CODEWITHSUNDEEP
pythonnode.jsencryptioncryptographypycrypto
Jos
Jos

Reputation: 69

AES Decryption using Pycrypto(python) not working. Getting correct decrypted code in crypto(Nodejs).

In node i am using the following code to get proper decrypted message:

//npm install --save-dev crypto-js
var CryptoJS = require("crypto-js");
var esp8266_msg = 'IqszviDrXw5juapvVrQ2Eh/H3TqBsPkSOYY25hOQzJck+ZWIg2QsgBqYQv6lWHcdOclvVLOSOouk3PmGfIXv//cURM8UBJkKF83fPawwuxg=';
var esp8266_iv  = 'Cqkbb7OxPGoXhk70DjGYjw==';

// The AES encryption/decryption key to be used.
var AESKey = '2B7E151628AED2A6ABF7158809CF4F3C';

var plain_iv =  new Buffer( esp8266_iv , 'base64').toString('hex');
var iv = CryptoJS.enc.Hex.parse( plain_iv );
var key= CryptoJS.enc.Hex.parse( AESKey );

console.log("Let's ");

// Decrypt
var bytes  = CryptoJS.AES.decrypt( esp8266_msg, key , { iv: iv} );
var plaintext = bytes.toString(CryptoJS.enc.Base64);
var decoded_b64msg =  new Buffer(plaintext , 'base64').toString('ascii');
var decoded_msg =     new Buffer( decoded_b64msg , 'base64').toString('ascii');

console.log("Decryptedage: ", decoded_msg);

But when i try to decrypt it in python i am not getting the proper decoded message.

esp8266_msg = 'IqszviDrXw5juapvVrQ2Eh/H3TqBsPkSOYY25hOQzJck+ZWIg2QsgBqYQv6lWHcdOclvVLOSOouk3PmGfIXv//cURM8UBJkKF83fPawwuxg='
esp8266_iv  = 'Cqkbb7OxPGoXhk70DjGYjw=='
key = '2B7E151628AED2A6ABF7158809CF4F3C'
iv = base64.b64decode(esp8266_iv)
message = base64.b64decode(esp8266_msg)
dec = AES.new(key=key, mode=AES.MODE_CBC, IV=iv)
value = dec.decrypt(message)
print(value)

I am getting the decoded message:

"ﾙﾶ+#ÊZûﾣﾧn*ÿÒá×G1ﾨﾷì;$-#fﾟãﾰk-ìØܳã-ﾄȒ~ﾇ8ﾍﾍ_ﾤ ﾝ?ﾤÑ:ÇäYﾑü'hﾕô<`

So i hope someone can show how it is done in python.

Upvotes: 2

Views: 728

Answers (1)

Artjom B.
Artjom B.

Reputation: 61952

You forgot to decode the key from Hex and remove the padding.

Full code:

from Crypto.Cipher import AES
import base64

unpad = lambda s : s[:-ord(s[len(s)-1:])]

esp8266_msg = 'IqszviDrXw5juapvVrQ2Eh/H3TqBsPkSOYY25hOQzJck+ZWIg2QsgBqYQv6lWHcdOclvVLOSOouk3PmGfIXv//cURM8UBJkKF83fPawwuxg='
esp8266_iv  = 'Cqkbb7OxPGoXhk70DjGYjw=='
key = '2B7E151628AED2A6ABF7158809CF4F3C'
iv = base64.b64decode(esp8266_iv)
message = base64.b64decode(esp8266_msg)
key = key.decode("hex")
dec = AES.new(key=key, mode=AES.MODE_CBC, IV=iv)
value = unpad(dec.decrypt(message))
print(value)
if len(value) % 4 is not 0:
    value += (4 - len(value) % 4) * "="
value = base64.b64decode(value)
print(value)

Output:

eyJkYXRhIjp7InZhbHVlIjozMDB9LCAiU0VRTiI6NzAwICwgIm1zZyI6IklUIFdPUktTISEiIH0
'{"data":{"value":300}, "SEQN":700 , "msg":"IT WORKS!!" }'

Security considerations

The IV must be unpredictable (read: random). Don't use a static IV, because that makes the cipher deterministic and therefore not semantically secure. An attacker who observes ciphertexts can determine when the same message prefix was sent before. The IV is not secret, so you can send it along with the ciphertext. Usually, it is simply prepended to the ciphertext and sliced off before decryption.

It is better to authenticate your ciphertexts so that attacks like a padding oracle attack are not possible. This can be done with authenticated modes like GCM or EAX, or with an encrypt-then-MAC scheme.

Upvotes: 5

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