ACTIVELOCK 1884

LOSS 27/10/2016 13:31:31
#468481

Boa tarde amigos,

Gostaria que alguém analisassem o código abaixo e descubra porque não gera a CHAVE de liberação.

Option Explicit
Dim mi_HashAlgorithm As Integer

Private Sub Command1_Click()
Dim F1 As Variant

Dim strKey As String
Dim strSoftwareName As String
Dim strSoftwarePassword As String
Dim strSoftwareCode As String
Dim intLiberationKeyLength As Integer

strSoftwarePassword = Me.Text(0)

strSoftwareName = Me.Text(4)

intLiberationKeyLength = Me.Text(1)

strSoftwareCode = Me.Text(2)

strKey = hash(strSoftwareName & strSoftwarePassword & strSoftwareCode)

strKey = UCase(Left(strKey, intLiberationKeyLength))

Text(3).Text = strKey

Text(3).SetFocus
Text(3).SelStart = 0
Text(3).SelLength = Len(Text(3).Text)

If Check1.Value = 1 Then
F1 = FreeFile
Open App.Path & [Ô]\[Ô] & Text(2).Text & [Ô].key[Ô] For Output As #F1
Print #F1, Text(3).Text
Close #F1
End If


SaveSetting [Ô]KeyGenerator[Ô], [Ô]Default[Ô], [Ô]SoftwareName[Ô], strSoftwareName
SaveSetting [Ô]KeyGenerator[Ô], [Ô]Default[Ô], [Ô]SoftwarePassword[Ô], strSoftwarePassword
SaveSetting [Ô]KeyGenerator[Ô], [Ô]Default[Ô], [Ô]LiberationKeyLength[Ô], intLiberationKeyLength
SaveSetting [Ô]KeyGenerator[Ô], [Ô]Default[Ô], [Ô]HashAlgorithm[Ô], mi_HashAlgorithm
End Sub

Private Function htSHA1AA1Hash(ByVal strHashThis As String)

End Function
Private Function htSHA1AA2Hash(ByVal strHashThis As String)

End Function
Private Function htMD5AA1Hash(ByVal strHashThis As String)

End Function
Private Function htMD5AA2Hash(ByVal strHashThis As String)

End Function
Private Function htMD5AB1Hash(ByVal strHashThis As String)

End Function
Private Function htMD5AB2Hash(ByVal strHashThis As String)

End Function
Private Function ht(ByVal strHashThis As String)

End Function

Private Function hash(ByVal strHashThis As String)

Select Case mi_HashAlgorithm
Case 0: hash = htSHA1AA1Hash(strHashThis)
Case 1: hash = htSHA1AA2Hash(strHashThis)
Case 2: hash = htMD5AA1Hash(strHashThis)
Case 3: hash = htMD5AA2Hash(strHashThis)
Case 4: hash = htMD5AB1Hash(strHashThis)
Case 5: hash = htMD5AB2Hash(strHashThis)
Case Else: hash = htSHA1AA1Hash(strHashThis) [ô]Default
End Select
End Function

Private Sub Command2_Click()
End
End Sub

Private Sub Form_Load()
Dim strSoftwareName As String
Dim strSoftwarePassword As String
Dim strSoftwareCode As String
Dim intLiberationKeyLength As Integer, intHashAlgorithm As Integer


lblPanel = [Ô]Ronaldo Rodrigues Alcântara - 55 (028) 9 9935-0095[Ô]


strSoftwareName = GetSetting([Ô]KeyGenerator[Ô], [Ô]Default[Ô], [Ô]SoftwareName[Ô], [Ô]AppName[Ô])
strSoftwarePassword = GetSetting([Ô]KeyGenerator[Ô], [Ô]Default[Ô], [Ô]SoftwarePassword[Ô], [Ô]password[Ô])
intLiberationKeyLength = GetSetting([Ô]KeyGenerator[Ô], [Ô]Default[Ô], [Ô]LiberationKeyLength[Ô], 16)
intHashAlgorithm = GetSetting([Ô]KeyGenerator[Ô], [Ô]Default[Ô], [Ô]HashAlgorithm[Ô], 0)

intLiberationKeyLength = 16

Me.Text(0).Text = strSoftwarePassword
Me.Text(1).Text = intLiberationKeyLength
optHashAlgorithm(intHashAlgorithm) = True

End Sub

Private Sub optHashAlgorithm_Click(Index As Integer)

mi_HashAlgorithm = optHashAlgorithm(Index).Index
End Sub

Private Sub optHashAlgorithm_KeyPress(Index As Integer, KeyAscii As Integer)

If KeyAscii = 13 Then
SendKeys [Ô]{TAB}[Ô]
KeyAscii = 0
End If
End Sub

Private Sub optHashAlgorithm_KeyUp(Index As Integer, KeyCode As Integer, Shift As Integer)

Command1_Click
End Sub

Private Sub optHashAlgorithm_MouseUp(Index As Integer, Button As Integer, Shift As Integer, X As Single, Y As Single)

Command1_Click
End Sub

Private Sub Text_GotFocus(Index As Integer)
Dim I

For I = 0 To Text.Count - 1
Text(I).SelStart = 0
Text(I).SelLength = Len(Text(I).Text)
Next I
End Sub

Private Sub Text_KeyPress(Index As Integer, KeyAscii As Integer)

If KeyAscii = 13 Then
SendKeys [Ô]{TAB}[Ô]
KeyAscii = 0
End If
End Sub