Generate Random String in java

Question

I'm trying to generate a string between capital A-Z in java using Secure Random. Currently I'm able to generate an alphanumeric string with special characters but I want a string with only upper case alphabets.

  public String createRandomCode(int codeLength, String id){   
     char[] chars = id.toCharArray();
        StringBuilder sb = new StringBuilder();
        Random random = new SecureRandom();
        for (int i = 0; i < codeLength; i++) {
            char c = chars[random.nextInt(chars.length)];
            sb.append(c);
        }
        String output = sb.toString();
        System.out.println(output);
        return output ;
    }

The input parameters are length of the output string & id whhich is alphanumeric string.Can't understand what modifications to make to the above code to generate only upper case alphabet string. Please help..

shmosel · Accepted Answer · 2016-08-30T09:03:29.877

Your method randomly selects characters out of the id argument. If you want those to only be uppercase letters, then pass a string with those characters:

String randomCode = createRandomCode(length, "ABCDEFGHIJKLMNOPQRSTUVWXYZ");

EDIT If you want to avoid duplicates, you can't just select characters at random. You'll want to shuffle them and pick out the first n characters:

public String createRandomCode(int codeLength, String id) {   
    List<Character> temp = id.chars()
            .mapToObj(i -> (char)i)
            .collect(Collectors.toList());
    Collections.shuffle(temp, new SecureRandom());
    return temp.stream()
            .map(Object::toString)
            .limit(codeLength)
            .collect(Collectors.joining());
}

EDIT 2 Just for fun, here's another way to implement the original random code generator (allowing duplicates):

public static String createRandomCode(int codeLength, String id) {
    return new SecureRandom()
            .ints(codeLength, 0, id.length())
            .mapToObj(id::charAt)
            .map(Object::toString)
            .collect(Collectors.joining());
}

What changes do I need to make in the above code so that the string contains no repeating alphabets?? — Lucy, Aug 30 '16 at 08:38

DimaSan · Answer 2 · 2016-08-30T08:15:24.530

Here is generator that I wrote and use:

public class RandomGenerator {
    private static final String characters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    public static String generateRandom(int length) {
        Random random = new SecureRandom();
        if (length <= 0) {
            throw new IllegalArgumentException("String length must be a positive integer");
        }

        StringBuilder sb = new StringBuilder(length);
        for (int i = 0; i < length; i++) {
            sb.append(characters.charAt(random.nextInt(characters.length())));
        }

        return sb.toString();
    }
}

in numChars string you can put any characters you want to be included. int length parameter is the length of generated random string.

score 0 · Answer 3 · edited May 23 '17 at 12:32

Here is an example method that uses the int range for characters A to Z (also this method avoids duplicate characters in the String) :

public String createRandomCode(final int codeLength) {

    int min = 65;// A
    int max = 90;// Z


    StringBuilder sb = new StringBuilder();
    Random random = new SecureRandom();

    for (int i = 0; i < codeLength; i++) {

        Character c;

        do {

            c = (char) (random.nextInt((max - min) + 1) + min);

        } while (sb.indexOf(c.toString()) > -1);

        sb.append(c);
    }

    String output = sb.toString();
    System.out.println(output);
    return output;
}

The range part comes from this topic : Generating random integers in a specific range

Generate Random String in java

3 Answers3