Link to home
Start Free TrialLog in
Avatar of Jeff Collins
Jeff CollinsFlag for Australia

asked on

VB6 to PHP Code Checksum

Hi,

I presently have a number of applications developed using VB6, that have a licence check attached to them based on a particular string, which is passed to a checksum function which returns an unlock access code.

Now I'm about to provide users with the ability to activate the applications online using the various inputs and what I must achieve is the same checksum access pin.

The VB6 code for the function is attached.

The website page is a PHP page, which i'm fairly new to, and i'm having difficulties in converting this particular critical function with byte variables and Xor operators from VB6 to PHP. The output string generally results in something like 3E696814 etc.. for the access pin.

Public Function getchecksum(sstring As String) As String
   
    Dim I As Integer
    Dim chk As Long
    Dim n1, n2, n3, n4, n5, n6, n7, n8 As Byte
    
    chk = 0
    For I = 1 To Len(sstring)
        chk = chk * 2
        
        ' Set bit 0 to whatever bit 16 is..
        If (chk And 65536) = 0 Then
            ' Bit 16 is clear, so clear bit 0
            chk = chk And 65534
        Else
            ' Bit 16 is set, so set bit 0
            chk = chk Or 1
        End If
        ' Clear bit 16..
        chk = chk And 65535
        
        ' XOR sum
        chk = chk Xor Asc(Mid$(buf, I, 1))
    Next I
    
    n8 = chk And 15
    chk = Int(chk / 4)
    n7 = chk And 15
    chk = Int(chk / 4)
    n6 = chk And 15
    chk = Int(chk / 4)
    n5 = chk And 15
    chk = Int(chk / 4)
    n4 = chk And 15
    chk = Int(chk / 4)
    n3 = chk And 15
    chk = Int(chk / 4)
    n2 = chk And 15
    chk = Int(chk / 4)
    n1 = chk And 15
    
    ' Add 48 to each one (ascii '0')
    n1 = n1 + 48
    
    If n1 > 57 Then
        n1 = 65 + (n1 - 57)
    End If
    
    n2 = n2 + 48
    
    If n2 > 57 Then
        n2 = 65 + (n2 - 57)
    End If
    
    n3 = n3 + 48
    
    If n3 > 57 Then
        n3 = 65 + (n3 - 57)
    End If
    
    n4 = n4 + 48
    If n4 > 57 Then
        n4 = 65 + (n4 - 57)
    End If
    
    n5 = n5 + 48
    If n5 > 57 Then
        n5 = 65 + (n5 - 57)
    End If
    
    n6 = n6 + 48
    If n6 > 57 Then
        n6 = 65 + (n6 - 57)
    End If
    
    n7 = n7 + 48
    If n7 > 57 Then
        n7 = 65 + (n7 - 57)
    End If
    
    n8 = n8 + 48
    If n8 > 57 Then
        n8 = 65 + (n8 - 57)
    End If
    
    
    getchecksum = Chr$(n1) + Chr$(n2) + Chr$(n3) + Chr$(n4) + Chr$(n5) + Chr$(n6) + Chr$(n7) + Chr$(n8)
End Function

Open in new window

Avatar of Brook Braswell
Brook Braswell
Flag of United States of America image

in PHP you use the ^ to represent Xor
Ord can be used for Asc
$chk = $chk ^ Ord(substr($buf,$I,$l))
ASKER CERTIFIED SOLUTION
Avatar of Brook Braswell
Brook Braswell
Flag of United States of America image

Link to home
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Start Free Trial
BTW Just noticed you var ( buf ) what is that from ?
Avatar of Jeff Collins

ASKER

Thanks heap Brook,

The Buf variable was actually a carry over from some testing, its is in fact the sstring variable.
Just cleaning up the above answer that needed some 'massaging' so other users find it heapful.

The final code for the checksum using PHP which is now working had the following amendments.

Key area of cleanup was;

1. Line 8:   If (($chk And 65536) === 0 ) changed to If (($chk And 65536) == 0 )

2. Line 13:  $chk = ($chk Or 1); changed to $chk = ($chk | 1);

3. Line 19: $chk = ($chk ^ Ord(substr($buf, $I, 1))); chaned to $chk = ($chk ^ Ord(substr($buf, $I-1, 1)));
 
4. Various Lines: All the And operators changed to &

Thanks Brooke1966 for pointing me in the right direction.
function getchecksum($sstring){
	    $chk = 0;
	    $len = strlen($sstring);	
	    for ($I=1; $I<=$len; $I++){
	        $chk = $chk * 2;
	        // Set bit 0 to whatever bit 16 is..
	        If (($chk & 65536) == 0 ){
	            // Bit 16 is clear, so clear bit 0
	            $chk = ($chk & 65534);
	        } else {
	            // Bit 16 is set, so set bit 0
	            $chk = ($chk | 1);
	        }
	        // Clear bit 16..
	        $chk = ($chk & 65535);
	        
	        // XOR sum
	        $chk = ($chk ^ Ord(substr($sstring, $I-1, 1)));
	    }
	    
	    $n8 = ($chk & 15);
	    $chk = intval($chk / 4);
	    $n7 = ($chk & 15);
	    $chk = intval($chk / 4);
	    $n6 = ($chk & 15);
	    $chk = intval($chk / 4);
	    $n5 = ($chk & 15);
	    $chk = intval($chk / 4);
	    $n4 = ($chk & 15);
	    $chk = intval($chk / 4);
	    $n3 = ($chk & 15);
	    $chk = intval($chk / 4);
	    $n2 = ($chk & 15);
	    $chk = intval($chk / 4);
	    $n1 = ($chk & 15);
	    
	    // Add 48 to each one (ascii '0')
	    $n1 = $n1 + 48;
	    
	    If ( $n1 > 57 ){
	        $n1 = 65 + ($n1 - 57);
	    }
	    
	    $n2 = $n2 + 48;
	    
	    If ( $n2 > 57 ){
	        $n2 = 65 + ($n2 - 57);
	    }
	    
	    $n3 = $n3 + 48;
	    
	    If ( $n3 > 57 ){
	       $n3 = 65 + ($n3 - 57);
	    }
	    
	    $n4 = $n4 + 48;
	    If ($n4 > 57){
	        $n4 = 65 + ($n4 - 57);
	    }
	    
	    $n5 = $n5 + 48;
	    If ($n5 > 57){
	        $n5 = 65 + ($n5 - 57);
	    }
	    
	    $n6 = $n6 + 48;
	    If ($n6 > 57){
	        $n6 = 65 + ($n6 - 57);
	    }
	    
	    $n7 = $n7 + 48;
	    If ($n7 > 57){
	        $n7 = 65 + ($n7 - 57);
	    }
	    
	    $n8 = $n8 + 48;
	    If ($n8 > 57 ){
	        $n8 = 65 + ($n8 - 57);
	    }

	    return chr($n1).chr($n2).chr($n3).chr($n4).chr($n5).chr($n6).chr($n7).chr($n8);
	}

Open in new window

Oh, and another quirky one was the original for ($I=1; strlen($sstring); $I++){ put the code into a constant loop for some reason, however when I declared the strlen() prior such as $len = strlen($sstring); and then used the for loop as for ($I=1; $I<=$len; $I++){ it worked.