How to generate a random string of 20 characters

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How to generate a random String in Java

I am wanting to generate a random string of 20 characters without using apache classes. I don't really care about whether is alphanumeric or not. Also, I am going to convert it to an array of bytes later FYI.

Thanks,

In one line:UUID.randomUUID().toString().replace("-","").substring(0,20) — Nolf, Nov 19 '15 at 15:14
These 3 single line codes are very much useful i guess.. `Long.toHexString(Double.doubleToLongBits(Math.random()));` `UUID.randomUUID().toString();` `RandomStringUtils.randomAlphanumeric(20);` — Manindar, Jun 08 '16 at 07:36

score 92 · Accepted Answer · edited Jan 24 '20 at 05:39

92

Here you go. Just specify the chars you want to allow on the first line.

char[] chars = "abcdefghijklmnopqrstuvwxyz".toCharArray();
StringBuilder sb = new StringBuilder(20);
Random random = new Random();
for (int i = 0; i < 20; i++) {
    char c = chars[random.nextInt(chars.length)];
    sb.append(c);
}
String output = sb.toString();
System.out.println(output);

If you are using this to generate something sensitive like a password reset URL or session ID cookie or temporary password reset, be sure to use java.security.SecureRandom instead. Values produced by java.util.Random and java.util.concurrent.ThreadLocalRandom are mathematically predictable.

edited Jan 24 '20 at 05:39

Jonathan Leitschuh

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answered Apr 15 '11 at 23:37

WhiteFang34

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this is not a good method...i tried getting hundred random nos. but very few were unique. I put this code in a while loop and iterated 0 to 99. many outputs were repeated.. – Nikhil Jun 20 '12 at 16:15
@Nikhil I suspect you did something wrong. If you need a better source of random numbers, try `SecureRandom` instead. That shouldn't matter for duplicates in this case though, looping over the above code I got these 100 unique strings: http://pastebin.com/ELZ950tk – WhiteFang34 Jun 21 '12 at 05:26
@WhiteFang34 Hi, I tried this code in a MIDlet. I assume this might be working fine on a desktop java application. You can check my results here :http://pastebin.com/yDcT41g8 . So when you use it in a MIDlet you get repeated strings. The only change i made was to use StringBuffer instead StringBuilder. – Nikhil Jun 21 '12 at 12:53
@WhiteFang34 Ok..just checked it in an usual java class...yes i get 100 unique strings...so this probably does not work well in midlet because the random class is prolly no so powerful...or something of that sort... – Nikhil Jun 21 '12 at 13:07
@Nikhil Ah, that's interesting. Perhaps the `Random` construction is getting the same seed (seems likely that it's using the system time and it's rounded). Try reusing the `Random` object by putting it outside of your loop. You should get the desired result since it'll keep generating different numbers from the same initial seed (use `SecureRandom` to prevent it from eventually repeating). – WhiteFang34 Jun 22 '12 at 00:54
2

@WhiteFang34 Yes..smooth. I put the random object out side of the loop and now the result is perfect. 100 randoms nos. is what i get. :) and white fang is a great book ;) ..i recommend call of the wild..just in case you haven't read it.. – Nikhil Jun 22 '12 at 03:22
another answer is here: http://stackoverflow.com/questions/41107/how-to-generate-a-random-alpha-numeric-string – bmichalik Mar 16 '14 at 18:07

score 12 · Answer 2 · edited Apr 15 '15 at 03:28

I'd use this approach:

String randomString(final int length) {
    Random r = new Random(); // perhaps make it a class variable so you don't make a new one every time
    StringBuilder sb = new StringBuilder();
    for(int i = 0; i < length; i++) {
        char c = (char)(r.nextInt((int)(Character.MAX_VALUE)));
        sb.append(c);
    }
    return sb.toString();
}

If you want a byte[] you can do this:

byte[] randomByteString(final int length) {
    Random r = new Random();
    byte[] result = new byte[length];
    for(int i = 0; i < length; i++) {
        result[i] = r.nextByte();
    }
    return result;
}

Or you could do this

byte[] randomByteString(final int length) {
    Random r = new Random();
    StringBuilder sb = new StringBuilder();
    for(int i = 0; i < length; i++) {
        char c = (char)(r.nextInt((int)(Character.MAX_VALUE)));
        sb.append(c);
    }
    return sb.toString().getBytes();
}

score 5 · Answer 3 · answered Apr 15 '11 at 23:40

5

You may use the class java.util.Random with method

char c = (char)(rnd.nextInt(128-32))+32

20x to get Bytes, which you interpret as ASCII. If you're fine with ASCII.

32 is the offset, from where the characters are printable in general.

answered Apr 15 '11 at 23:40

user unknown

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score 3 · Answer 4 · answered Apr 15 '11 at 23:39

3

public String randomString(String chars, int length) {
  Random rand = new Random();
  StringBuilder buf = new StringBuilder();
  for (int i=0; i<length; i++) {
    buf.append(chars.charAt(rand.nextInt(chars.length())));
  }
  return buf.toString();
}

answered Apr 15 '11 at 23:39

maerics

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How to generate a random string of 20 characters

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