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| This makes little sense. Here's the REAL way on getting the password out of this script. Just follow the
authentication procedure backwards. It's that simple. This script tries to confuse the reader with bit
shifts and math functions and constants, but if you were to just work them out it's really quite simple.
For instance we have the following line of code:
phase1 = Math.ceil(Math.random()) - 6 + (2<<2)
But, 0 < Math.random() < 1 and 2<<2 = 8, so Math.ceil(...) = 1 and therefore phase1 = 8. It's just that
simple. Anyway, the password is always in pass[3] and pass[4][0] is the length of the password,
not lim. So in your example:
pass[0]="J1MtnBQAfoxi1WO"
pass[1]="gFViHrEGzELGXK"
pass[2]="ALoTt4CuswQGzah"
pass[3]="wyVAKUCMEIVg"
pass[4]="9LGQAm5YIHf6WCvJ"
pass[5]="53LGQAm5YIHf6WCv"
The "encrypted" password is wyVAKUCME of length 9. The number 53 (pass[5][0] and pass[5][1]) is the "key" so to say. In the function testit(), this number is XOR'd with the index of numletter[] where the letter of the encrypted password occurs. (If you don't know what XOR is and why it work in this script, you really have no business being a programmer in my opinion... Know your bitwise operators!). So we have 'w' which is at index 32 in numletter. 32 XOR 53 = 100000 XOR 110101 = 010101 = 21. So the first letter of the password is numletter[21] = l. Continuing in the fashion give you password. By the way "blanchard" is not the password in this example.
The better way to password protect is just have the password be the name of webpage (as is here) and skip the validation all together. If the password is wrong, they get a 404, and if it's right they're in. |
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| | Correction, phase1 = 3... Can't add today! |
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| What do you think is the password here, then?? XD Guess what it is!
var pass=new Array(
var t3=""
var lim=7
pass[0]="ncHc57SrAJ0v2Nc"
pass[1]="XPsUBB3U8zmXHzf"
pass[2]="yxiz8LlAhTqLfV"
pass[3]="6bo5rliF19kO2hl"
pass[4]="7l8zkfDkOly1HGGa"
pass[5]="21l8zkfDkOly1HGG"
email me at potlog2carlo @ yahoo dot com
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| | anonymous, you can try it...if you want to crack a page, the least you can do is to find the password by yourself!! XDD |
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| Just make a small Java program like this:
public static void main(String[] args) {
int lim = 9;
String [] pass = new String[6];
String numletter="0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
pass[0]="OUR4O2ZpDtLP7qI";
pass[1]="yp1qUglYgOyK1bu";
pass[2]="sNdHBTPgeK8Ht0b";
pass[3]="jm3fileafMGvhFS";
pass[4]="9mixyaT7TrOd1M5W";
pass[5]="24mixyaT7TrOd1M5";
String answer = getPass(pass, lim, numletter);
System.out.println("\nThe password is: '" + answer + "'");
}
public static String getPass(String [] pass, int lim, String numletter) {
String decoded = "";
String encoded = pass[3].substring(0, lim);
for (int i=0; i<lim; i++) {
int p = numletter.indexOf(encoded.charAt(i));
int s = Integer.parseInt(pass[5].substring(0, 2));
decoded += numletter.charAt(p^s);
}
return decoded;
}
Change the lim, pass array, and numletter to whatever is on the website. Run the program, get password. |
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Copyright Open Source Institute, 2006 