Question

我一直在尝试用Python实现AES CBC解密。由于加密文本不是16字节的倍数，因此填充是必要的。没有填充，此错误浮出水面

“TypeError：Odd-length string”

但我找不到在PyCrypto Python中实现PKCS5的正确参考。有没有命令来实现这个？感谢

在调查马库斯的建议后，我做了这个。

我的目标实际上是使用此代码解密十六进制消息（128字节）。但是，输出是“？：”，它非常小，unpad命令正在删除这些字节。这是代码。

from Crypto.Cipher import AES
BS = 16
pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS) 
unpad = lambda s : s[0:-ord(s[-1])]

class AESCipher:
    def __init__( self, key ):
    self.key = key 

    def encrypt( self, raw ):
        raw = pad(raw)
        iv = raw[:16]
        raw=raw[16:]
        #iv = Random.new().read( AES.block_size )
        cipher = AES.new( self.key, AES.MODE_CBC, iv )
        return ( iv + cipher.encrypt( raw ) ).encode("hex")

    def decrypt( self, enc ):
        iv = enc[:16]
        enc= enc[16:]
        cipher = AES.new(self.key, AES.MODE_CBC, iv )
        return unpad(cipher.decrypt( enc))

mode = AES.MODE_CBC
key = "140b41b22a29beb4061bda66b6747e14"
ciphertext = "4ca00ff4c898d61e1edbf1800618fb2828a226d160dad07883d04e008a7897ee2e4b7465d5290d0c0e6c6822236e1daafb94ffe0c5da05d9476be028ad7c1d81";
key=key[:32]
decryptor = AESCipher(key)
decryptor.__init__(key)
plaintext = decryptor.decrypt(ciphertext)
print plaintext

Answer 1

您需要在解密之前解码十六进制编码值。如果您想使用十六进制编码密钥，也可以对其进行解码。

在这里，这应该有用。

from Crypto.Cipher import AES
from Crypto import Random

BS = 16
pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS) 
unpad = lambda s : s[0:-ord(s[-1])]

class AESCipher:
    def __init__( self, key ):
        """
        Requires hex encoded param as a key
        """
        self.key = key.decode("hex")

    def encrypt( self, raw ):
        """
        Returns hex encoded encrypted value!
        """
        raw = pad(raw)
        iv = Random.new().read(AES.block_size);
        cipher = AES.new( self.key, AES.MODE_CBC, iv )
        return ( iv + cipher.encrypt( raw ) ).encode("hex")

    def decrypt( self, enc ):
        """
        Requires hex encoded param to decrypt
        """
        enc = enc.decode("hex")
        iv = enc[:16]
        enc= enc[16:]
        cipher = AES.new(self.key, AES.MODE_CBC, iv )
        return unpad(cipher.decrypt( enc))

if __name__== "__main__":
    key = "140b41b22a29beb4061bda66b6747e14"
    ciphertext = "4ca00ff4c898d61e1edbf1800618fb2828a226d160dad07883d04e008a7897ee2e4b7465d5290d0c0e6c6822236e1daafb94ffe0c5da05d9476be028ad7c1d81"
    key=key[:32]
    decryptor = AESCipher(key)
    plaintext = decryptor.decrypt(ciphertext)
    print "%s" % plaintext

使用PKCS5 Python进行AES解密填充

1 个答案: