如何解码ruby中的Rijndael(编码在VB.net中)

I am using Rinjael to encode in VB.NET and need to decode in Ruby. My VB.NET encryption class looks like this:

我正在用Rinjael来编码VB。需要在Ruby中解码。我的VB。NET加密类如下所示:

Private Class Encryptor
        Private symmetricKey As System.Security.Cryptography.RijndaelManaged
        Private iVector As Byte()
        Private Key As Byte()
        Public Function encrypt(ByVal data As String) As String
            Try
                Dim plainTextBytes As Byte() = System.Text.Encoding.ASCII.GetBytes(data)
                Dim encryptor As System.Security.Cryptography.ICryptoTransform = symmetricKey.CreateEncryptor(Key, iVector)
                Dim memoryStream As New System.IO.MemoryStream
                Dim cryptoStream As System.Security.Cryptography.CryptoStream = New System.Security.Cryptography.CryptoStream(memoryStream, encryptor, System.Security.Cryptography.CryptoStreamMode.Write)
                cryptoStream.Write(plainTextBytes, 0, plainTextBytes.Length)
                cryptoStream.FlushFinalBlock()
                Dim cipherTextBytes As Byte() = memoryStream.ToArray()
                memoryStream.Close()
                cryptoStream.Close()
                Return Convert.ToBase64String(cipherTextBytes)
            Catch
                Return ""
            End Try
        End Function
        Public Function decrypt(ByVal data As String) As String
            Try
                Dim crypted As Byte() = Convert.FromBase64String(data)
                Dim decryptor As System.Security.Cryptography.ICryptoTransform = symmetricKey.CreateDecryptor(Key, iVector)
                Dim memoryStream As New System.IO.MemoryStream(crypted)
                Dim cryptoStream As System.Security.Cryptography.CryptoStream = New System.Security.Cryptography.CryptoStream(memoryStream, decryptor, System.Security.Cryptography.CryptoStreamMode.Read)
                Dim plain(crypted.Length) As Byte
                Dim count As Integer = cryptoStream.Read(plain, 0, plain.Length)
                memoryStream.Close()
                cryptoStream.Close()
                Return System.Text.Encoding.UTF8.GetString(plain, 0, count)
            Catch
                Return ""
            End Try
        End Function

        Public Sub New(ByVal clientkey As String)
            iVector = System.Text.Encoding.ASCII.GetBytes("1234567890123456")
            Key = System.Text.Encoding.ASCII.GetBytes(clientkey)
            symmetricKey = New System.Security.Cryptography.RijndaelManaged
            symmetricKey.Mode = System.Security.Cryptography.CipherMode.CBC
        End Sub
    End Class

this works ok and I am able to decrypt in java using the AES/CBC/PKCS5Padding. Now, my password and iv are 16 characters long (16*16bit = 256). When I try to decrypt in Ruby, it complains that my password is to short... I assume that it is using 8bit chars. I use this class for decryption in ruby:

这工作正常，我可以使用AES/CBC/ pkcs5填充在java中解密。现在，我的密码和iv是16个字符长(16*16bit = 256)。当我尝试用Ruby解密时，它会抱怨我的密码太短了……我假设它使用的是8位字符。我用这个类来解密ruby:

    require 'openssl'

module Crypt
  # Decrypts a block of data (encrypted_data) given an encryption key
  # and an initialization vector (iv).  Keys, iv's, and the data 
  # returned are all binary strings.  Cipher_type should be
  # "AES-256-CBC", "AES-256-ECB", or any of the cipher types
  # supported by OpenSSL.  Pass nil for the iv if the encryption type
  # doesn't use iv's (like ECB).
  #:return: => String
  #:arg: encrypted_data => String 
  #:arg: key => String
  #:arg: iv => String
  #:arg: cipher_type => String
  def Crypt.decrypt(encrypted_data, key, iv, cipher_type)
    aes = OpenSSL::Cipher::Cipher.new(cipher_type)
    aes.decrypt
    aes.key = key
    aes.iv = iv if iv != nil
    aes.update(encrypted_data) + aes.final  
  end

  # Encrypts a block of data given an encryption key and an 
  # initialization vector (iv).  Keys, iv's, and the data returned 
  # are all binary strings.  Cipher_type should be "AES-256-CBC",
  # "AES-256-ECB", or any of the cipher types supported by OpenSSL.  
  # Pass nil for the iv if the encryption type doesn't use iv's (like
  # ECB).
  #:return: => String
  #:arg: data => String 
  #:arg: key => String
  #:arg: iv => String
  #:arg: cipher_type => String  
  def Crypt.encrypt(data, key, iv, cipher_type)
    aes = OpenSSL::Cipher::Cipher.new(cipher_type)
    aes.encrypt
    aes.key = key
    aes.iv = iv if iv != nil
    aes.update(data) + aes.final      
  end
end

Now. By trying to decrypt using Crypt.decrypt(data,key,iv, "AES-CBC-256") i am certain that preliminary string/byte transformations have to be done on my data,key,iv to work correctly.

现在。通过使用Crypt.decrypt(data,key,iv，“AES-CBC-256”)进行解密，我确信必须对我的数据、密钥、iv进行初始字符串/字节转换。

How shall I call Crypt.decrypt using key = "passwordpassword" and iv="1234567890123456"? Do I need to base64 decode my data?

如何使用key =" passwordpassword"和iv="1234567890123456"调用Crypt.decrypt ?我需要base64解码我的数据吗?

Here's my decryption call, which doesn't seem to work (trying padding with zeroes):

下面是我的解密调用，它似乎不起作用(尝试用0填充):

   text = Base64.decode64(text)
   pass = Digest::SHA1.hexdigest("#{@pass}0000000000000000").unpack('a2'*32).map{|x| x.hex}.pack('c'*32)
   iv = Digest::SHA1.hexdigest("12345678901234560000000000000000").unpack('a2'*32).map{|x| x.hex}.pack('c'*32)
   return Crypt.decrypt(text,pass,iv,"AES-256-CBC")

2 个解决方案

#1

Solution. It was a 128bit encryption (wrongly thinking that the default in .NET managed was 256). Hence, this code works (notice "AES-128-CBC"):

解决方案。它是128位加密(错误地认为。net管理的默认值是256)。因此，本代码有效(注意“AES-128-CBC”):

   text = Base64.decode64(text)
   pass = "passwordpassword"
   iv = "1234567890123456"
   return Crypt.decrypt(text,pass,iv,"AES-128-CBC")

#2

Because the VB app encodes the result as base64, it does look like the ruby script will need to use the Base64 module to decode it first.

因为VB应用程序将结果编码为base64，所以看起来ruby脚本需要先使用base64模块对其进行解码。

I believe the key given to the Ruby AES key must be 256 bits. So your password would need to be 32 bytes long exactly in that case. It would be better to use a scheme such as this.

我认为给Ruby AES密钥的密钥必须是256位。所以在这种情况下，你的密码需要32字节长。最好采用这样的方案。

#1