IsNumber Problems

[Unc3nZureD 13]

I'm currently trying to make a serial-system for my software, and it would be the following logic:

Creat a Serial:

$Number = Random(10, 99999, 1)
   $Encrypted = _StringEncrypt(1, $Number, "Test_Code", 2)
   ClipPut($Encrypted)

Okay, it works well.

But when I want to read an input and decrypt it, my program fails. Decrypting the code would give me a number.

Decryption:

$Given = GUICtrlRead($Input1)
   $Decrypted = _StringEncrypt(0, $Given, "Test_Code", 2)
   If IsNumber($Encrypted) = 1 Then
    MsgBox(64, "", "Awesome! Your Key is perfect." & @CRLF & "Your username is: " & $Decrypted)
   Else
    MsgBox(16, "", "Key is wrong." & @CRLF & "Incorrect username is: " & $Decrypted)
   EndIf

No, it's not a number... Is there any way to fix it?

Edited April 27, 2012 by Unc3nZureD

[guinness 1,459]

Are you sure that's returning a number? Look at StringIsNumber anyway.

Edited April 27, 2012 by guinness

[UEZ 1,082]

Try this:

If StringIsInt($Decrypted) Then
    MsgBox(64, "", "Awesome! Your Key is perfect." & @CRLF & "Your username is: " & $Decrypted)
   Else
    MsgBox(16, "", "Key is wrong." & @CRLF & "Incorrect username is: " & $Decrypted)
   EndIf

Br,

UEZ

[Unc3nZureD 13]

Hmm, StringIsNumber doesn't exist.

And UEZ:

Someway it still isn't OK.

Fo example the "3F1A25F26D18BC1729D27D6A552069E45304812F" Returns: "65927"

And it doesn't accept it.

I tried a non generated code just a simplified one:

$qwe = "123"
$asd = 123
MsgBox(0, "", IsNumber($qwe) & @CRLF & IsNumber($asd))

The first one is not a number but second one is a number!

Edited April 27, 2012 by Unc3nZureD

[guinness 1,459]

Whoops, maybe StringIsXDigit.

[Unc3nZureD 13]

It's getting closer but still not good :S

Real code: 3F1A22826D18C26129D27A6C552069E15304845F

Fake code: asd

Both are accepted.

Username at real one: 43273

Username at fake one: Is an ascii character for the symbol of women.

It is not a valid username =D

[UEZ 1,082]

You mixed up a variable ->

If IsNumber($Encrypted) Then

it must be

If IsNumber($Decrypted) Then

This example works for me.

#include <string.au3>


$Number = Random(10, 99999, 1)
ConsoleWrite('@@ Debug(' & @ScriptLineNumber & ') : $Number = ' & $Number & @crlf & '>Error code: ' & @error & @crlf) ;### Debug Console

$Encrypted = _StringEncrypt(1, $Number, "Test_Code", 2)
ConsoleWrite('@@ Debug(' & @ScriptLineNumber & ') : $Encrypted = ' & $Encrypted & @crlf & '>Error code: ' & @error & @crlf) ;### Debug Console

$Given = $Encrypted
$Decrypted = _StringEncrypt(0, $Given, "Test_Code", 2)

If StringIsInt($Decrypted) Then
    MsgBox(64, "", "Awesome! Your Key is perfect." & @CRLF & "Your username is: " & $Decrypted)
Else
    MsgBox(16, "", "Key is wrong." & @CRLF & "Incorrect username is: " & $Decrypted)
EndIf

Br,

UEZ

Edited April 27, 2012 by UEZ

[Unc3nZureD 13]

Lol. Epic Fail. Thanks