Unique numeric IDs derived from string value

Question

If I have an unique String IDs that is made up of numbers letters and hyphens, i.e.

3c40df7f-1192-9fc9-ba43-5a2ffb833633

is there any way in Java to generate a numeric representations of these IDs and ensure that will also be unique.

Are all letters and numbers in such a string restricted to hex digits, i.e. [0-9a-f]? — rgettman
– rgettman, Commented Oct 16, 2013 at 16:04

Paul Vargas · Accepted Answer · 2013-10-16 16:06:06Z

2

With 32-bit integers no guarantee that they are unique. Possibly with BigInteger.

public static void main(String Args[]) {
    String id = "3c40df7f-1192-9fc9-ba43-5a2ffb833633";
    BigInteger number = new BigInteger(id.replace("-", ""), Character.MAX_RADIX);
    System.out.println(number);
}

Output:

5870285826737482651911256837071133277773559673999

The problem is the result will be the same for the following:

3c40df7f-1192-9fc9-ba43-5a2ffb833633
3c40df7f1192-9fc9-ba43-5a2ffb83-3633

answered Oct 16, 2013 at 16:06

Paul Vargas

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OldCurmudgeon · Accepted Answer · 2013-10-16 16:26:22Z

1

You can roll the bytes of the string directly into a BigInteger.

public void test() {
  String id1 = "3c40df7f-1192-9fc9-ba43-5a2ffb833633";
  BigInteger b1 = new BigInteger(id1.getBytes());
  System.out.println("B1="+b1+ "("+b1.toString(16)+")");
  String id2 = "3c40df7f1192-9fc9-ba43-5a2ffb833633";
  BigInteger b2 = new BigInteger(id2.getBytes());
  System.out.println("B2="+b2+ "("+b2.toString(16)+")");
  String id3 = "Gruntbuggly I implore thee";
  BigInteger b3 = new BigInteger(id3.getBytes());
  System.out.println("B3="+b3+ "("+b3.toString(16)+")");
  // You can even recover the original.
  String s = new String(b3.toByteArray());
  System.out.println("s="+s);
}

prints

B1=99828927016901697435065009039863622178352177789384078556155000206819390954492243882803(33633430646637662d313139322d396663392d626134332d356132666662383333363333)
B2=389956746159772255607368246163988513318828061453840042257565172951767502233603027763(3363343064663766313139322d396663392d626134332d356132666662383333363333)
B3=114811070151326385608028676923400900586729364355854547418637669(4772756e74627567676c79204920696d706c6f72652074686565)
s=Gruntbuggly I implore thee

I suppose you now need to explain what you mean by Number.

edited Oct 16, 2013 at 16:26

answered Oct 16, 2013 at 16:20

OldCurmudgeon

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1 Comment

Paul Vargas Over a year ago

Surely this is a better solution than mine.

ovunccetin · Accepted Answer · 2013-10-16 16:11:24Z

-1

You can convert a hexadecimal string to a number as follows:

String hexKey = "3c40df7f-1192-9fc9-ba43-5a2ffb833633";
long longKey = new BigInteger(hexKey.replace("-", ""), 16).longValue();

The generated long is also unique as long as the string without hyphens is unique.

edited Oct 16, 2013 at 16:11

answered Oct 16, 2013 at 16:05

ovunccetin

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1 Comment

Étienne Miret Over a year ago

Actually, only the 16 last characters (not including hyphens) of the string are taken into account with your solution. Since the example string given by the OP is 32 characters long (plus some hyphens), I don’t think this solution would fit the OP.

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