Java random number with given length

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;
}

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();

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

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));
	}

try this:

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

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

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

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:

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

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

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