Question

我已经在加密和解密字符串

的答案中实施了这些方法

对于大多数字符串，这似乎是加密和解密的，直到字符串有两个或更多空格。

即

＆＃39;巴斯＆＃39; - 加密/解密（缓冲区/长度= 16）
＆＃39; Buzz Aldrin＆＃39; - 加密/解密（缓冲区/长度= 16）
＆＃39; Buzz Aldrin宇航员＆＃39; - 加密精细/解密错误（缓冲区/长度= 31）

System.Security.Cryptography.CryptographicException：＆＃39;要解密的数据长度无效。＆＃39;

 Public Shared Function AES_Decrypt(ByVal ciphertext As String, ByVal key As String) As String
 Dim AES As New System.Security.Cryptography.RijndaelManaged
            Dim SHA256 As New System.Security.Cryptography.SHA256Cng
            Dim plaintext As String = ""
            Dim iv As String = ""
            Try
                Dim ivct = ciphertext.Split(CChar("="))
                iv = ivct(0) & "=="
                ciphertext = ivct(2) & "=="

                AES.Key = SHA256.ComputeHash(System.Text.ASCIIEncoding.ASCII.GetBytes(key))
                AES.IV = Convert.FromBase64String(iv)
                AES.Mode = Security.Cryptography.CipherMode.CBC
                Dim DESDecrypter As System.Security.Cryptography.ICryptoTransform = AES.CreateDecryptor
                Dim Buffer As Byte() = Convert.FromBase64String(ciphertext)

Exception  ---->   plaintext = System.Text.ASCIIEncoding.ASCII.GetString(DESDecrypter.TransformFinalBlock(Buffer, 0, Buffer.Length))
                Return plaintext
Catch ex As system.Exception
                Return ex.Message
            End Try
        End Function

任何想法我做错了什么或我如何纠正它？

示例更新

Try
        Dim s1, s2, s3 As String
        s1 = Crypto.AES_Encrypt("Buzz", "Password")
        s2 = Crypto.AES_Encrypt("Buzz Aldrin", "Password")
        s3 = Crypto.AES_Encrypt("Buzz Aldrin Astronaut", "Password")
        Debug.Print("Buzz : " & s1 & " : " & Crypto.AES_Decrypt(s1, "Password"))
        Debug.Print("Buzz Aldrin : " & s2 & " : " & Crypto.AES_Decrypt(s2, "Password"))
        Debug.Print("Buzz Aldrin Astronaut : " & s3 & " : " & Crypto.AES_Decrypt(s3, "Password"))
    Catch ex As System.Exception
        Debug.Print(ex.Message.ToString())
    End Try

Debug.Print输出
  Buzz：aTBh1U0OFqW7 + 266LiC7Vg == GC6bUY5pK10L2KgQzpAtgg ==：Buzz
  Buzz Aldrin：80fmD0z57R8jmmCkKhCsXg == dixi7bqheBzKhXcT1UEpWQ ==：Buzz Aldrin
  抛出异常：＆＃39; System.Security.Cryptography.CryptographicException＆＃39;在mscorlib.dll中   要解密的数据长度无效。

Answer 1

Buzz Aldrin宇航员：/ 1RInYgi / XPCpKYKxCCQLg == NgtahaolZmtyRKqG5d3XdWbnTP3o782hoyg7jp6VVAA =

这就是我运行你的例子。

您的上一次String结尾只有一个=，因此此行不正确并生成此错误

ciphertext = ivct(2) & "=="

替换以下行

Dim ivct = ciphertext.Split(CChar("="))
iv = ivct(0) & "=="
ciphertext = ivct(2) & "=="

通过此代码

Dim ivct = ciphertext.Split({"=="}, StringSplitOptions.None)
iv = ivct(0) & "=="
ciphertext = If(ivct.Length = 3, ivct(1) & "==", ivct(1))

这应该运行得很好。

希望这有帮助。

Answer 2

用于拆分IV和密文的代码实际上会通过附加==来破解密文。这导致了一个破坏的Base64编码，VB.Net由于某种原因没有问题。

添加

ciphertext = ciphertext.Substring(0, ciphertext.Length - ciphertext.Length Mod 4)

后

ciphertext = ivct(2) & "=="

此行修复了Base64编码。

Answer 3

您还可以更改我的实现，以便加密算法将IV与密文连接在一起，并且解密将从那里拆分并删除＃。对每个人来说都应该更方便。抱歉，最初的不便之处。

System.Security.Cryptography.CryptographicException：＆＃39;要解密的数据长度无效。＆＃39;字符串双空格

3 个答案: