[RESOLVED] Why is it so hackable?

why is this code so easy to hack? my friend told me if i do this to may app ppl can just find out whats the password or username very easy?

VB Code:

Private Sub cmdlogin_Click()
If txtusername.Text = "TestUser" And txtpassword.Text = "TestPassword" Then
frmloged.Show
Unload Me
Else
MsgBox "The UserName Or Password is not correct!", vbExclamation + vbOKOnly, "Error!!!"
End If
End Sub

Because you have the user name and password in strings which will stay as pure text even after you have compiled your program.

Best to have an external file where the password is saved only encrypted.

how do i get it kind of like encrypted?

On the short note - if you compile your app and then change file extension from EXE to say TXT and open it in Notepad you will see something like the following:
... cmdlogin 
 Command1 p°¿ï
 ÿ*  txtpassword ` h
¿ï
 Text2 
ÿ*
 txtusername Hh
¿ï
 Text1  ÿ ...

It would take probably under a minute a relatively inexperienced amature to decode that garbage...

Anyway, you never hardcode any values especially User Name/Password. Instead you store it say database but ecrypted. There are many encryption algorythms exist so you will have to find something that fits your level of expertise by searching our forums or elsewhere.

Good luck.

Quote:

---

Originally Posted by Joacim Andersson

Because you have the user name and password in strings which will stay as pure text even after you have compiled your program.

---

you can see

wow WIZ126 its so easy to get the username or password!!!!

how do i make it like encrypted in someway?

@RhinoBull, you dont need to change the file extension from exe to txt. Just open Notepad and drag ' drop your exe from Explorer into the notepad text area and you will get the same result. ;)

@RobDog: that is too much work - changing ext works better for me. ;)

You create your own algorithm that changes the ascii bytes of the characters of your password into a complete mess that someone cannot understand, and you can use this algorithm to convert it to the real password ;)

That is a big NO-NO, Jacob - it can be decrypted within a few seconds (by experienced guy :) ). 64/128 bit encryption is the way to go (if you can afford it...)

Quote:

---

Originally Posted by Jacob Roman

You create your own algorithm that changes the ascii bytes of the characters of your password into a complete mess that someone cannot understand, and you can use this algorithm to convert it to the real password ;)

---

any good links on how to make it?

Quote:

---

Originally Posted by RhinoBull

That is a big NO-NO, Jacob - it can be decrypted within a few seconds (by experienced guy :) ). 64/128 bit encryption is the way to go (if you can afford it...)

---

I didn't say how complicated it had to be or how it had to be encrypted. Of course the real big government agencies use very large and complex Calculus based algorithms for their passwords (wasn't that on the movie Mercury Rising?) But in his case since he's a beginner, any normal algorithm will suffice.

If you need to hard code a username/password into your app (which, as others have already said is not a good idea), you can do some really simple stuff which will make it difficult for the casual hacker. For example,

VB Code:

Option Explicit
 
Dim TestUser As String
Dim TestPass As String
 
Private Sub cmdLogin_Click()
 
    If (txtUserName.Text = TestUser) And (txtPassword.Text = TestPass) Then
        MsgBox "Wooo hooo... you'[color=black]re in!!"[/color]
    Else
        MsgBox "Incorrect username/password"
        Unload Me
    End If
    
End Sub
 
Private Sub Form_Load()
 
    TestUser = Chr$(Asc("T")) & Chr$(Asc("e")) & Chr$(Asc("s")) & Chr$(Asc("t"))
    TestUser = TestUser & Chr$(Asc("U")) & Chr$(Asc("s")) & Chr$(Asc("e")) & Chr$(Asc("r"))
    TestPass = Chr$(Asc("T")) & Chr$(Asc("e")) & Chr$(Asc("s")) & Chr$(Asc("t"))
    TestPass = TestPass & Chr$(Asc("P")) & Chr$(Asc("a")) & Chr$(Asc("s")) & Chr$(Asc("s"))
        
End Sub

That's pretty simple but fairly effective without resorting to complex encryption algorithms. Obviously you'd make your test username & password a bit more obscure.

BTW Did I mention that hardcoding a username & password is a bad idea? :eek2:

Quote:

---

Originally Posted by pnish

VB Code:

If (txtUserName.Text = TestUser) And (txtPassword.Text = TestPass) Then
        MsgBox "Wooo hooo... you'[color=black]re in!!"[/color]
    Else
        MsgBox "Incorrect username/password"
        Unload Me
    End If

---

Actually, if I can throw my 2 cents in on a resolved thread, this type of structure is also trivial to hack. All you have to do is trace back the "Wooo hooo..." string to it's test and then make the jump instruction non-conditional. This type of string basically provides a big "crack me here" sign in your code. These two links (one from a cracking site) give some really good advice on protecting software:


http://www.woodmann.com/fravia/protec.htm
http://lastbit.com/vitas/antihack.asp

here is some code I have used before works well

VB Code:

Public Function Encode(Data As String) As String
    
    Randomize
    
    Dim Key() As Long
    ReDim Key(Len(Data))
    
    Dim i As Long
    Dim LenData As Long
    Dim Coded As String
    Coded = ""
    
    LenData = Len(Data)
    For i = 1 To LenData
        Key(i) = (Rnd() * 50 + 1) + 20 'Define keys for each character
    Next
    
    For i = 1 To LenData
        'Adding the key to each character's ascii
        If Asc(Mid$(Data, i, 1)) + Key(i) > 255 Then
            'If the new ascii value exceeds 255(Highest char ascii), then count upwards from 0
            Coded = Coded & Chr$(Key(i)) & Chr$(Asc(Mid$(Data, i, 1)) + Key(i) - 255)
        Else
            Coded = Coded & Chr$(Key(i)) & Chr$(Asc(Mid$(Data, i, 1)) + Key(i))
        End If
    Next
    'Return encoded value
    'Debug.Print Len(Coded)
    Encode = Coded
    
End Function
 
Public Function Decode(Data As String) As String
    
    Dim Key() As Long
    ReDim Key(Len(Data))
    
    Dim i As Long
    Dim Decoded As String
    Dim CodedString As String
    Dim LenData As Long
    Dim LenCodedString As Long
    Dim NextChr As Long
    
    Decoded = ""
    CodedString = ""
    
    'Seperate the key from the actual code
    LenData = Len(Data)
    For i = 1 To LenData
        If (i / 2) = Int(i / 2) Then
            CodedString = CodedString & Mid$(Data, i, 1)
        Else
            Key(((i - 1) / 2) + 1) = Asc(Mid$(Data, i, 1))
        End If
    Next
    
    'Minus the key from each character
    LenCodedString = Len(CodedString)
    For i = 1 To LenCodedString
    NextChr = Asc(Mid$(CodedString, i, 1)) - Key(i)
    'If the new ascii is below 0, then count backwards from 255
    If NextChr <= 0 Then
    NextChr = NextChr + 255
End If
'Add to decoded string
Decoded = Decoded + Chr$(NextChr)
Next
'Return Decoded value
Decode = Decoded
 
End Function