Question

我的任务是在项目上实施AES加密。参考代码是用Java编写的 - 它需要转换为Python。在组织我的笔记写一个SO问题时，我不小心偶然发现了答案！希望其他人认为这有用，我将在这里提到我的笔记作为“分享你的知识”的问题。

要求是使用具有给定密钥的AES加密消息。以下是参考代码（Java中）的简要介绍，

import javax.crypto.Cipher;
import javax.crypto.spec.SecretKeySpec;
import org.apache.commons.codec.binary.Base64;
import sun.misc.BASE64Encoder;


public class EncryptAES {

    private static String toHexString(byte[] data) {        
        StringBuffer buf = new StringBuffer();
        for (int i = 0; i < data.length; ++i) {
            String s = Integer.toHexString(data[i] & 0XFF);
            buf.append((s.length() == 1) ? ("0" + s) : s);
        }
        return buf.toString();
    }

    public static String encrypt(String input, String key) {
        byte[] crypted = null;
        try {
            SecretKeySpec skey = new SecretKeySpec(key.getBytes(), "AES");
            Cipher cipher = Cipher.getInstance("AES/ECB/PKCS5Padding");
            cipher.init(Cipher.ENCRYPT_MODE, skey);
            crypted = cipher.doFinal(input.getBytes());
            final String encryptedString = toHexString(Base64.encodeBase64(crypted));
            return encryptedString;
        } catch (Exception e) {
            System.out.println(e.toString());
        }
        return new String(new BASE64Encoder().encode(crypted));
    }

    public static void main(String[] args) {
        String key = args[0];
        String plaintext = args[1];
        System.out.println("KEY = " + key);
        System.out.println("PLAINTEXT = " + plaintext);
        System.out.println("CIPHER = " + EncryptAES.encrypt(plaintext, key));
    }
}

如果将上述内容保存为“EncryptAES.java”并将库文件commons-codec-1.7.jar保存在同一目录中，则可以使用以下命令对其进行编译，

$ javac EncryptAES.java -cp commons-codec-1.7.jar

这是运行程序几次时的输出，

$ java -cp "commons-codec-1.7.jar:." EncryptAES ddddffffeeeerrrr message
KEY = ddddffffeeeerrrr
MESSAGE = message
CRYPTO = 397a59594d35524e6b6a463253706f41467668646b773d3d
$

$ java -cp "commons-codec-1.7.jar:." EncryptAES qqqqwwwweeeerrrr ThisIsAVeryImportantMessage
KEY = qqqqwwwweeeerrrr
PLAINTEXT = ThisIsAVeryImportantMessage
CIPHER = 56536a384d667736756b595a394e396b6d504d736231444673375250736d5639596f637072792f6e4b424d3d
$

环顾四周，我找到了Python Crypto库。这是我必须复制上述输出的早期尝试之一，

#!/usr/bin/python

import sys
from Crypto.Cipher import AES

if __name__ == '__main__':
    key = sys.argv[1]
    plaintext = sys.argv[2]
    print 'KEY = ' + key
    print 'PLAINTEXT = ' + plaintext

    encobj = AES.new(key, AES.MODE_ECB)
    ciphertext = encobj.encrypt(plaintext)
    print 'CIPHER = ' + ciphertext.encode('hex')

这不太符合我的需要。相反，我收到一条关于输入字符串的错误消息，该字符串需要是16的倍数。这让我接下来的尝试，

#!/usr/bin/python

import sys
from Crypto.Cipher import AES

# ref: https://gist.github.com/crmccreary/5610068
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)
        cipher = AES.new(self.key, AES.MODE_ECB)
        return cipher.encrypt(raw).encode("hex")


if __name__ == '__main__':
    key = sys.argv[1]
    plaintext = sys.argv[2]
    print 'KEY = ' + key
    print 'PLAINTEXT = ' + plaintext

    # ref: http://stackoverflow.com/a/16882092
    hex_key = "".join("{:02x}".format(ord(c)) for c in key)

    encryptor = AESCipher(hex_key)
    ciphertext = encryptor.encrypt(plaintext)
    print 'CIPHER = ' + ciphertext

说实话，我不确定输出是什么，

$ python EncryptAES2.py ddddffffeeeerrrr message
KEY = ddddffffeeeerrrr
PLAINTEXT = message
CIPHER = f7361833944d9231764a9a0016f85d93
$

我尝试了很多东西 - 不同的加密模式，博客，SO问题，并放弃了自己寻找解决方案。就在这时，我决定收集我的笔记并在这里提出一个问题。现在，如果我没有列出我的尝试，那就不会有太多意义了，所以我开始将它们组织在一个文件夹中并标记为EncryptAES.py，EncryptAES2.py ..等等。

Answer 1

当我准备列表时，灵感震惊了我，为了我的最后一次尝试，我决定用十六进制重新格式化输出。令我惊喜的是，它奏效了！这是获奖代码，

#!/usr/bin/python

import sys
import base64
from Crypto.Cipher import AES

# ref: http://stackoverflow.com/a/12525165
BS = 16
pad = lambda s: s + (BS - len(s) % BS) * chr(BS - len(s) % BS) 

class AESCipher:
    def __init__( self, key ):
        self.key = key

    def encrypt( self, raw ):
        raw = pad(raw)
        cipher = AES.new(self.key, AES.MODE_ECB)
        return base64.b64encode(cipher.encrypt(raw)) 


if __name__ == '__main__':
    key = sys.argv[1]
    plaintext = sys.argv[2]
    print 'KEY = ' + key
    print 'PLAINTEXT = ' + plaintext

    encryptor = AESCipher(key)
    ciphertext = encryptor.encrypt(plaintext)

    hex_ciphertext = "".join("{:02x}".format(ord(c)) for c in ciphertext)

    print 'CIPHER = ' + hex_ciphertext

这里的参考是使用我用于Java示例的早期输入的输出，

$ python EncryptAES3.py ddddffffeeeerrrr message
KEY = ddddffffeeeerrrr
PLAINTEXT = message
CIPHER = 397a59594d35524e6b6a463253706f41467668646b773d3d
$

$ python EncryptAES3.py qqqqwwwweeeerrrr ThisIsAVeryImportantMessage
KEY = qqqqwwwweeeerrrr
PLAINTEXT = ThisIsAVeryImportantMessage
CIPHER = 56536a384d667736756b595a394e396b6d504d736231444673375250736d5639596f637072792f6e4b424d3d
$

获得此解决方案带给我很多反复试验。如果有一种更加严谨的方法将Java转换为Python，我很乐意听到它！

Answer 2

采用@chronodekar 的答案并将其更改为适用于 Python 3.9。

Python 3 的 PyCryptoDome 版本引发了一个类型错误：对象类型不能传递给 C 代码。为了解决这个问题，我将密钥和原始明文转换成字节数组。

    class simple_AES:

    def __init__(self, key):
        self.key = bytearray(key.encode())

    def encrypt_AES(self, raw):
        raw = bytearray(pad(raw).encode())
        cipher = AES.new(self.key, AES.MODE_ECB)
        return base64.b64encode(cipher.encrypt(raw))

但是，我也遇到了输出问题。我找到了将输出转换为十六进制 python base64 to hex 的答案。

    hex_ciphertext = base64.b64decode(ciphertext).hex()
    print('CIPHER = ' + hex_ciphertext)

Answer 3

Set rng = ThisWorkbook.Sheets("Edit Entry").Range("M31:M41")

For each cell in rng 
    If cell.interior.Colorindex = 15 then 
        If Not Intersect(Target, Range("M31:M41")) Is Nothing then 
            Msgbox "NOT AN EDITABLE FIELD.", vbCritical + vbOkOnly, "NO DATA ENTRY"

            With Application 
                .EnableEvents = False 
                .Undo 
                .EnableEvents = True 
            End With 
        End If 
    End If 
Next Cell

使用AES / ECB / PKSC5加密，将Java迁移到Python

3 个答案: