PKCS5 Python的AES解密填充

问题描述

我一直在尝试在Python中实现AES CBC解密。由于密文不是16字节的倍数，因此必须进行填充。没有填充，此错误浮出水面

I have been trying to implement AES CBC decryption in Python. Since the ciphered text is not a multiple of 16bytes, padding was necessary. Without padding, this error surfaced

TypeError：奇数长度的字符串

"TypeError: Odd-length string"

但是我找不到在PyCrypto Python中实现PKCS5的正确参考。
是否有任何命令可以实现？
谢谢

But I could not find a proper reference for implementing PKCS5 in PyCrypto Python.Are there any commands to implement this?Thanks

研究了马库斯的建议后，我做到了。

After looking into Marcus's suggestion I did this.

我的目标实际上是解密使用此代码的十六进制消息（128字节）。但是，输出为？：，它很小，unpad命令正在删除这些字节。这是代码。

My goal actually is to decrypt a hex message(128bytes) using this code. However, the output is " ?:" which is very small and the unpad command is deleting those bytes. This is the code.

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

推荐答案

在解密之前，您需要解码十六进制编码的值。如果要使用十六进制编码的密钥，也需要对其进行解码。.

You need to decode your hex encoded value before decryption. If you want to work with hex encoded keys, decode it as well..

在这里，这应该可以工作。

Here, this should work.

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

这篇关于PKCS5 Python的AES解密填充的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！