How to create a random number with range?

Question

I am trying to create a random phone number with a range. The format being (xxx)-xxx-xxx and the area code not starting with 0,8, or 9 and the next set of three being in a range from 100-742 and then the last set of 4 can be any digit. How would i create the first two parts? Any help would be appreciated. Thanks!

import java.util.*;
import java.text.*;

public class PhoneNumber{
    public static void main(String[] arg){
        Random ranNum = new Random();
        //int areaCode = 0;
        //int secSet = 0;
        //int lastSet = 0;
        DecimalFormat areaCode = new DecimalFormat("(000)");
        DecimalFormat secSet = new DecimalFormat("-000");
        DecimalFormat lastSet = new DecimalFormat("-0000");
        //DecimalFormat phoneNumber = new DecimalFormat("(###)-###-####");
        int i = 0;
        //areaCode = (ranNum.nextInt()); //cant start with 0,8,9
        //secSet = (ranNum.nextInt()); // not greater than 742 and less than 100
        //lastSet = (ranNum.nextInt(999)) + 1; // can be any digits


        i = ranNum.nextInt();
        System.out.print(areaCode.format(i));
        i = ranNum.nextInt();
        System.out.print(secSet.format(i));
        i = ranNum.nextInt();
        System.out.print(lastSet.format(i));

    }
}

nextInt takes an argument: http://docs.oracle.com/javase/7/docs/api/java/util/Random.html#nextInt(int) `areaCode = nextInt(700) + 100;` to start you off. — Radiodef, Oct 28 '13 at 22:57

kiruwka · Answer 1 · 2013-10-28T23:42:14.177

1

well, basically, you need to generate numbers in two ranges

[1; 7]
[100; 742]

To have random integer in range [m; n] you could write:
updated (remove Math.random())

int numberInRange = m + new Random().nextInt(n - m + 1);

HTH

edited Oct 28 '13 at 23:42

answered Oct 28 '13 at 23:02

kiruwka

9,250
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1

don't do Math.random() * 48. That's buggy. Always use Math.nextInt(48) instead. – iluxa Oct 28 '13 at 23:33
Check this out for more info: http://stackoverflow.com/questions/738629/math-random-versus-random-nextintint – iluxa Oct 28 '13 at 23:42
@iluxa Yep, read it alredy. It's a shame i stayed ignorant of nextInt(). thanks – kiruwka Oct 28 '13 at 23:43

Mazzy · Answer 2 · 2013-10-28T23:42:18.880

1

1,2,3,4,5,6,7 makes 7 different values

    ranNum.nextInt(7)+1; //So 1 is your lowest number and 7 is the number of different solutions

nexInt will range between 0 and intPassed exclusive,
So ranNum.nextInt(7) will run between 0 and 6, + 1 makes 1 .. 7
This will range between 1 and 7
You can take the same principal for the second range

edited Oct 28 '13 at 23:42

answered Oct 28 '13 at 23:04

Mazzy

1,901
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Paul Vargas · Answer 3 · 2013-10-28T23:12:03.323

0

You can try the next:

int sec = java.util.concurrent.ThreadLocalRandom.current().nextInt(100, 743);

And so on with the other parts of the phone number.

This method returns a pseudorandom, uniformly distributed value between the given least value (inclusive) and bound (exclusive).

edited Oct 28 '13 at 23:12

answered Oct 28 '13 at 23:06

Paul Vargas

41,222
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CandyCane · Answer 4 · 2013-10-28T23:24:04.713

Just make a random number greater than 99 and less than 800, then the next would be about the same way.

 Random rand = new Random();

// add 1 to make it inclusive
                                   max   min
int firstRandomSet = rand.nextInt((799 - 100) + 1) + 100; 
//none starts with 0,8, or 9


int secondRandomSet = rand.nextInt((742 - 100) + 1) + 100; 
//produces anything from 100-742

to get the numbers 0001 - 9999 you'll have to be creative.

int maxValues= 9999;

       int thirdRandomSet = rand.nextInt(maxValues);

       System.out.printf("%04d\n", thirdRandomSet);

How to create a random number with range?

4 Answers4