Answers for "how to generate a random number string java"

8

generate random string in java

import java.util.Random;

public class Generator {
	public static String generateRandomPassword(int len) {
		String chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijk"
          +"lmnopqrstuvwxyz!@#$%&";
		Random rnd = new Random();
		StringBuilder sb = new StringBuilder(len);
		for (int i = 0; i < len; i++)
			sb.append(chars.charAt(rnd.nextInt(chars.length())));
		return sb.toString();
	}
}
Posted by: Guest on July-16-2020
0

java generate random string and number

import java.util.Random;

public class passGen{
    // Version 1.0
    private static final String dCase = "abcdefghijklmnopqrstuvwxyz";
    private static final String uCase = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    private static final String sChar = "!@#$%^&*";
    private static final String intChar = "0123456789";
    private static Random r = new Random();
    private static StringBuilder pass = new StringBuilder();

    public static void main (String[] args) {
        System.out.println ("Generating pass...");
        while (pass.length () != 16){
            int rPick = r.nextInt(4);
            if (rPick == 0){
                int spot = r.nextInt(26);
                pass.append(dCase.charAt(spot));
            } else if (rPick == 1) {
                int spot = r.nextInt(26);
                pass.append(uCase.charAt(spot));
            } else if (rPick == 2) {
                int spot = r.nextInt(8);
                pass.append(sChar.charAt(spot));
            } else {
                int spot = r.nextInt(10);
                pass.append(intChar.charAt(spot));
            }
        }
        System.out.println ("Generated Pass: " + pass.toString());
    }
}
Posted by: Guest on February-26-2022

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