Generate Random String in java [duplicate]

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 · 2016-08-30 09:03:29Z

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??

DimaSan · Accepted Answer · 2016-08-30 08:15:24Z

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.

Community · Accepted Answer · 2017-05-23 12:32:47Z

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

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