Question

我有一个十六进制编码的字符串，需要使用RSA_OAEP算法解密。下面是我的代码段：

final_token = decoded_req['token']
print(final_token)
print("Converting actual token into bytes")
## Hex encoded string to bytes
token_to_bytes = bytes.fromhex(final_token)
print(token_to_bytes)

## Read the private key
with open('private.rsa', 'r') as pvt_key:
    miPvt = pvt_key.read()

## Decrypt the token using RSA_OAEP
print("Decrypting the token")
rsakey_obj = RSA.importKey(miPvt)
cipher = PKCS1_OAEP.new(rsakey_obj)
dec_token = cipher.decrypt(token_to_bytes)
print(dec_token)

下面是命令行输出：

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

Converting actual token into bytes

b'ly\x85\\\xa1Z:\xc00\x8d\x17\xdb\x97\xa1\xc5v\xb6\xf3^K\xb60\xda"\x86}\r\x08\x1dU\xf0^\x1b\x1cd\r~{\xd8\xc8\xde\x06\xb6\xa0>.\xf7D\x9e1\xfa\x9c\x0f\x0cg^\xbb\xf88\xbb\x88\nLn0\x93\x91D\x1e\x1cA\xa0\xc4/\xbe\x11\xea\xf60k\xb3\x9e{\xbb\xff\xbe-\xbc\xed\x8c\x9b\xb0vy\xff\x18\xce\xf3H\xb2\xb7\xce\x08\xaa\x05\x02\x8f\xda\x81\x8a\xc4\xce\x08\xad\x84$l\x94\xaf\xbc\xac@]\xb5%\x8cS3\xf6aH\x00{\xe8\xfaS\x84\xfax\xa0\xd5L\xcf&\x02\x85q\x18\x07#\xb2\xa6\x1f\xe1\x8e\xbd\x01$\xf7\xe0\x08\x90.y\xb4\x8aS\xd2\x99S\xff\xde\x94\x98x\xb4\xeb\x99\xd3v\xcc\xf6\x83A\xb0\xec,\xeb\x03\xb0P\xe7\xc2`\xf8\x90]\xcc\xdd\xb5\xe5\x0c\xb31\xa3\xbbk\xc04\xfc\x06\xdf\xafr/\xd8\xfa\xd8\xcbz1\x1b\x02\x0b\xe8\xb1\xa9\x11Ry\x17;\x18\xf4l\xcc\x0e(\xc7\x1f\nog\xe86*\x10?\xa4r\x90W\xd8\x92N\xf0\xa6\xfbZ3\xee\xb5\xb4,\xdcvO\xdeS\xb9\xd63\x84r\xfcs\xe8\x0c\xf7\x85\xebT\xff\xb4\xac\x15\xa6\xbfc\xdb\x19\xba\xcaXe\x08\xb8\x16\x0b\xf7=\xba\xd3""\x1d\xdbc/Z\xe8\x04\x0e\xf8h\x05\xa4\x08\xfc}\xf6\xf4\xa8\xee\\zDJZ\x13\xdb\xdc\xb8\xb9X\xc5*\x8b\xdcC\x94\xe5\xde\xaa\xbc\xe2\x03\xa3\xff\x86u\x07\x9ef\x18\xac5X\xcd\xaf\\\xe6\xdaw\xcaf{\xd54\x9b\xa8,\xc1\xad?q\xa6b\xa2\xb7\xc8\xb4|\xc5H \x92d\xa4Ze\xb3\xf6\xf2_\xed\xca\xd8D\x96\x99\xa1\x9e)\x10\xd2L\x13\xb4N>\x94E\xa6/\x88!3F\xe0\x00HH\x02\x18\x99j\xde~\x01\x15\x1b~\x03\x8f\xbc\x8a_\xf0;\xfaQ\x11&v=Eq\xde\xc6\xce\x0fQ\x830*\x99\xee\xe6/\xac\xab\xe0\x92\x11\xf3\xa6\x1b\x8d\x15J8\xf0\xee\x9a&G\x99\x8e.\xc1\xb8\xee\x96\xb5,D?@\x8c\xec$\x14\x088aly\xd8,\xbaKw\x17\x1bb\x1f&\x159\xe29'

Decrypting the token

Traceback (most recent call last):
  File "script.py", line 80, in <module>
    dec_token = cipher.decrypt(token_to_bytes)
  File "/home/venv/lib/python3.6/site-packages/Crypto/Cipher/PKCS1_OAEP.py", line 201, in decrypt
    raise ValueError("Incorrect decryption.")
ValueError: Incorrect decryption.

有人可以帮助我解决这个问题吗？

Answer 1

我怀疑您没有正确传输加密令牌。但是没什么可继续的。

尽管我使用了示例代码来帮助我编写下面的代码，但似乎还是行得通的（谢谢），或者至少它解密了另一个实现中的值。

在我的情况下，这是一个节点JS实现

var nodeRsa = require("node-rsa");
const key = new nodeRsa( "-----BEGIN PUBLIC KEY-----\n\
                              .
                              .
                              .
                          -----END PUBLIC KEY-----");
key.encrypt("passwd","base64");

如果我将上面的输出放入一个名为“密文”的文件中，则以下python正确解密了该消息。

import Crypto.Cipher.PKCS1_OAEP as rsaenc
import Crypto.PublicKey.RSA as RSA
import codecs

## Load the private key and the base 64 ciphertext.
with open("key.pem") as f:
    keystring = f.read()

with open("ciphertext","rb") as f:
    base64msg = f.read()

# Create the binary ciphertext
binmsg=codecs.decode(base64msg,'base64')

# Setup the RSA objetcs
k = RSA.importKey(keystring)
cipher = rsaenc.new(k)

plaintext = cipher.decrypt(binmsg)
print (plaintext)

Python中的RSA OAEP解密

1 个答案: