by BarryDunne (3 Submissions)
Category: Encryption
Compatability: Visual Basic 3.0
Difficulty: Beginner
Date Added: Wed 3rd February 2021
Rating: 
(7 Votes)
This is a module that you can add to your project to encrypt/decrypt using the CryptoAPI. This is the standard API used regardless of who provides the dll which actually does the encryption. Microsoft give you one such dll as standard with windows or NT, but the API ensures that you have the same interface to anyone elses, or even write your own. These different encryption dlls are called Cryptographic Service Providers (CSP's) and the standard Microsoft one is called "Microsoft Base Cryptographic Provider v1.0". To use a different CSP all you have to do is change a constant in this module. This module ensures that there are no carriage returns or line feeds in the encrypted value so that you can easily write it to an ini file for example. This version contains a fix to the original version.
Inputs
There are two main functions:
'
'Function EncryptData(ByVal Data As String, ByVal Password As String) As String
' Where Data is the String to encrypt and password is used to encrypt it
'
'Function DecryptData(ByVal Data As String, ByVal Password As String) As String
' Where Data is the encrypted String and password is used to decrypt it
VBcodersOption Explicit
'This is based on the 2 MSDN articles
' "Example C Program: Using CryptAcquireContext"
' "Example C Program: Encrypting a File"
'Example usage:
'
' Private Const MY_PASSWORD As String = "isdflkaatdfuhwfnasdf"
'
' Public Sub Main()
' MsgBox EncryptData("hello world", MY_PASSWORD)
' MsgBox DecryptData(EncryptData("hello world", MY_PASSWORD), MY_PASSWORD)
' End Sub
Private Declare Function CryptAcquireContext Lib "advapi32.dll" Alias "CryptAcquireContextA" _
(ByRef phProv As Long, _
ByVal pszContainer As String, _
ByVal pszProvider As String, _
ByVal dwProvType As Long, _
ByVal dwFlags As Long) As Long
Private Declare Function CryptGetProvParam Lib "advapi32.dll" _
(ByVal hProv As Long, _
ByVal dwParam As Long, _
ByRef pbData As Any, _
ByRef pdwDataLen As Long, _
ByVal dwFlags As Long) As Long
Private Declare Function CryptCreateHash Lib "advapi32.dll" _
(ByVal hProv As Long, _
ByVal Algid As Long, _
ByVal hKey As Long, _
ByVal dwFlags As Long, _
ByRef phHash As Long) As Long
Private Declare Function CryptHashData Lib "advapi32.dll" _
(ByVal hHash As Long, _
ByVal pbData As String, _
ByVal dwDataLen As Long, _
ByVal dwFlags As Long) As Long
Private Declare Function CryptDeriveKey Lib "advapi32.dll" _
(ByVal hProv As Long, _
ByVal Algid As Long, _
ByVal hBaseData As Long, _
ByVal dwFlags As Long, _
ByRef phKey As Long) As Long
Private Declare Function CryptDestroyHash Lib "advapi32.dll" _
(ByVal hHash As Long) As Long
Private Declare Function CryptEncrypt Lib "advapi32.dll" _
(ByVal hKey As Long, _
ByVal hHash As Long, _
ByVal Final As Long, _
ByVal dwFlags As Long, _
ByVal pbData As String, _
ByRef pdwDataLen As Long, _
ByVal dwBufLen As Long) As Long
Private Declare Function CryptDestroyKey Lib "advapi32.dll" _
(ByVal hKey As Long) As Long
Private Declare Function CryptReleaseContext Lib "advapi32.dll" _
(ByVal hProv As Long, _
ByVal dwFlags As Long) As Long
Private Declare Function CryptDecrypt Lib "advapi32.dll" _
(ByVal hKey As Long, _
ByVal hHash As Long, _
ByVal Final As Long, _
ByVal dwFlags As Long, _
ByVal pbData As String, _
ByRef pdwDataLen As Long) As Long
Private Const SERVICE_PROVIDER As String = "Microsoft Base Cryptographic Provider v1.0"
Private Const KEY_CONTAINER As String = "Metallica"
Private Const PROV_RSA_FULL As Long = 1
Private Const PP_NAME As Long = 4
Private Const PP_CONTAINER As Long = 6
Private Const CRYPT_NEWKEYSET As Long = 8
Private Const ALG_CLASS_DATA_ENCRYPT As Long = 24576
Private Const ALG_CLASS_HASH As Long = 32768
Private Const ALG_TYPE_ANY As Long = 0
Private Const ALG_TYPE_STREAM As Long = 2048
Private Const ALG_SID_RC4 As Long = 1
Private Const ALG_SID_MD5 As Long = 3
Private Const CALG_MD5 As Long = ((ALG_CLASS_HASH Or ALG_TYPE_ANY) Or ALG_SID_MD5)
Private Const CALG_RC4 As Long = ((ALG_CLASS_DATA_ENCRYPT Or ALG_TYPE_STREAM) Or ALG_SID_RC4)
Private Const ENCRYPT_ALGORITHM As Long = CALG_RC4
Private Const NUMBER_ENCRYPT_PASSWORD As String = "´o¸sçPQ]"
Public Function EncryptData(ByVal Data As String, ByVal Password As String) As String
Dim sEncrypted As String
Dim lEncryptionCount As Long
Dim sTempPassword As String
'It is possible that the normal encryption will give you a string
'containing cr or lf characters which make it difficult to write to files
'Do a loop changing the password and keep encrypting until the result is ok
'To be able to decrypt we need to also store the number of loops in the result
'Try first encryption
lEncryptionCount = 0
sTempPassword = Password & lEncryptionCount
sEncrypted = EncryptDecrypt(Data, sTempPassword, True)
'Loop if this contained a bad character
Do While (InStr(1, sEncrypted, vbCr) > 0) _
Or (InStr(1, sEncrypted, vbLf) > 0) _
Or (InStr(1, sEncrypted, Chr$(0)) > 0) _
Or (InStr(1, sEncrypted, vbTab) > 0)
'Try the next password
lEncryptionCount = lEncryptionCount + 1
sTempPassword = Password & lEncryptionCount
sEncrypted = EncryptDecrypt(Data, sTempPassword, True)
'Don't go on for ever, 1 billion attempts should be plenty
If lEncryptionCount = 99999999 Then
Err.Raise vbObjectError + 999, "EncryptData", "This data cannot be successfully encrypted"
EncryptData = ""
Exit Function
End If
Loop
'Build encrypted string, starting with number of encryption iterations
EncryptData = EncryptNumber(lEncryptionCount) & sEncrypted
End Function
Public Function DecryptData(ByVal Data As String, ByVal Password As String) As String
Dim lEncryptionCount As Long
Dim sDecrypted As String
Dim sTempPassword As String
'When encrypting we may have gone through a number of iterations
'How many did we go through?
lEncryptionCount = DecryptNumber(Mid$(Data, 1, 8))
'start with the last password and work back
sTempPassword = Password & lEncryptionCount
sDecrypted = EncryptDecrypt(Mid$(Data, 9), sTempPassword, False)
DecryptData = sDecrypted
End Function
Public Function GetCSPDetails() As String
Dim hCryptProv As Long
Dim lLength As Long
Dim yContainer() As Byte
'Get handle to CSP
If CryptAcquireContext(hCryptProv, KEY_CONTAINER, SERVICE_PROVIDER, PROV_RSA_FULL, CRYPT_NEWKEYSET) = 0 Then
HandleError "Error during CryptAcquireContext for a new key container." & vbCrLf & _
"A container with this name probably already exists."
Exit Function
End If
'For developer info, show what the CSP & container name is
lLength = 1000
ReDim yContainer(lLength)
If CryptGetProvParam(hCryptProv, PP_NAME, yContainer(0), lLength, 0) <> 0 Then
GetCSPDetails = "Cryptographic Service Provider name: " & ByteToStr(yContainer, lLength)
End If
lLength = 1000
ReDim yContainer(lLength)
If CryptGetProvParam(hCryptProv, PP_CONTAINER, yContainer(0), lLength, 0) <> 0 Then
GetCSPDetails = GetCSPDetails & vbCrLf & "Key Container name: " & ByteToStr(yContainer, lLength)
End If
'Release provider handle.
If hCryptProv <> 0 Then
CryptReleaseContext hCryptProv, 0
End If
End Function
Private Function EncryptDecrypt(ByVal Data As String, ByVal Password As String, ByVal Encrypt As Boolean) As String
Dim hCryptProv As Long
Dim lLength As Long
Dim sTemp As String
Dim hHash As Long
Dim hKey As Long
'Get handle to CSP
If CryptAcquireContext(hCryptProv, KEY_CONTAINER, SERVICE_PROVIDER, PROV_RSA_FULL, CRYPT_NEWKEYSET) = 0 Then
If CryptAcquireContext(hCryptProv, KEY_CONTAINER, SERVICE_PROVIDER, PROV_RSA_FULL, 0) = 0 Then
HandleError "Error during CryptAcquireContext for a new key container." & vbCrLf & _
"A container with this name probably already exists."
Exit Function
End If
End If
'--------------------------------------------------------------------
'The data will be encrypted with a session key derived from the
'password.
'The session key will be recreated when the data is decrypted
'only if the password used to create the key is available.
'--------------------------------------------------------------------
'Create a hash object.
If CryptCreateHash(hCryptProv, CALG_MD5, 0, 0, hHash) = 0 Then
HandleError "Error during CryptCreateHash!"
End If
'Hash the password.
If CryptHashData(hHash, Password, Len(Password), 0) = 0 Then
HandleError "Error during CryptHashData."
End If
'Derive a session key from the hash object.
If CryptDeriveKey(hCryptProv, ENCRYPT_ALGORITHM, hHash, 0, hKey) = 0 Then
HandleError "Error during CryptDeriveKey!"
End If
'Do the work
sTemp = Data
lLength = Len(Data)
If Encrypt Then
'Encrypt data.
If CryptEncrypt(hKey, 0, 1, 0, sTemp, lLength, lLength) = 0 Then
HandleError "Error during CryptEncrypt."
End If
Else
'Encrypt data.
If CryptDecrypt(hKey, 0, 1, 0, sTemp, lLength) = 0 Then
HandleError "Error during CryptDecrypt."
End If
End If
'This is what we return.
EncryptDecrypt = Mid$(sTemp, 1, lLength)
'Destroy session key.
If hKey <> 0 Then
CryptDestroyKey hKey
End If
'Destroy hash object.
If hHash <> 0 Then
CryptDestroyHash hHash
End If
'Release provider handle.
If hCryptProv <> 0 Then
CryptReleaseContext hCryptProv, 0
End If
End Function
Private Sub HandleError(ByVal Error As String)
'You could write the error to the screen or to a file
Debug.Print Error
End Sub
Private Function ByteToStr(ByRef ByteArray() As Byte, ByVal lLength As Long) As String
Dim i As Long
For i = LBound(ByteArray) To (LBound(ByteArray) + lLength)
ByteToStr = ByteToStr & Chr$(ByteArray(i))
Next i
End Function
Private Function EncryptNumber(ByVal lNumber As Long) As String
Dim i As Long
Dim sNumber As String
sNumber = Format$(lNumber, "00000000")
For i = 1 To 8
EncryptNumber = EncryptNumber & Chr$(Asc(Mid$(NUMBER_ENCRYPT_PASSWORD, i, 1)) + Val(Mid$(sNumber, i, 1)))
Next i
End Function
Private Function DecryptNumber(ByVal sNumber As String) As Long
Dim i As Long
For i = 1 To 8
DecryptNumber = (10 * DecryptNumber) + (Asc(Mid$(sNumber, i, 1)) - Asc(Mid$(NUMBER_ENCRYPT_PASSWORD, i, 1)))
Next i
End Function
