How to generate a random alpha-numeric string

Question

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".

Answer 1

Algorithm

To generate a random string, concatenate characters drawn randomly from the set of acceptable symbols until the string reaches the desired length.

Implementation

Here's some fairly simple and very flexible code for generating random identifiers. Read the information that follows for important application notes.

public class RandomString {

    /**
     * Generate a random string.
     */
    public String nextString() {
        for (int idx = 0; idx < buf.length; ++idx)
            buf[idx] = symbols[random.nextInt(symbols.length)];
        return new String(buf);
    }

    public static final String upper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    public static final String lower = upper.toLowerCase(Locale.ROOT);

    public static final String digits = "0123456789";

    public static final String alphanum = upper + lower + digits;

    private final Random random;

    private final char[] symbols;

    private final char[] buf;

    public RandomString(int length, Random random, String symbols) {
        if (length < 1) throw new IllegalArgumentException();
        if (symbols.length() < 2) throw new IllegalArgumentException();
        this.random = Objects.requireNonNull(random);
        this.symbols = symbols.toCharArray();
        this.buf = new char[length];
    }

    /**
     * Create an alphanumeric string generator.
     */
    public RandomString(int length, Random random) {
        this(length, random, alphanum);
    }

    /**
     * Create an alphanumeric strings from a secure generator.
     */
    public RandomString(int length) {
        this(length, new SecureRandom());
    }

    /**
     * Create session identifiers.
     */
    public RandomString() {
        this(21);
    }

}

Usage examples

Create an insecure generator for 8-character identifiers:

RandomString gen = new RandomString(8, ThreadLocalRandom.current());

Create a secure generator for session identifiers:

RandomString session = new RandomString();

Create a generator with easy-to-read codes for printing. The strings are longer than full alphanumeric strings to compensate for using fewer symbols:

String easy = RandomString.digits + "ACEFGHJKLMNPQRUVWXYabcdefhijkprstuvwx";
RandomString tickets = new RandomString(23, new SecureRandom(), easy);

Use as session identifiers

Generating session identifiers that are likely to be unique is not good enough, or you could just use a simple counter. Attackers hijack sessions when predictable identifiers are used.

There is tension between length and security. Shorter identifiers are easier to guess, because there are fewer possibilities. But longer identifiers consume more storage and bandwidth. A larger set of symbols helps, but might cause encoding problems if identifiers are included in URLs or re-entered by hand.

The underlying source of randomness, or entropy, for session identifiers should come from a random number generator designed for cryptography. However, initializing these generators can sometimes be computationally expensive or slow, so effort should be made to re-use them when possible.

Use as object identifiers

Not every application requires security. Random assignment can be an efficient way for multiple entities to generate identifiers in a shared space without any coordination or partitioning. Coordination can be slow, especially in a clustered or distributed environment, and splitting up a space causes problems when entities end up with shares that are too small or too big.

Identifiers generated without taking measures to make them unpredictable should be protected by other means if an attacker might be able to view and manipulate them, as happens in most web applications. There should be a separate authorization system that protects objects whose identifier can be guessed by an attacker without access permission.

Care must be also be taken to use identifiers that are long enough to make collisions unlikely given the anticipated total number of identifiers. This is referred to as "the birthday paradox." The probability of a collision, p, is approximately n²/(2q^x), where n is the number of identifiers actually generated, q is the number of distinct symbols in the alphabet, and x is the length of the identifiers. This should be a very small number, like 2^‑50 or less.

Working this out shows that the chance of collision among 500k 15-character identifiers is about 2^‑52, which is probably less likely than undetected errors from cosmic rays, etc.

Comparison with UUIDs

According to their specification, UUIDs are not designed to be unpredictable, and should not be used as session identifiers.

UUIDs in their standard format take a lot of space: 36 characters for only 122 bits of entropy. (Not all bits of a "random" UUID are selected randomly.) A randomly chosen alphanumeric string packs more entropy in just 21 characters.

UUIDs are not flexible; they have a standardized structure and layout. This is their chief virtue as well as their main weakness. When collaborating with an outside party, the standardization offered by UUIDs may be helpful. For purely internal use, they can be inefficient.

Answer 2

Java supplies a way of doing this directly. If you don't want the dashes, they are easy to strip out. Just use uuid.replace("-", "")

import java.util.UUID;

public class randomStringGenerator {
    public static void main(String[] args) {
        System.out.println(generateString());
    }

    public static String generateString() {
        String uuid = UUID.randomUUID().toString();
        return "uuid = " + uuid;
    }
}

Output

uuid = 2d7428a6-b58c-4008-8575-f05549f16316

Answer 3

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();
}

Answer 4

If you're happy to use Apache classes, you could use org.apache.commons.text.RandomStringGenerator (Apache Commons Text).

Example:

RandomStringGenerator randomStringGenerator =
        new RandomStringGenerator.Builder()
                .withinRange('0', 'z')
                .filteredBy(CharacterPredicates.LETTERS, CharacterPredicates.DIGITS)
                .build();
randomStringGenerator.generate(12); // toUpperCase() if you want

Since Apache Commons Lang 3.6, RandomStringUtils is deprecated.

Answer 5

You can use an Apache Commons library for this, RandomStringUtils:

RandomStringUtils.randomAlphanumeric(20).toUpperCase();

Answer 6

In one line:

Long.toHexString(Double.doubleToLongBits(Math.random()));

Source: Java - generating a random string

Answer 7

This is easily achievable without any external libraries.

1. Cryptographic Pseudo Random Data Generation (PRNG)

First you need a cryptographic PRNG. Java has SecureRandom for that and typically uses the best entropy source on the machine (e.g. /dev/random). Read more here.

SecureRandom rnd = new SecureRandom();
byte[] token = new byte[byteLength];
rnd.nextBytes(token);

Note: SecureRandom is the slowest, but most secure way in Java of generating random bytes. I do however recommend not considering performance here since it usually has no real impact on your application unless you have to generate millions of tokens per second.

2. Required Space of Possible Values

Next you have to decide "how unique" your token needs to be. The whole and only point of considering entropy is to make sure that the system can resist brute force attacks: the space of possible values must be so large that any attacker could only try a negligible proportion of the values in non-ludicrous time¹.

Unique identifiers such as random UUID have 122 bit of entropy (i.e., 2^122 = 5.3x10^36) - the chance of collision is "*(...) for there to be a one in a billion chance of duplication, 103 trillion version 4 UUIDs must be generated²". We will choose 128 bits since it fits exactly into 16 bytes and is seen as highly sufficient for being unique for basically every, but the most extreme, use cases and you don't have to think about duplicates. Here is a simple comparison table of entropy including simple analysis of the birthday problem.

^{For simple requirements, 8 or 12 byte length might suffice, but with 16 bytes you are on the "safe side".}

And that's basically it. The last thing is to think about encoding so it can be represented as a printable text (read, a String).

3. Binary to Text Encoding

Typical encodings include:

• Base64 every character encodes 6 bit, creating a 33% overhead. Fortunately there are standard implementations in Java 8+ and Android. With older Java you can use any of the numerous third-party libraries. If you want your tokens to be URL safe use the URL-safe version of RFC4648 (which usually is supported by most implementations). Example encoding 16 bytes with padding: XfJhfv3C0P6ag7y9VQxSbw==

• Base32 every character encodes 5 bit, creating a 40% overhead. This will use A-Z and 2-7, making it reasonably space efficient while being case-insensitive alpha-numeric. There isn't any standard implementation in the JDK. Example encoding 16 bytes without padding: WUPIL5DQTZGMF4D3NX5L7LNFOY

• Base16 (hexadecimal) every character encodes four bit, requiring two characters per byte (i.e., 16 bytes create a string of length 32). Therefore hexadecimal is less space efficient than Base32, but it is safe to use in most cases (URL) since it only uses 0-9 and A to F. Example encoding 16 bytes: 4fa3dd0f57cb3bf331441ed285b27735. See a Stack Overflow discussion about converting to hexadecimal here.

Additional encodings like Base85 and the exotic Base122 exist with better/worse space efficiency. You can create your own encoding (which basically most answers in this thread do), but I would advise against it, if you don't have very specific requirements. See more encoding schemes in the Wikipedia article.

4. Summary and Example

• Use SecureRandom
• Use at least 16 bytes (2^128) of possible values
• Encode according to your requirements (usually hex or base32 if you need it to be alpha-numeric)

Don't

• ... use your home brew encoding: better maintainable and readable for others if they see what standard encoding you use instead of weird for loops creating characters at a time.
• ... use UUID: it has no guarantees on randomness; you are wasting 6 bits of entropy and have a verbose string representation

Example: Hexadecimal Token Generator

public static String generateRandomHexToken(int byteLength) {
    SecureRandom secureRandom = new SecureRandom();
    byte[] token = new byte[byteLength];
    secureRandom.nextBytes(token);
    return new BigInteger(1, token).toString(16); // Hexadecimal encoding
}

//generateRandomHexToken(16) -> 2189df7475e96aa3982dbeab266497cd

Example: Base64 Token Generator (URL Safe)

public static String generateRandomBase64Token(int byteLength) {
    SecureRandom secureRandom = new SecureRandom();
    byte[] token = new byte[byteLength];
    secureRandom.nextBytes(token);
    return Base64.getUrlEncoder().withoutPadding().encodeToString(token); //base64 encoding
}

//generateRandomBase64Token(16) -> EEcCCAYuUcQk7IuzdaPzrg

Example: Java CLI Tool

If you want a ready-to-use CLI tool you may use dice:

Example: Related issue - Protect Your Current Ids

If you already have an id you can use (e.g., a synthetic long in your entity), but don't want to publish the internal value, you can use this library to encrypt it and obfuscate it: https://github.com/patrickfav/id-mask

IdMask<Long> idMask = IdMasks.forLongIds(Config.builder(key).build());
String maskedId = idMask.mask(id);
// Example: NPSBolhMyabUBdTyanrbqT8
long originalId = idMask.unmask(maskedId);

Answer 8

Using Dollar should be as simple as:

// "0123456789" + "ABCDE...Z"
String validCharacters = $('0', '9').join() + $('A', 'Z').join();

String randomString(int length) {
    return $(validCharacters).shuffle().slice(length).toString();
}

@Test
public void buildFiveRandomStrings() {
    for (int i : $(5)) {
        System.out.println(randomString(12));
    }
}

It outputs something like this:

DKL1SBH9UJWC
JH7P0IT21EA5
5DTI72EO6SFU
HQUMJTEBNF7Y
1HCR6SKYWGT7

Answer 9

Here it is in Java:

import static java.lang.Math.round;
import static java.lang.Math.random;
import static java.lang.Math.pow;
import static java.lang.Math.abs;
import static java.lang.Math.min;
import static org.apache.commons.lang.StringUtils.leftPad

public class RandomAlphaNum {
  public static String gen(int length) {
    StringBuffer sb = new StringBuffer();
    for (int i = length; i > 0; i -= 12) {
      int n = min(12, abs(i));
      sb.append(leftPad(Long.toString(round(random() * pow(36, n)), 36), n, '0'));
    }
    return sb.toString();
  }
}

Here's a sample run:

scala> RandomAlphaNum.gen(42)
res3: java.lang.String = uja6snx21bswf9t89s00bxssu8g6qlu16ffzqaxxoy

Answer 10

A short and easy solution, but it uses only lowercase and numerics:

Random r = new java.util.Random ();
String s = Long.toString (r.nextLong () & Long.MAX_VALUE, 36);

The size is about 12 digits to base 36 and can't be improved further, that way. Of course you can append multiple instances.

Answer 11

Surprising, no one here has suggested it, but:

import java.util.UUID

UUID.randomUUID().toString();

Easy.

The benefit of this is UUIDs are nice, long, and guaranteed to be almost impossible to collide.

Wikipedia has a good explanation of it:

" ...only after generating 1 billion UUIDs every second for the next 100 years, the probability of creating just one duplicate would be about 50%."

The first four bits are the version type and two for the variant, so you get 122 bits of random. So if you want to, you can truncate from the end to reduce the size of the UUID. It's not recommended, but you still have loads of randomness, enough for your 500k records easy.

Answer 12

An alternative in Java 8 is:

static final Random random = new Random(); // Or SecureRandom
static final int startChar = (int) '!';
static final int endChar = (int) '~';

static String randomString(final int maxLength) {
  final int length = random.nextInt(maxLength + 1);
  return random.ints(length, startChar, endChar + 1)
        .collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append)
        .toString();
}

Answer 13

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());
    }
}

This just adds the password into the string and... yeah, it works well. Check it out... It is very simple; I wrote it.

Answer 14

public static String generateSessionKey(int length){
    String alphabet =
        new String("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"); // 9

    int n = alphabet.length(); // 10

    String result = new String();
    Random r = new Random(); // 11

    for (int i=0; i<length; i++) // 12
        result = result + alphabet.charAt(r.nextInt(n)); //13

    return result;
}

Answer 15

Using UUIDs is insecure, because parts of the UUID aren't random at all. The procedure of erickson is very neat, but it 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 choose 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 not a 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 "unnecessary" / "longer" numbers. Therefore 2^5 is the better option, even if we lose four 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 its required length prepending zeros.

Answer 16

I found this solution that generates a random hex encoded string. The provided unit test seems to hold up to my primary use case. Although, it is slightly more complex than some of the other answers provided.

/**
 * Generate a random hex encoded string token of the specified length
 *  
 * @param length
 * @return random hex string
 */
public static synchronized String generateUniqueToken(Integer length){ 
    byte random[] = new byte[length];
    Random randomGenerator = new Random();
    StringBuffer buffer = new StringBuffer();

    randomGenerator.nextBytes(random);

    for (int j = 0; j < random.length; j++) {
        byte b1 = (byte) ((random[j] & 0xf0) >> 4);
        byte b2 = (byte) (random[j] & 0x0f);
        if (b1 < 10)
            buffer.append((char) ('0' + b1));
        else
            buffer.append((char) ('A' + (b1 - 10)));
        if (b2 < 10)
            buffer.append((char) ('0' + b2));
        else
            buffer.append((char) ('A' + (b2 - 10)));
    }
    return (buffer.toString());
}

@Test
public void testGenerateUniqueToken(){
    Set set = new HashSet();
    String token = null;
    int size = 16;

    /* Seems like we should be able to generate 500K tokens 
     * without a duplicate 
     */
    for (int i=0; i<500000; i++){
        token = Utility.generateUniqueToken(size);

        if (token.length() != size * 2){
            fail("Incorrect length");
        } else if (set.contains(token)) {
            fail("Duplicate token generated");
        } else{
            set.add(token);
        }
    }
}

Answer 17

1. Change String characters as per as your requirements.

2. String is immutable. Here StringBuilder.append is more efficient than string concatenation.

public static String getRandomString(int length) {
    final String characters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJLMNOPQRSTUVWXYZ1234567890!@#$%^&*()_+";
    StringBuilder result = new StringBuilder();

    while(length > 0) {
        Random rand = new Random();
        result.append(characters.charAt(rand.nextInt(characters.length())));
        length--;
    }
    return result.toString();
}

Answer 18

import java.util.Date;
import java.util.Random;

public class RandomGenerator {

  private static Random random = new Random((new Date()).getTime());

    public static String generateRandomString(int length) {
      char[] values = {'a','b','c','d','e','f','g','h','i','j',
               'k','l','m','n','o','p','q','r','s','t',
               'u','v','w','x','y','z','0','1','2','3',
               '4','5','6','7','8','9'};

      String out = "";

      for (int i=0;i<length;i++) {
          int idx=random.nextInt(values.length);
          out += values[idx];
      }
      return out;
    }
}

Answer 19

I don't really like any of these answers regarding a "simple" solution :S

I would go for a simple ;), pure Java, one liner (entropy is based on random string length and the given character set):

public String randomString(int length, String characterSet) {
    return IntStream.range(0, length).map(i -> new SecureRandom().nextInt(characterSet.length())).mapToObj(randomInt -> characterSet.substring(randomInt, randomInt + 1)).collect(Collectors.joining());
}

@Test
public void buildFiveRandomStrings() {
    for (int q = 0; q < 5; q++) {
        System.out.println(randomString(10, "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")); // The character set can basically be anything
    }
}

Or (a bit more readable old way)

public String randomString(int length, String characterSet) {
    StringBuilder sb = new StringBuilder(); // Consider using StringBuffer if needed
    for (int i = 0; i < length; i++) {
        int randomInt = new SecureRandom().nextInt(characterSet.length());
        sb.append(characterSet.substring(randomInt, randomInt + 1));
    }
    return sb.toString();
}

@Test
public void buildFiveRandomStrings() {
    for (int q = 0; q < 5; q++) {
        System.out.println(randomString(10, "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")); // The character set can basically be anything
    }
}

But on the other hand you could also go with UUID which has a pretty good entropy:

UUID.randomUUID().toString().replace("-", "")

Answer 20

I'm using a library from Apache Commons to generate an alphanumeric string:

import org.apache.commons.lang3.RandomStringUtils;

String keyLength = 20;
RandomStringUtils.randomAlphanumeric(keylength);

It's fast and simple!

Answer 21

You mention "simple", but just in case anyone else is looking for something that meets more stringent security requirements, you might want to take a look at jpwgen. jpwgen is modeled after pwgen in Unix, and is very configurable.

Answer 22

import java.util.*;
import javax.swing.*;

public class alphanumeric {
    public static void main(String args[]) {
        String nval, lenval;
        int n, len;

        nval = JOptionPane.showInputDialog("Enter number of codes you require: ");
        n = Integer.parseInt(nval);

        lenval = JOptionPane.showInputDialog("Enter code length you require: ");
        len = Integer.parseInt(lenval);

        find(n, len);
    }

    public static void find(int n, int length) {
        String str1 = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        StringBuilder sb = new StringBuilder(length);
        Random r = new Random();

        System.out.println("\n\t Unique codes are \n\n");
        for(int i=0; i<n; i++) {
            for(int j=0; j<length; j++) {
                sb.append(str1.charAt(r.nextInt(str1.length())));
            }
            System.out.println("  " + sb.toString());
            sb.delete(0, length);
        }
    }
}

Answer 23

You can use the following code, if your password mandatory contains numbers and alphabetic special characters:

private static final String NUMBERS = "0123456789";
private static final String UPPER_ALPHABETS = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
private static final String LOWER_ALPHABETS = "abcdefghijklmnopqrstuvwxyz";
private static final String SPECIALCHARACTERS = "@#$%&*";
private static final int MINLENGTHOFPASSWORD = 8;

public static String getRandomPassword() {
    StringBuilder password = new StringBuilder();
    int j = 0;
    for (int i = 0; i < MINLENGTHOFPASSWORD; i++) {
        password.append(getRandomPasswordCharacters(j));
        j++;
        if (j == 3) {
            j = 0;
        }
    }
    return password.toString();
}

private static String getRandomPasswordCharacters(int pos) {
    Random randomNum = new Random();
    StringBuilder randomChar = new StringBuilder();
    switch (pos) {
        case 0:
            randomChar.append(NUMBERS.charAt(randomNum.nextInt(NUMBERS.length() - 1)));
            break;
        case 1:
            randomChar.append(UPPER_ALPHABETS.charAt(randomNum.nextInt(UPPER_ALPHABETS.length() - 1)));
            break;
        case 2:
            randomChar.append(SPECIALCHARACTERS.charAt(randomNum.nextInt(SPECIALCHARACTERS.length() - 1)));
            break;
        case 3:
            randomChar.append(LOWER_ALPHABETS.charAt(randomNum.nextInt(LOWER_ALPHABETS.length() - 1)));
            break;
    }
    return randomChar.toString();
}

Answer 24

You can use the UUID class with its getLeastSignificantBits() message to get 64 bit of random data, and then convert it to a radix 36 number (i.e. a string consisting of 0-9,A-Z):

Long.toString(Math.abs( UUID.randomUUID().getLeastSignificantBits(), 36));

This yields a string up to 13 characters long. We use Math.abs() to make sure there isn't a minus sign sneaking in.

Answer 25

Here is the one-liner by AbacusUtil:

String.valueOf(CharStream.random('0', 'z').filter(c -> N.isLetterOrDigit(c)).limit(12).toArray())

Random doesn't mean it must be unique. To get unique strings, use:

N.uuid() // E.g.: "e812e749-cf4c-4959-8ee1-57829a69a80f". length is 36.
N.guid() // E.g.: "0678ce04e18945559ba82ddeccaabfcd". length is 32 without '-'

Answer 26

Here it is a Scala solution:

(for (i <- 0 until rnd.nextInt(64)) yield { 
  ('0' + rnd.nextInt(64)).asInstanceOf[Char] 
}) mkString("")

Answer 27

Using an Apache Commons library, it can be done in one line:

import org.apache.commons.lang.RandomStringUtils;
RandomStringUtils.randomAlphanumeric(64);

Documentation

Answer 28

public static String randomSeriesForThreeCharacter() {
    Random r = new Random();
    String value = "";
    char random_Char ;
    for(int i=0; i<10; i++)
    {
        random_Char = (char) (48 + r.nextInt(74));
        value = value + random_char;
    }
    return value;
}

Answer 29

I think this is the smallest solution here, or nearly one of the smallest:

 public String generateRandomString(int length) {
    String randomString = "";

    final char[] chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz01234567890".toCharArray();
    final Random random = new Random();
    for (int i = 0; i < length; i++) {
        randomString = randomString + chars[random.nextInt(chars.length)];
    }

    return randomString;
}

The code works just fine. If you are using this method, I recommend you to use more than 10 characters. A collision happens at 5 characters / 30362 iterations. This took 9 seconds.

Answer 30

public static String getRandomString(int length)
{
    String randomStr = UUID.randomUUID().toString();
    while(randomStr.length() < length) {
        randomStr += UUID.randomUUID().toString();
    }
    return randomStr.substring(0, length);
}