is the a function to anonymise a String i e to only keep the kotlinlang #announcements

fun anonymise(s: String, n: Int): String {
        val replaceLength = kotlin.math.min(s.length, n)
        val replaceChars = String(CharArray(replaceLength, {'*'} ))
        return "$replaceChars${s.substring(replaceLength)}"
    }
    
    @Test
    fun `anonymise`() {
        anonymise("blah", 3).shouldBe("***h")
        anonymise("blah", 4).shouldBe("****")
        anonymise("blah", 5).shouldBe("****")
    }

👍 1

maybe someone can suggest something neater

reik.schatz

05/17/2019, 10:11 AM

thanks 🙂

👍 1

Daniel Garibaldi

05/17/2019, 10:32 AM

maybe an extension function is better

something like

fun String.anonymise(numberOfCharactersToHide: Int = this.length - 1) =
        replaceRange(0,
                kotlin.math.min(this.length -1, numberOfCharactersToHide),
                "*")

it's untested, by default will replace all the characters but the last one

Jonathan Mew

05/17/2019, 10:46 AM

I hadn't come across

replaceRange

, totally the right tool for the job! kotlinlang is such a great slack channel

Ah, except that replaces the entire range with a single asterisk

fun Char.times(n: Int) = String(CharArray(n, {this} ))

    fun String.anonymise(numberOfCharactersToHide: Int = this.length - 1): String {
        val resolvedCharacterToHide = kotlin.math.min(this.length, numberOfCharactersToHide)

        return replaceRange(
            0,
            resolvedCharacterToHide,
            '*'.times(resolvedCharacterToHide)
        )
    }

    @Test
    fun `anonymise`() {
        val s = "blah"
        s.anonymise(3).shouldBe("***h")
        s.anonymise(4).shouldBe("****")
        s.anonymise(5).shouldBe("****")
    }

antonis

05/17/2019, 2:11 PM

Something like this should work https://gist.github.com/antonis/cb95f56a85c703bc015f6d1c02348493

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