from Crypto.Cipher import AES
class AESCipher(object):
def __init__(self, key):
self.bs = 32
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))
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:])]
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:])
Crypto
和PyCryptodomex
库,并且运行速度很快...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=16
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)
当输入长度不是BLOCK_SIZE的倍数时,可能需要以下两个函数来填充(加密时)和取消填充(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:])]
所以您要询问密钥的长度?您可以使用密钥的md5sum而不是直接使用它。
更多,根据我使用PyCrypto的经验,在输入相同的情况下,IV用于混合加密输出,因此将IV选择为随机字符串,并将其用作加密输出的一部分,然后使用它解密邮件。
这是我的实现,希望它对您有用:
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:] ))
[让我解决您有关“模式”的问题。 AES256是一种分组密码
NOT
PyCryptodomex