Generate secure random number with SecureRandom

Question

How can I generate a 6 digit integer using SecureRandom class of Java?

I am trying the following code to generate random numbers :

SecureRandom secureRandom = new SecureRandom();
int secureNumber = secureRandom.nextInt();

It is generating random numbers with any length including negative number. I don't find any method in SecureRandom class to provide a range of numbers. I want to generate a 6 digits positive random number

Answer 1

simply done it with an array

int[] arr = new int[6];
Random rand = new SecureRandom();
for(int i= 0; i< 6 ; i++){
    // 0 to 9
    arr[i] = rand.nextInt(10);   
}

Dont know if you need it in another type (if you want int look here: How to convert int array to int?

Answer 2

This will create a string of 6 random digits.

SecureRandom test = new SecureRandom();
int result = test.nextInt(1000000);
String resultStr = result + "";
if (resultStr.length() != 6) 
    for (int x = resultStr.length(); x < 6; x++) resultStr = "0" + resultStr;
System.out.println(resultStr); // prints the 6 digit number

Should work quite nicely. Tested so that it would spit out 1000 numbers and all of them were 6 digits long including leading zeros for when the random number initially chosen was less than 100,000.

Answer 3

You can do something like this:

static String generate(int len){
   SecureRandom sr = new SecureRandom();
   String result = (sr.nextInt(9)+1) +"";
   for(int i=0; i<len-2; i++) result += sr.nextInt(10);
   result += (sr.nextInt(9)+1);
   return result;
}

Test:

public static void main(String[] argv){
   for(int i=3; i<25; i++) System.out.println(generate(i));
}

Output:

639
1617
84489
440757
9982141
28220183
734679206
1501896787
29547455245
417101844095
9997440470982
24273208689568
235051176494856
9515304245005008
73519153118911442
665598930463570609
9030671114119582966
96572353350467673335
450430466373510518561
9664395407658562145827
26651025927755496179441
421157180102739403860678

Answer 4

This worked for me.

new SecureRandom().nextBytes(values);
int number = Math.abs(ByteBuffer.wrap(values).getInt());
while (number >= 1000000) {
     number /= 10;
}

Answer 5

This is how you would do it using Java IntStream

private final SecureRandom secureRandom = new SecureRandom();

public String generateCode() {
    return IntStream.iterate(0, i -> secureRandom.nextInt(10))
            .limit(6)
            .collect(StringBuilder::new, StringBuilder::append, StringBuilder::append).toString();
}

Answer 6

Maybe using this code?

String.format("%06d", secureRandom.nextInt(999999));