Convert String to Integer List in Kotlin

Question

I want to make function for typeconverter room. The original data is in List and i store it as String in database.

My code and the output is like this:

fun main() {
    val genreIds = listOf(28, 80)
    val stringFromList = fromList(genreIds)
    val stringToList = toList(stringFromList)
    
    println(genreIds) // output: [28, 80]
    println(stringFromList) // output: [28, 80]
    println(stringToList) // output: [91, 50, 56, 44, 32, 56, 48, 93]
}

fun fromList(genreIds: List<Int>): String = genreIds.toString()

fun toList(genreIdString: String): List<Int> {
    return genreIdString.map { it.toInt() }
}

I did try to use split(",") but it always give me error because of the "[" and "]". I want the output of toList function is [28, 80] too. Thanks for help.

Aplet123 · Accepted Answer · 2021-01-03 14:20:52Z

2

You can use substring to remove the first and last character:

fun toList(genreIdString: String): List<Int> =
    genreIdString
        .substring(1, genreIdString.length - 1)
        .split(", ")
        .map { it.toInt() }

answered Jan 3, 2021 at 14:20

Aplet123

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Comments

Konstantin Raspopov · Accepted Answer · 2021-01-03 15:06:48Z

1

It doesn't seem to be a good idea to parse the input manually. I propose to use Kotlin Serialization library for this purpose.

Using it you can write as simple as:

import kotlinx.serialization.decodeFromString
import kotlinx.serialization.encodeToString
import kotlinx.serialization.json.Json

fun main() {
    val list = listOf(1, 2, 3)
    val str = Json.encodeToString(list)
    println(str) // "[1,2,3]"
    val list2: List<Int> = Json.decodeFromString(str)
    println(list2) // [1, 2, 3]
}

answered Jan 3, 2021 at 15:06

Konstantin Raspopov

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3 Comments

wiryadev Over a year ago

Thanks for the answer. Can you explain why it is better other than writing?

Konstantin Raspopov Over a year ago

@wiryadev Using the toString() method you don't control the output - the result will depend on the method implementation which might change in the future. You can use toString() for debug purposes but if you need to store some data it's better to choose the approach which will give you predictable output.

FWDekker Over a year ago

I like Konstantin's argument of defensive programming against implementation changes. As an alternative solution, if you know that the strings that your toList method receives were created by the fromList method, then you could change your fromList method to genreIds.joinToString(", ", "[", "]"). Combine this with the solution by Aplet123 or by me, and you won't have to depend on the implementation anymore. Then again, I don't think the toString implementation is likely to change, so I would suggest choosing the solution that is easiest to read.

FWDekker · Accepted Answer · 2021-01-04 15:21:14Z

1

Use the removeSurrounding method to remove delimiters:

fun toList(genreIdString: String): List<Int> =
    genreIdString
        .removeSurrounding("[", "]")
        .split(", ")
        .map { it.toInt() }

I think this approach is better than using substring because it more clearly conveys your intent.

answered Jan 4, 2021 at 15:21

FWDekker

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