Kotlin 'when' 语句与 Java 'switch'

问题描述 投票:0回答:12

Kotlin 中的模式匹配很好,而且在 90% 的用例中它不执行下一个模式匹配这一事实是很好的。

在 Android 中,当数据库更新时,如果我们不放置中断以使代码看起来像这样,我们将使用 Java switch 属性继续下一种情况:

switch (oldVersion) {
    case 1: upgradeFromV1();
    case 2: upgradeFromV2(); 
    case 3: upgradeFromV3();
}

因此,如果有人拥有带有 DB 版本 1 的应用程序,但错过了带有 DB v2 的应用程序版本,他将获得执行的所有所需升级代码。

转换为 Kotlin,我们会遇到这样的混乱:

when (oldVersion) {
    1 -> {
        upgradeFromV1()
        upgradeFromV2()
        upgradeFromV3()
    }
    2 -> {
        upgradeFromV2()
        upgradeFromV3()
    }
    3 -> {
        upgradeFromV3()
    }
}

这里我们只有 3 个版本,想象一下当 DB 达到版本 19 时。

无论如何要以相同的方式行动然后切换?我尝试继续但没有运气。

java switch-statement kotlin
12个回答
77
投票

简单但啰嗦的解决方案是:

if (oldVersion <= 1) upgradeFromV1()
if (oldVersion <= 2) upgradeFromV2()
if (oldVersion <= 3) upgradeFromV3()

另一种可能的解决方案,带有函数引用

fun upgradeFromV0() {}
fun upgradeFromV1() {}
fun upgradeFromV2() {}
fun upgradeFromV3() {}

val upgrades = arrayOf(::upgradeFromV0, ::upgradeFromV1, ::upgradeFromV2, ::upgradeFromV3)

fun upgradeFrom(oldVersion: Int) {
    for (i in oldVersion..upgrades.lastIndex) {
        upgrades[i]()
    }
}

18
投票

编辑:下面的原始回复。这是我目前正在做的事情:

fun upgrade() {
    fun upgradeFromV1() { /* Do stuff */ }
    fun upgradeFromV3() { /* Do stuff */ }

    tailrec fun upgradeFrom(version: Int): Unit = when (version) {
        LATEST_VERSION -> {
            Config.version = version
        } 1 -> {
            upgradeFromV1()
            upgradeFrom(2)
        } in 2..3 -> {
            upgradeFromV3()
            upgradeFrom(4)
        } else -> {
            Log("Uncaught upgrade from $version")
            upgradeFrom(version+1)
    }

    upgradeFrom(Config.version)
}

这是答案@C.A.B 的一个变体。给了:

fun upgrade(oldVersion: Int) {
    when (oldVersion) {
        latestVersion -> return
        1 -> upgradeFromV1()
        2 -> upgradeFromV2()
        3 -> upgradeFromV3()
    }
    upgrade(oldVersion + 1)
}

14
投票

这个怎么样:

fun upgradeFromV3() {/* some code */}
fun upgradeFromV2() {/* some code */ upgradeFromV3()}
fun upgradeFromV1() {/* some code */ upgradeFromV2()}
fun upgradeFromV0() {/* some code */ upgradeFromV1()}

fun upgrade(oldVersion: Int) {
    when (oldVersion) {
        1 -> upgradeFromV1()
        2 -> upgradeFromV2()
        3 -> upgradeFromV3()
    }
}

补充:

我喜欢 @lukle 将升级路径定义为列表的想法。这允许为不同的初始阶段定义不同的升级路径。例如:

  1. 从发布版本到最新发布版本的简单快速路径
  2. 从热修复版本开始的追赶路径(可能连续几个),从以前的完整版本到下一个完整版本时不应该应用

为此,我们需要知道要应用列表中的哪些元素。

fun <Vs, V> Pair<Vs, V>.apply(upgrade: () -> Unit): (V) -> V {
    return { current: V ->
        if (first == current) {
            upgrade()
            second
        } else {
            current
        }
    }
}

val upgradePath = listOf(
        (0 to 10).apply  { /* do something */ },
        (5 to 15).apply  { /* do something */ },
        (10 to 20).apply { /* do something */ },
        (15 to 20).apply { /* do something */ },
        (20 to 30).apply { /* do something */ },
        (30 to 40).apply { /* do something */ }
)

fun upgrade(oldVersion: Int) {
    var current = oldVersion
    upgradePath.forEach { current = it(current) }
}

在此代码中,Vs 可以与 V 相同或具有重写的

equals(other: Any?): Boolean
方法的某种 V 值集合。


12
投票

你可以只使用for循环和when。

for (version in oldVersion..newVersion) when (version) {
    1 -> upgradeFromV1()
    2 -> upgradeFromV2()
    3 -> upgradeFromV3()
}

3
投票

用于自定义实现的 Kotlin DSL 怎么样?像这样的方法:

class SwitchTest {

    @Test
    fun switchTest() {

        switch {
            case(true) {
                println("case 1")
            }
            case(true) {
                println("case 2")
            }
            case(false) {
                println("case 3")
            }
            caseBreak(true) {
                println("case 4")
            }
            case(true) {
                println("case 5")
            }
//          default { //TODO implement
//
//          }
        }
    }
}

class Switch {
    private var wasBroken: Boolean = false

    fun case(condition: Boolean = false, block: () -> Unit) {
        if (wasBroken) return
        if (condition)
            block()
    }

    fun caseBreak(condition: Boolean = false, block: () -> Unit) {
        if (condition) {
            block()
            wasBroken = true
        }
    }
}

fun switch(block: Switch.() -> Unit): Switch {
    val switch = Switch()
    switch.block()
    return switch
}

打印:


case 1
case 2
case 4
UPD:一些重构和输出示例。


2
投票

这绝对是可能的 引用官方参考:控制流:if、when、for、while

If many cases should be handled in the same way, the branch conditions may be combined with a comma:

when (x) {
    0, 1 -> print("x == 0 or x == 1")
    else -> print("otherwise")
}

因此,如果相同的条件列表很短,那么您可以用逗号分隔它们,或者使用其他答案中所述的范围,例如 1..10 中的条件


2
投票

OP 答案的另一个变体:

override fun onUpgrade(db: SQLiteDatabase, oldVersion: Int, newVersion: Int) {
    when (oldVersion) {
        newVersion -> return
        1 -> TODO("upgrade from v1 to v2")
        2 -> TODO("upgrade from v2 to v3")
    }
    oldVersion++
    onUpgrade(db, oldVersion, newVersion)
}

0
投票

这是 bashor 的两个答案的混合,带有一点功能糖:

fun upgradeFromV0() {}
fun upgradeFromV1() {}
fun upgradeFromV2() {}
fun upgradeFromV3() {}

val upgrades = arrayOf(::upgradeFromV0, ::upgradeFromV1, ::upgradeFromV2, ::upgradeFromV3)

fun upgradeFrom(oldVersion: Int) {
    upgrades.filterIndexed { index, kFunction0 -> oldVersion <= index }
            .forEach { it() }
}

0
投票

如果您不关心运行这些函数的顺序,您可以制作自己的伪开关,例如:

function PretendSwitch() {
  if(oldVersion>3) return
  upgradeFromV3();
  if(oldVersion==3) return
  upgradeFromV2()
  if(oldVersion==2) return
  upgradeFromV1()
  if(oldVersion==1) return
}

没有什么比使用开关更干净的了。不幸的是,Kotlin 缺少 switch 语句,因此无法优雅地执行此操作。


-1
投票
val orders = arrayListOf(
            { upgradeFromV1()},
            { upgradeFromV2()},
            { upgradeFromV3()}
)

orders.drop(oldVersion).forEach { it() }

-2
投票

val 旧版本 = 6 val 新版本 = 10

for (version in oldVersion until newVersion) {
    when (oldVersion) {
        1 -> upgradeFromV1()
        2 -> upgradeFromV2()
        3 -> upgradeFromV3()
        4 -> upgradeFromV4()
        5 -> upgradeFromV5()
        6 -> upgradeFromV6()
        7 -> upgradeFromV7()
        8 -> upgradeFromV8()
        9 -> upgradeFromV9()
    }
    println("~~~")
}

-4
投票

Kotlin 使用不同的流程控制,称为when。

你的代码,当使用它时,可以是这样的。

显然代码可能不同,但我知道你的问题只是关于 switch 的使用。

fun main(args: Array<String>) {
val month = 8

val monthString = when(month) {
    1 -> "Janeiro"
    2 -> "February"
    3 -> "March"
    4 -> "April"
    5 -> "May"
    6 -> "June"
    7 -> "July"
    8 -> "August"
    9 -> "September"
    12 -> "October"
    11 -> "November"
    10 -> "December"
    else -> "Invalid month"      
}

println(monthString);
}
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