"当"声明与Java“切换”声明 [英] "When" statement vs Java "switch" statement

问题描述

Kotlin中的模式匹配很好，并且它在90％的用例中都没有执行下一个模式匹配。

Pattern matching in Kotlin is nice and the fact it does not execute next pattern match is good in 90% of use cases.

在Android中，当数据库更新时，我们使用Java开关属性继续下一个案例如果我们不休息让代码看起来像这样：

In Android, when database is updated, we use Java switch property to go on next case if we do not put a break to have code looking like that:

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

因此，如果有人拥有数据库版本1的应用并错过了应用使用DB v2的版本，他将获得所有需要的升级代码。

So if someone has an app with version 1 of the DB and missed the app version with DB v2, he will get all the needed upgrade code executed.

转换为Kotlin，我们得到一个混乱：

Converted to Kotlin, we get a mess like:

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

我们这里有只有3个版本，想象当DB到达版本19时：/

Here we have only 3 version, imagine when DB reaches version 19 :/

无论如何以与开关相同的方式进行操作？我试着继续没有运气。

Anyway to makes when acting in the same way than switch? I tried continue without luck.

推荐答案

简单而又冗长的解决方案是：

Simple but wordy solution is:

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

另一种可能的解决方案函数引用：

Another possible solution with function references:

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]()
    }
}

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