Rails 使用 AES 加密，过于复杂

Question

我在加密来自我正在使用的第三方供应商的值时遇到问题。

他们的指令如下：

1) Convert the encryption password to a byte array.
2) Convert the value to be encrypted to a byte array.
3) The entire length of the array is inserted as the first four bytes onto the front 
   of the first block of the resultant byte array before encryption.
4) Encrypt the value using AES with:
        1. 256-bit key size,
        2. 256-bit block size, 
        3. Encryption Mode ECB, and
        4. an EMPTY initialization vector.
5) After encryption, you should now have a byte array that holds the encrypted value. 
6) Convert each byte to a HEX format and string all the HEX values together.
7) The final result is a string of HEX values. This is the final encrypted value to be passed. 
   The length of the final value will always be an even number.

EXAMPLE:
Given the following input values:
plainText: 2017/02/07 22:46
secretKey: ABCD1234FGHI5678
The following string will be produced:
D6281D5BE6CD6E79BB41C039F4DD020FBEC9D290AD631B2598A6DFF55C68AD04

到目前为止我尝试了什么...

plain_text = "2017/02/07 22:46"
secret_key = "ABCD1234FGHI5678"

plain_text_byte_array = plain_text.bytes
plain_text_byte_array.unshift(0).unshift(0).unshift(0).unshift(16) # I found a Java example in their documentation and this is what they do. They prepend their byte array with 16, 0, 0, 0
secret_byte_array = secret_key.bytes
secret_byte_array = secret_byte_array.concat([0, 0, 0,...]) # also from their java example, they append the secret_byte array with 16 0's in order to get its length to 32

cipher = OpenSSL::Cipher::AES256.new(:ECB)
cipher.key = secret_byte_array.pack("C*")
encrypted = cipher.update(plain_text_byte_array.pack("C*")) + cipher.final

p encrypted.unpack("H*").first.to_s.upcase

# Result is: 
#    "84A0E5DCA7D704C41332F86E707DDAC244A1A87C38A906145DE4060D2BC5C8F4"

如您所见，我的结果与实际结果不符，应该是 "D6281D5BE6CD6E79BB41C039F4DD020FBEC9D290AD631B2598A6DFF55C68AD04"

有谁知道我是否遗漏了什么或做了什么奇怪的事情。他们的指示对我来说很难解析，所以也许我遗漏了一些东西。感谢您提供的任何帮助！（我已经尝试了很多你在上面看到的不同的变体）。我只是需要一些指导，或者至少需要有人告诉我，我没有因为不理解他们的指示而发疯。

Answer 1

我设法重现了他们的结果——他们使用的过程极其复杂，远非优雅。我附上了对实现结果所需步骤的更具描述性的解释，以及我曾经这样做的 C# 源代码。

将密码转换为字节数组。字节数组必须长度为32字节，并且，如果密码不够长，右补0字节。因此他们的密码，十六进制编码，变成 4142434431323334464748493536373800000000000000000000000000000000.
将要加密的值转成byte数组这个很简单，用UTF-8编码就可以了
数组的整个长度作为前四个字节插入到加密前生成的字节数组的第一个块的前面。这是愚蠢且毫无用处，但将步骤 2 中的字节数组的长度作为 无符号 32 位整数 并转换为小端字节数组。将其作为步骤 2 中的数组的前缀。
使用 AES 加密值。嗯。不，不要那样做。使用 256 位块大小、256 位密钥大小、ECB 模式和零填充，使用 Rijndael 加密值。
剩下的就简单了，把加密后的结果转成十六进制就可以了

我用来实现这个结果的代码如下，在 C# 中。我不太了解 Ruby 抱歉。

    // 1. Convert the encryption password to a byte array.
    byte[] passwordBytesOriginal = Encoding.UTF8.GetBytes("ABCD1234FGHI5678");
    byte[] passwordBytes = new byte[32];
    Array.Copy(passwordBytesOriginal, 0, passwordBytes, 0, passwordBytesOriginal.Length);


    // 2. Convert the value to be encrypted to a byte array.
    byte[] valueBytes = Encoding.UTF8.GetBytes("2017/02/07 22:46");

    // 3. The entire length of the array is inserted as the first four bytes onto the front 
    // of the first block of the resultant byte array before encryption.
    byte[] valueLengthAsBytes = BitConverter.GetBytes((uint)valueBytes.Length);
    byte[] finalPlaintext = new byte[valueBytes.Length + valueLengthAsBytes.Length];
    Array.Copy(valueLengthAsBytes, 0, finalPlaintext, 0, valueLengthAsBytes.Length);
    Array.Copy(valueBytes, 0, finalPlaintext, valueLengthAsBytes.Length, valueBytes.Length);

    // 4. Encrypt the value using AES...
    byte[] ciphertext;
    using (RijndaelManaged rijn = new RijndaelManaged())
    {
        rijn.BlockSize = 256;
        rijn.KeySize = 256;
        rijn.Key = passwordBytes;
        rijn.Mode = CipherMode.ECB;
        rijn.Padding = PaddingMode.Zeros;

        var encryptor = rijn.CreateEncryptor();
        ciphertext = encryptor.TransformFinalBlock(finalPlaintext, 0, finalPlaintext.Length);
    }

    // 5., 6., 7...
    string result = BitConverter.ToString(ciphertext).Replace("-", "").ToUpper();
    Console.WriteLine(result); // D6281D5BE6CD6E79BB41C039F4DD020FBEC9D290AD631B2598A6DFF55C68AD04

Answer 2

基于 here is the Ruby version. I had to use the ruby-mcryptgem并使用brew install libmcrypt在本地安装mcrypt库。

正如Luke的回答所指出的那样，密钥应该用0右补。这是我的代码：

   plain_text = "2017/02/07 22:46"
   secret_text = "ABCD1234FGHI5678"
   answer = "D6281D5BE6CD6E79BB41C039F4DD020FBEC9D290AD631B2598A6DFF55C68AD04"

   def format_byte_arrays(plain, secret)
     zero_byte_array = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
     length_array = [16, 0, 0, 0]

     plain_bytes = length_array.concat(plain.bytes)
     secret_bytes = secret.bytes.concat(zero_byte_array)

     [plain_bytes, secret_bytes]
   end

   plain_bytes, secret_bytes = format_byte_arrays(plain_text, secret_text)
   final_plain, final_secret = [plain_bytes.pack("C*"), secret_bytes.pack("C*")]

   cipher = Mcrypt.new("rijndael-256", :ecb, final_secret, nil, :zeros)
   encrypted = cipher.encrypt(final_plain)
   result = encrypted.unpack("H*").first.to_s.upcase

结果将是正确答案。

Rails 使用 AES 加密，过于复杂

Rails Encryption using AES, overcomplicated

ruby

encryption

openssl

aes

ecb