Java random number with given length

Question

I need to genarate a random number with exactly 6 digits in Java. I know i could loop 6 times over a randomicer but is there a nother way to do this in the standard Java SE ?

EDIT - Follow up question:

Now that I can generate my 6 digits i got a new problem, the whole ID I'm trying to create is of the syntax 123456-A1B45. So how do i randomice the last 5 chars that can be either A-Z or 0-9? I'm thinking of using the char value and randomice a number between 48 - 90 and simply drop any value that gets the numbers that represent 58-64. Is this the way to go or is there a better solution?

EDIT 2:

This is my final solution. Thanks for all the help guys!

protected String createRandomRegistryId(String handleId)
{
    // syntax we would like to generate is DIA123456-A1B34      
    String val = "DI";      

    // char (1), random A-Z
    int ranChar = 65 + (new Random()).nextInt(90-65);
    char ch = (char)ranChar;        
    val += ch;      

    // numbers (6), random 0-9
    Random r = new Random();
    int numbers = 100000 + (int)(r.nextFloat() * 899900);
    val += String.valueOf(numbers);

    val += "-";
    // char or numbers (5), random 0-9 A-Z
    for(int i = 0; i<6;){
        int ranAny = 48 + (new Random()).nextInt(90-65);

        if(!(57 < ranAny && ranAny<= 65)){
        char c = (char)ranAny;      
        val += c;
        i++;
        }

    }

    return val;
}

http://stackoverflow.com/questions/363681/java-generating-random-number-in-a-range — Feanor, Mar 22 '11 at 14:21
Voting to reopen. Random number in *range* is slightly different from random number with given *length*. — aioobe, Jul 16 '19 at 14:22

aioobe · Answer 1 · 2016-08-13T16:39:13.457

To generate a 6-digit number:

Use Random and nextInt as follows:

Random rnd = new Random();
int n = 100000 + rnd.nextInt(900000);

Note that n will never be 7 digits (1000000) since nextInt(900000) can at most return 899999.

So how do I randomize the last 5 chars that can be either A-Z or 0-9?

Here's a simple solution:

// Generate random id, for example 283952-V8M32
char[] chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789".toCharArray();
Random rnd = new Random();
StringBuilder sb = new StringBuilder((100000 + rnd.nextInt(900000)) + "-");
for (int i = 0; i < 5; i++)
    sb.append(chars[rnd.nextInt(chars.length)]);

return sb.toString();

score 25 · Accepted Answer · edited Apr 15 '20 at 07:13

25

Generate a number in the range from 100000 to 999999.

// pseudo code
int n = 100000 + random_float() * 900000;

For more details see the documentation for Random

edited Apr 15 '20 at 07:13

Jean-François Fabre

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answered Mar 22 '11 at 14:20

Bombe

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random_float() is a pseudo-code function (i.e. it does not exist) that returns a random floating-point number between 0 (inclusive) and 1 (exclusive). – Bombe Mar 22 '11 at 14:26
Thank you this worked great! Got any solution for my follow up q also? – Marthin Mar 22 '11 at 14:43
16

Bad pseudo code and bad attitude – smeeb Apr 14 '17 at 00:54

Chand Priyankara · Answer 3 · 2013-02-21T09:44:28.003

10

If you need to specify the exact charactor length, we have to avoid values with 0 in-front.

Final String representation must have that exact character length.

String GenerateRandomNumber(int charLength) {
        return String.valueOf(charLength < 1 ? 0 : new Random()
                .nextInt((9 * (int) Math.pow(10, charLength - 1)) - 1)
                + (int) Math.pow(10, charLength - 1));
    }

edited Feb 21 '13 at 09:44

answered Feb 21 '13 at 09:31

Chand Priyankara

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Have to be careful with Integer overflow. Overall this is a clean solution. Thank you. – Devendra Lattu Mar 13 '23 at 18:15

score 6 · Answer 4 · edited Jul 03 '13 at 13:19

6

try this:

public int getRandomNumber(int min, int max) {
    return (int) Math.floor(Math.random() * (max - min + 1)) + min;
}

edited Jul 03 '13 at 13:19

Code Enthusiastic

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answered Mar 22 '11 at 14:22

Tobias

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T.K. · Answer 5 · 2011-03-22T14:32:54.730

1

int rand = (new Random()).getNextInt(900000) + 100000;

EDIT: Fixed off-by-1 error and removed invalid solution.

edited Mar 22 '11 at 14:32

answered Mar 22 '11 at 14:21

T.K.

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Mod a huge number by 899999 won't necessarily result in a uniform randomization. – aioobe Mar 22 '11 at 14:27
Should be `nextInt()` instead of `getNextInt()` – Aleksander Lech May 13 '20 at 07:45

score 0 · Answer 6 · edited Feb 22 '16 at 11:10

0

For the follow-up question, you can get a number between 36^5 and 36^6 and convert it in base 36

UPDATED:

using this code

http://javaconfessions.com/2008/09/convert-between-base-10-and-base-62-in_28.html

It's written BaseConverterUtil.toBase36(60466176+r.nextInt(2116316160))

but in your use case, it can be optimized by using a StringBuilder and having the number in the reverse order ie 71 should be converted in Z1 instead of 1Z

EDITED:

edited Feb 22 '16 at 11:10

Shashank Agrawal

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answered Mar 22 '11 at 15:00

Xavier Combelle

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Thanks for yore answer but I'm not sure how you mean? Coudl you please explain a bit more? – Marthin Mar 22 '11 at 15:22

score -1 · Answer 7 · answered Mar 22 '11 at 14:24

-1

Would that work for you?

public class Main {

public static void main(String[] args) {
    Random r = new Random(System.currentTimeMillis());
    System.out.println(r.nextInt(100000) * 0.000001);
}

}

result e.g. 0.019007

answered Mar 22 '11 at 14:24

monty

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seems I was a little bit late :) – monty Mar 22 '11 at 14:26
You can skip the argument to `currentTimeMillis`. The `Random()` constructor has a better approach. – aioobe Mar 22 '11 at 14:35

score -3 · Answer 8 · answered Mar 22 '11 at 14:21

-3

Generate a random number (which is always between 0-1) and multiply by 1000000

Math.round(Math.random()*1000000);

answered Mar 22 '11 at 14:21

Manoj

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Java random number with given length

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