Java random number with given length

This question already has an answer here:

How do I generate random integers within a specific range in Java? 57 answers

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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 eg Random and can figure out the rest yourself.

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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 AZ 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();

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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));
    }

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