如何在Python中使用客户端加密

Question

我尝试在将敏感数据移动到S3上的云存储并将其转移到redshift之前使用客户端加密对其进行加密。我试着使用AWS提供的示例代码，在使用了它之后，我让它运行而没有返回错误，但是，它没有做任何我能知道的事情，因为没有任何东西可以打印出来。

def cycle_string(key_arn, source_plaintext, botocore_session=None):
    """Encrypts and then decrypts a string using a KMS customer master key (CMK)

    :param str key_arn: [encryption key]
    (http://docs.aws.amazon.com/kms/latest/developerguide/viewing-keys.html)
    :param bytes source_plaintext: 
    :param botocore_session: Existing botocore session
    :type botocore_session: botocore.session.Session
    """

    # Create a KMS master key provider
    kms_kwargs = dict(key_ids=[key_arn])
    if botocore_session is not None:
        kms_kwargs['botocore_session'] = botocore_session
    master_key_provider = 
    aws_encryption_sdk.KMSMasterKeyProvider(**kms_kwargs)

    # Encrypt the plaintext source data
    ciphertext, encryptor_header = aws_encryption_sdk.encrypt(
        source=source_plaintext,
        key_provider=master_key_provider
    )
    print('Ciphertext: ', ciphertext)

    # Decrypt the ciphertext
    cycled_plaintext, decrypted_header = aws_encryption_sdk.decrypt(
        source=ciphertext,
        key_provider=master_key_provider
    )

    # Verify that the "cycled" (encrypted, then decrypted) plaintext is identical to the source
    # plaintext
    assert cycled_plaintext == source_plaintext

    # Verify that the encryption context used in the decrypt operation includes all key pairs from
    # the encrypt operation. (The SDK can add pairs, so don't require an exact match.)
    #
    # In production, always use a meaningful encryption context. In this sample, we omit the
    # encryption context (no key pairs).
    assert all(
        pair in decrypted_header.encryption_context.items()
        for pair in encryptor_header.encryption_context.items()
    )

    print('Decrypted: ', cycled_plaintext)

我是Python和加密的新手，所以我可能缺少一些语法，或者只是缺乏对其工作原理的了解。这是在python中对AWS使用客户端加密的最佳方法吗？如果是这样，为什么这个代码不返回任何内容？

更新： 我用了一种稍微不同的方法

^{pr2}$

现在它打印出来了，但我不知道如何判断数据是否真的加密了

Answer 1

您可以使用pycrypto：

#!/usr/bin/env python
# -*- coding: UTF-8 -*-

import hashlib
import base64
from Crypto import Random
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 = hashlib.sha256(key.encode('utf-8')).digest()

    def encrypt( self, raw ):
        raw = pad(raw)
        iv = Random.new().read( AES.block_size )
        cipher = AES.new( self.key, AES.MODE_CBC, iv )
        return base64.b64encode( iv + cipher.encrypt( raw ) )

    def decrypt( self, enc ):
        enc = base64.b64decode(enc)
        iv = enc[:16]
        cipher = AES.new(self.key, AES.MODE_CBC, iv )
        return unpad(cipher.decrypt( enc[16:] ))


#password
password="mypassword"

#content
global_report="Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. It has survived not only five centuries, but also the leap into electronic typesetting, remaining essentially unchanged. It was popularised in the 1960s with the release of Letraset sheets containing Lorem Ipsum passages, and more recently with desktop publishing software like Aldus PageMaker including versions of Lorem Ipsum."

#generate cipher
cipher = AESCipher(bytes(password))

#encrypt
encrypted = cipher.encrypt(bytes(global_report))

#show encrypted
print encrypted

#decrypt
decrypted = cipher.decrypt(encrypted)

#show decrypted
print decrypted

哪个返回：

^{pr2}$

Answer 2

该函数不返回任何内容，因为它不包含返回调用。在

该函数的目的是演示如何使用库对明文进行加密，然后对生成的密文消息进行解密，从而证明该循环产生了相同的明文。在

如果您在实践中使用这种方法，那么在任何给定的时间都需要该周期的一半（即：加密或解密，但不能同时使用）。在