#### Java random number with given length

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Hi Expert!

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;
}

1. ##### Re: Java random number with given length

Hi Goti Bandhu!

Generate a number in the range from 100000 to 999999.

// pseudo code
int n = 100000 + random_float() * 900000;
I’m pretty sure you have already read the documentation for e.g. Random and can figure out the rest yourself.