从Python 2 md5库正确迁移到Python 3 hashlib [英] Correctly migrate from Python 2 md5 library to Python 3 hashlib

问题描述

我正在尝试在Django 1.11，Python 3.5.2中集成第三方支付网关（ CCAvenue ）

I'm trying to integrate a 3rd party payment gateway (CCAvenue) in Django 1.11, Python 3.5.2

第三方提供的参考代码使用了已弃用的库md5来加密文本。

The reference code provided by the 3rd party uses the deprecated library md5 to encrypt texts.

from Crypto.Cipher import AES
import md5

def pad(data):
    length = 16 - (len(data) % 16)
    data += chr(length)*length
    return data

def encrypt(plainText,workingKey):
    iv = '\x00\x01\x02\x03\x04\x05\x06\x07\x08\x09\x0a\x0b\x0c\x0d\x0e\x0f'
    plainText = pad(plainText)
    encDigest = md5.new ()
    encDigest.update(workingKey)
    enc_cipher = AES.new(encDigest.digest(), AES.MODE_CBC, iv)
    encryptedText = enc_cipher.encrypt(plainText).encode('hex')
    return encryptedText

如何使用 hashlib使上面的 encrypt（）方法Python 3兼容/ code> Python 3库？您可以发布整个方法吗？

How do I make the above encrypt() method Python 3 compatible using the hashlib library of Python 3? Can you post the whole method?

推荐答案

这是Python 3兼容的方法：

Here's a Python 3 compatible way of doing the same:

from binascii import hexlify, unhexlify
from Crypto.Cipher import AES
from hashlib import md5

from django.utils.encoding import force_bytes

iv = '\x00\x01\x02\x03\x04\x05\x06\x07\x08\x09\x0a\x0b\x0c\x0d\x0e\x0f'

BS = 16


def _pad(data):
    return data + (BS - len(data) % BS) * chr(BS - len(data) % BS)


def _unpad(data):
    return data[0:-ord(data[-1])]


def encrypt(plain_text, working_key):
    plain_text = _pad(plain_text)
    enc_cipher = AES.new(md5(force_bytes(working_key)).digest(), AES.MODE_CBC, _iv)
    return hexlify(enc_cipher.encrypt(plain_text)).decode('utf-8')


def decrypt(cipher_text, working_key):
    encrypted_text = unhexlify(cipher_text)
    dec_cipher = AES.new(md5(force_bytes(working_key)).digest(), AES.MODE_CBC, _iv)
    return _unpad(dec_cipher.decrypt(encrypted_text).decode('utf-8'))

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