Wsrtjtyk

Lazy folks would just do

java.util.UUID.randomUUID().toString()

You can not restrict the character range here, but I guess it's fine in many situations anyway.

Without JDK8:

fun ClosedRange<Char>.randomString(lenght: Int) = 

    (1..lenght)

        .map { (Random().nextInt(endInclusive.toInt() - start.toInt()) + start.toInt()).toChar() }

        .joinToString("")

usage:

('a'..'z').randomString(6)

Since Kotlin 1.3 you can do this:

fun getRandomString(length: Int) : String {

    val allowedChars = "ABCDEFGHIJKLMNOPQRSTUVWXTZabcdefghiklmnopqrstuvwxyz"

    return (1..length)

        .map { allowedChars.random() }

        .joinToString("")

}

Or use coroutine API for the true Kotlin spirit:

buildSequence { val r = Random(); while(true) yield(r.nextInt(24)) }

   .take(10)

   .map{(it+ 65).toChar()}

   .joinToString("")

('A'..'z').map { it }.shuffled().subList(0, 4).joinToString("")

Building off the answer from Paul Hicks, I wanted a custom string as input. In my case, upper and lower-case alphanumeric characters. Random().ints(...) also wasn't working for me, as it required an API level of 24 on Android to use it.

This is how I'm doing it with Kotlin's Random abstract class:

import kotlin.random.Random



object IdHelper {



    private val ALPHA_NUMERIC = ('0'..'9') + ('A'..'Z') + ('a'..'z')

    private const val LENGTH = 20



    fun generateId(): String {

        return List(LENGTH) { Random.nextInt(0, ALPHA_NUMERIC.size) }

                .map { ALPHA_NUMERIC[it] }

                .joinToString(separator = "")

    }

}

The process and how this works is similar to a lot of the other answers already posted here:

1. Generate a list of numbers of length LENGTH that correspond to the index values of the source string, which in this case is ALPHA_NUMERIC
   


2. Map those numbers to the source string, converting each numeric index to the character value


3. Convert the resulting list of characters to a string, joining them with the empty string as the separator character.


4. Return the resulting string.

Usage is easy, just call it like a static function: IdHelper.generateId()

Using Kotlin 1.3:

This method uses an input of your desired string length desiredStrLength as an Integer and returns a random alphanumeric String of your desired string length.

fun randomAlphaNumericString(desiredStrLength: Int): String {

    val charPool: List<Char> = ('a'..'z') + ('A'..'Z') + ('0'..'9')



    return (1..desiredStrLength)

        .map{ kotlin.random.Random.nextInt(0, charPool.size) }

        .map(charPool::get)

        .joinToString("")

}

If you prefer unknown length of alphanumeric (or at least a decently long string length like 36 in my example below), this method can be used:

fun randomAlphanumericString(): String {

    val charPool: List<Char> = ('a'..'z') + ('A'..'Z') + ('0'..'9')

    val outputStrLength = (1..36).shuffled().first()



    return (1..outputStrLength)

        .map{ kotlin.random.Random.nextInt(0, charPool.size) }

        .map(charPool::get)

        .joinToString("")

}

fun randomAlphaNumericString(@IntRange(from = 1, to = 62) lenght: Int): String {

    val alphaNumeric = ('a'..'z') + ('A'..'Z') + ('0'..'9')

    return alphaNumeric.shuffled().take(lenght).joinToString("")

}

The best way I think:

fun generateID(size: Int): String {

    val source = "A1BCDEF4G0H8IJKLM7NOPQ3RST9UVWX52YZab1cd60ef2ghij3klmn49opq5rst6uvw7xyz8"

    return (source).map { it }.shuffled().subList(0, size).joinToString("")

}