我正在尝试将此C#代码转换为Python(2.5,GAE)。问题是每次加密(在同一个字符串上)时,python脚本中的加密字符串都不同。
string Encrypt(string textToEncrypt, string passphrase)
{
RijndaelManaged rijndaelCipher = new RijndaelManaged();
rijndaelCipher.Mode = CipherMode.CBC;
rijndaelCipher.Padding = PaddingMode.PKCS7;
rijndaelCipher.KeySize = 128;
rijndaelCipher.BlockSize = 128;
byte[] pwdBytes = Encoding.UTF8.GetBytes(passphrase);
byte[] keyBytes = new byte[16];
int len = pwdBytes.Length;
if (len > keyBytes.Length)
{
len = keyBytes.Length;
}
Array.Copy(pwdBytes, keyBytes, len);
rijndaelCipher.Key = keyBytes;
rijndaelCipher.IV = new byte[16];
ICryptoTransform transform = rijndaelCipher.CreateEncryptor();
byte[] plainText = Encoding.UTF8.GetBytes(textToEncrypt);
return Convert.ToBase64String(transform.TransformFinalBlock(plainText, 0, plainText.Length));
}
Python代码:(PKCS7Encoder:http://japrogbits.blogspot.com/2011/02/using-encrypted-data-between-python-and.html)
from Crypto.Cipher import AES
from pkcs7 import PKCS7Encoder
#declared outside of all functions
key = '####'
mode = AES.MODE_CBC
iv = '\x00' * 16
encryptor = AES.new(key, mode, iv)
encoder = PKCS7Encoder()
def function(self):
text = self.request.get('passwordTextBox')
pad_text = encoder.encode(text)
cipher = encryptor.encrypt(pad_text)
enc_cipher = base64.b64encode(cipher)
C#代码是继承的。 Python代码必须以相同的方式加密和解密,以便C#代码可以正确解码值。
注意:我是python的菜鸟:)
编辑:抱歉。应该区分出有一个函数被调用。
谢谢!
答案 0 :(得分:3)
将python代码更改为:
from Crypto.Cipher import AES
from pkcs7 import PKCS7Encoder
#declared outside of all functions
key = '####'
mode = AES.MODE_CBC
iv = '\x00' * 16
encoder = PKCS7Encoder()
def function(self):
encryptor = AES.new(key, mode, iv)**
text = self.request.get('passwordTextBox')
pad_text = encoder.encode(text)
cipher = encryptor.encrypt(pad_text)
enc_cipher = base64.b64encode(cipher)
如果有人通过谷歌访问此页面
答案 1 :(得分:3)
您的C#代码无效。
Encrypt
函数将密码短语视为string passphrase
,但后来尝试在此行byte[] pwdBytes = Encoding.UTF8.GetBytes(key);
中引用密码
将key
更改为passphrase
。
这两个函数现在为我产生相同的结果:
<强>的Python 强>
secret_text = 'The rooster crows at midnight!'
key = 'A16ByteKey......'
mode = AES.MODE_CBC
iv = '\x00' * 16
encoder = PKCS7Encoder()
padded_text = encoder.encode(secret_text)
e = AES.new(key, mode, iv)
cipher_text = e.encrypt(padded_text)
print(base64.b64encode(cipher_text))
# e = AES.new(key, mode, iv)
# cipher_text = e.encrypt(padded_text)
# print(base64.b64encode(cipher_text))
C#(上面提到的拼写错误修复)
Console.WriteLine(Encrypt("The rooster crows at midnight!", "A16ByteKey......"));
Python结果
XAW5KXVbItrc3WF0xW175UJoiAfonuf + s54w2iEs + 7A =
C#结果
XAW5KXVbItrc3WF0xW175UJoiAfonuf + s54w2iEs + 7A =
我怀疑你多次在你的python代码中重复使用'e'。如果你取消注释我的python脚本的最后两行,你会看到输出现在是不同的。但如果你取消注释最后三行,你会看到输出是相同的。正如Foon所说,这是由于how CBC works。
CBC(密码块链接)在加密块中的字节序列时起作用。第一个块通过将IV与明文的第一个字节(“公鸡......”)合并来加密。第二个块使用第一个操作的结果而不是IV。
当您再次呼叫e.encrypt()
时(例如,通过取消对python脚本的最后两行的解除),您将从中断的位置开始。它将使用最后一个加密块的输出,而不是在加密第一个块时使用IV。这就是结果看起来不同的原因。通过解开python脚本的最后三行,您初始化一个新的加密器,它将使用IV作为其第一个块,从而使您获得相同的结果。
答案 2 :(得分:1)
这种esotic PKCS7编码器是一个功能,可以填充静态长度。 所以我用一个代码片段实现了它
#!/usr/bin/env python
from Crypto.Cipher import AES
import base64
# the block size for the cipher object; must be 16, 24, or 32 for AES
BLOCK_SIZE = 16
# the character used for padding--with a block cipher such as AES, the value
# you encrypt must be a multiple of BLOCK_SIZE in length. This character is
# used to ensure that your value is always a multiple of BLOCK_SIZE
# PKCS7 method
PADDING = '\x06'
mode = AES.MODE_CBC
iv = '\x08' * 16 # static vector: dangerous for security. This could be changed periodically
#
# one-liner to sufficiently pad the text to be encrypted
pad = lambda s: s + (BLOCK_SIZE - len(s) % BLOCK_SIZE) * PADDING
# one-liners to encrypt/encode and decrypt/decode a string
# encrypt with AES, encode with base64
EncodeAES = lambda c, s: base64.b64encode(c.encrypt(pad(s)))
DecodeAES = lambda c, e: c.decrypt(base64.b64decode(e)).rstrip(PADDING)
def CryptIt(password, secret):
cipher = AES.new(secret, mode, iv)
encoded = EncodeAES(cipher, password)
return encoded
def DeCryptIt(encoded, secret):
cipher = AES.new(secret, mode, iv)
decoded = DecodeAES(cipher, encoded)
return decoded
我希望这可以提供帮助。 干杯
答案 3 :(得分:0)
微软对PKCS7的实现与Python有点不同。
这篇文章帮助我解决了这个问题: http://japrogbits.blogspot.com/2011/02/using-encrypted-data-between-python-and.html
他的pkcs7编码和解码代码在github上: https://github.com/janglin/crypto-pkcs7-example
使用该PKCS7库,此代码对我有用:
from Crypto.Cipher import AES
aes = AES.new(shared_key, AES.MODE_CBC, IV)
aes.encrypt(PKCS7Encoder().encode(data))