8 chars unique String in Java

Question

Is there any way to generate 8 characters long random and unique String in Java?

String id = getRandomString();

and then id will be for example wf41Av5g

Answer 1

As none of the answers above ensures uniqueness, I decided to add another answer.

First we define all valid options in an array:

char[] chars = new char[]{ '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',
        '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',
        '1', '2', '3', '4', '5', '6', '7', '8', '9', '0' };

Now we can mutate the algorithm to Random shuffling of an array, to get the first 8 elements:

Random rnd = new Random();
int charsLength = chars.length;
int passLength = 8;
StringBuilder password = new StringBuilder();
for (int i = 0; i < passLength; i++)
{
    int index = rnd.nextInt(charsLength - i - 1);
    // Simple swap
    char a = chars[i + index];
    chars[i + index] = chars[i];
    chars[i] = a;
    password.append(a);
}

System.out.print(password);

It ensures uniqueness by that once a char is in the 8 first places of the array, it is not a valid choice for the next round.

Few outputs that I got:

ifJjuYrR
NpY3TfIU
LGFm8Ng9
sV4Gctb7
4fyhlSuQ

Answer 2

By now you must've gotten your answer, but for those who reach this page, try this code:

public class RandomValueXML {

    public static void main(String[] args) {
        
        System.out.println(org.apache.commons.lang3.RandomStringUtils.random(6, true, true)); 

    }
    
    static Set<String> getRandomUniqueStrings(int count, int length, boolean letters, boolean numbers){
        Set<String> rus = new HashSet<>();
        
        while (rus.size() < count){
            rus.add(RandomStringUtils.random(length, letters, numbers));
        }
        
        return rus;
    } 
}

Here the use of set ensures the uniqueness while use of HashSet ensures better performance while reading and writing.

Answer 3

I have got no idea whether there is a method that can produce 8 digit unique random string but I would create an offline key-generator-service that constantly generates random keys and insert them to key table if the key does not exist in the table and whenever your online app needs a key then it can fetch from one of the pre-produced keys.

Answer 4

I know this is an old post but this may help someone.

This one worked for me:

System.out.println(UniqueIdGenerator.fromBase10(System.currentTimeMillis()));

should be unique in every millisecond on the same JVM.

with:

public class UniqueIdGenerator {

 private static final String MIX_STRING = "abcdefghijklmnopqrstuvwxyz1234567890";

 private static final int MIX_STRING_LENGTH = MIX_STRING.length();

 public static String fromBase10(final long base10) {
    if (base10 == 0)
        return "0";

    long temp = base10;
    final StringBuilder sb = new StringBuilder();

    while (temp > 0) {
        temp = fromBase10(temp, sb);
    }
    return sb.reverse().toString();
 }

 private static Long fromBase10(final long base10, final StringBuilder sb) {
    final int rem = (int) (base10 % MIX_STRING_LENGTH);
    sb.append(MIX_STRING.charAt(rem));
    return base10 / MIX_STRING_LENGTH;
 }
}

you can also change the char order in MIX_STRING, even add upperCase letters or symbols.

Answer 5

You can use RandomStringUtils

import org.apache.commons.lang.RandomStringUtils;
public class RandomStringUtilsTrial {
  public static void main(String[] args) {
    System.out.print("8 char string  >>>");
    System.out.println(RandomStringUtils.random(8, true, true));

  }
}

Answer 6

long unixTime = System.currentTimeMillis() / 1000L;
String unique_no=Long.toHexString(unixTime).toUpperCase();

gives you unique 8 chars String in HEX format.

Answer 7

You can create this string on base of the current time.

For example:

2014 06 04 16 21 20 79 ---> 14 06 04 16 21 20 79

Then you can translate every two digits (except last two) for an alphanumeric character using table ( http://en.wikipedia.org/wiki/Base64 - note that all of them are smaller than 59)

then you will receive

14 - O
06 - G
04 - E
16 - Q
21 - V
20 - U

OGEQVU79 <- exactly 8 characters. A little bit complicated, but it works if you would not call your function often than 0.01 s

Answer 8

If the uniqueness is important, you can't simply randomly generate strings. There is no way to avoid a collision. Even UUIDs can collide, although it is quite unlikely.

You could keep a record of the strings you've used, and when you generate a new random string, check the record to see if it's a duplicate, and if so, discard it and try again.

However, what i would suggest is that you don't generate random numbers. Instead, keep a counter, and encrypt its output to produce random-looking numbers. These are guaranteed never to collide if you do this properly. See an earlier answer of mine, or one by Rossum

Answer 9

Please try following code:

private String uniqid() {
    Random random = new Random();
    String tag = Long.toString(Math.abs(random.nextLong()), 36);
    return tag.substring(0, 8);
}

Answer 10

Quick and dirty solution using java.util.UUID:

public String getRandomString(int length) {
    return UUID.randomUUID().toString().subSequence(0, length);
}

Answer 11

public String getRandomString(int length, char[] characterSet) {
    StringBuilder sb = new StringBuilder();

    for (int loop = 0; loop < length; loop++) {
        int index = new Random().nextInt(characterSet.length);
        sb.append(characterSet[index]);
    }

    String nonce = sb.toString();
    return nonce;
}

Answer 12

The uniqueness property depends on the scope in which you use it. Java can certainly generate random strings, although if you want a universally unique identifier you can use UUID class.

String unique = UUID.randomUUID().toString();