我试图使用PyCrypto构建两个函数,它们接受两个参数:消息和密钥,然后加密/解密消息。
我在网上找到了几个链接来帮助我,但每一个都有缺陷:
codekoala的这个使用os。这是PyCrypto不鼓励的。
此外,我给函数的键不能保证具有预期的确切长度。我要怎么做才能做到呢?
还有,有几种模式,推荐哪种模式?我不知道用什么:/
最后,静脉注射到底是什么?我是否可以提供不同的IV来加密和解密,或者这将返回不同的结果?
我试图使用PyCrypto构建两个函数,它们接受两个参数:消息和密钥,然后加密/解密消息。
我在网上找到了几个链接来帮助我,但每一个都有缺陷:
codekoala的这个使用os。这是PyCrypto不鼓励的。
此外,我给函数的键不能保证具有预期的确切长度。我要怎么做才能做到呢?
还有,有几种模式,推荐哪种模式?我不知道用什么:/
最后,静脉注射到底是什么?我是否可以提供不同的IV来加密和解密,或者这将返回不同的结果?
当前回答
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。
其他回答
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:])
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')
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:])]
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。