How to use Kotlin when to find if a string is numeric?

问题

I'd like to use a when() expression in Kotlin to return different values from a function. The input is a String, but it might be parsable to an Int, so I'd like to return the parsed Int if possible, or a String if it is not. Since the input is a String I can not use the is type check expression.

Is there any idiomatic way to achieve that?

Edit: My problem is how the when() expression should look like, not about the return type.

回答1:

Version 1 (using toIntOrNull and when when as requested)

fun String.intOrString(): Any {
    val v = toIntOrNull()
    return when(v) {
        null -> this
        else -> v
    }
}

"4".intOrString() // 4
"x".intOrString() // x

Version 2 (using toIntOrNull and the elvis operator ?:)

when is actually not the optimal way to handle this, I only used when because you explicitely asked for it. This would be more appropriate:

fun String.intOrString() = toIntOrNull() ?: this

Version 3 (using exception handling):

fun String.intOrString() = try { // returns Any
   toInt()
} catch(e: NumberFormatException) {
   this
}

回答2:

The toIntOrNull function in the kotlin.text package (in kotlin-stdlib) is probably what you're looking for:

toIntOrNull
fun String.toIntOrNull(): Int? (source)
Platform and version requirements: Kotlin 1.1

Parses the string as an Int number and returns the result or null if the string is not a valid representation of a number.
fun String.toIntOrNull(radix: Int): Int? (source)
Platform and version requirements: Kotlin 1.1

Parses the string as an Int number and returns the result or null if the string is not a valid representation of a number.

More information: https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.text/to-int-or-null.html

回答3:

Using let for one

fun isInteger(str: String?) = str?.toIntOrNull()?.let { true } ?: false

来源：https://stackoverflow.com/questions/48116753/how-to-use-kotlin-when-to-find-if-a-string-is-numeric

标签

kotlin