System.Security.Cryptography.CryptographicException: 'Length of the data to decrypt is invalid.' String double spaces

Question

I have implemented the methods in this answer to encrypt and decrypt a string

This appears to be encrypt and decrypt fine for most strings until the string has two or more spaces.

I.e.

'Buzz' - encrypt / decrypt fine (Buffer /Length = 16)
'Buzz Aldrin' - encrypt / decrypt fine (Buffer /Length = 16)
'Buzz Aldrin Astronaut' - encrypt fine / decrypt error (Buffer /Length = 31)

System.Security.Cryptography.CryptographicException: 'Length of the data to decrypt is invalid.'

 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

Any ideas what I am doing wrong or how I might correct it?

Example Update

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 Output
Buzz : aTBh1U0OFqW7+266LiC7Vg==GC6bUY5pK10L2KgQzpAtgg== : Buzz
Buzz Aldrin : 80fmD0z57R8jmmCkKhCsXg==dixi7bqheBzKhXcT1UEpWQ== : Buzz Aldrin
Exception thrown: 'System.Security.Cryptography.CryptographicException' in mscorlib.dll
Length of the data to decrypt is invalid.

@ŇɏssaPøngjǣrdenlarp, Hashing can not be used to store passwords into databases for login process. Passwords used in the Login process can ONLY be Encrypted. — evry1falls, Aug 29 '23 at 22:49
Fun fact if you can decrypt or decode the password, then bad actors can as well. Such things are a limitation of many auto didacts — Ňɏssa Pøngjǣrdenlarp, Aug 31 '23 at 14:50

score 1 · Accepted Answer · answered Aug 22 '17 at 12:25

Buzz Aldrin Astronaut : /1RInYgi/XPCpKYKxCCQLg==NgtahaolZmtyRKqG5d3XdWbnTP3o782hoyg7jp6VVAA=

This is what I get running your example.

Your last String end with only one = so this line isn't correct and generates this error

ciphertext = ivct(2) & "=="

Replace the following lines

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

by this code

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

and this should run just fine.

Hope this helps.

score 0 · Answer 2 · answered Aug 22 '17 at 12:38

The code that is used for splitting the IV and ciphertext actually breaks the ciphertext by always appending ==. This leads to a broken Base64 encoding which VB.Net for some reason doesn't have a problem with.

Add

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

after

ciphertext = ivct(2) & "=="

This line fixes the Base64 encoding.

score 0 · Answer 3 · answered Aug 23 '17 at 12:02

You can also change my implementation so that the encryption algorithm will concatenate the IV with ciphertext with a # character in between and the decryption will split it from there and remove #. It should be more convenient for everyone. Sorry for the initial inconvenience.

System.Security.Cryptography.CryptographicException: 'Length of the data to decrypt is invalid.' String double spaces

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