How to generate a random alpha-numeric string?

By : Todd

I've been looking for a simple Java algorithm to generate a pseudo-random alpha-numeric string. In my situation it would be used as a unique session/key identifier that would "likely" be unique over 500K+ generation (my needs don't really require anything much more sophisticated). Ideally, I would be able to specify a length depending on my uniqueness needs. For example, a generated string of length 12 might look something like "AEYGF7K0DM1X".

By : Todd


Maybe this is helpful

package password.generater;

import java.util.Random;

 * @author dell
public class PasswordGenerater {

     * @param args the command line arguments
    public static void main(String[] args) {
        int length= 11;

        // TODO code application logic here
    static char[] generatePswd(int len){
        System.out.println("Your Password ");
        String Chars="abcdefghijklmnopqrstuvwxyz";
        String nums="0123456789";
        String symbols="[email protected]#$%^&*()_+-=.,/';:?><~*/-+";
        String passSymbols=charsCaps + Chars + nums +symbols;
        Random rnd=new Random();
        char[] password=new char[len];
        int index=0;
        for(int i=0; i<<len;i++){
      return password;


Using UUIDs is insecure, because parts of the UUID arn't random at all. The procedure of @erickson is very neat, but does not create strings of the same length. The following snippet should be sufficient:

 * The random generator used by this class to create random keys.
 * In a holder class to defer initialization until needed.
private static class RandomHolder {
    static final Random random = new SecureRandom();
    public static String randomKey(int length) {
        return String.format("%"+length+"s", new BigInteger(length*5/*base 32,2^5*/, random)
            .toString(32)).replace('\u0020', '0');

Why choosing length*5. Let's assume the simple case of a random string of length 1, so one random character. To get a random character containing all digits 0-9 and characters a-z, we would need a random number between 0 and 35 to get one of each character. BigInteger provides a constructor to generate a random number, uniformly distributed over the range 0 to (2^numBits - 1). Unfortunately 35 is no number which can be received by 2^numBits - 1. So we have two options: Either go with 2^5-1=31 or 2^6-1=63. If we would choose 2^6 we would get a lot of "unnecesarry" / "longer" numbers. Therefore 2^5 is the better option, even if we loose 4 characters (w-z). To now generate a string of a certain length, we can simply use a 2^(length*numBits)-1 number. The last problem, if we want a string with a certain length, random could generate a small number, so the length is not met, so we have to pad the string to it's required length prepending zeros.

static final String AB = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
static SecureRandom rnd = new SecureRandom();

String randomString( int len ){
   StringBuilder sb = new StringBuilder( len );
   for( int i = 0; i < len; i++ ) 
      sb.append( AB.charAt( rnd.nextInt(AB.length()) ) );
   return sb.toString();
By : maxp

This video can help you solving your question :)
By: admin