javascript - TypeScript:在 Enum 上执行 switch-case 来设置变量的惯用方法

Question

示例问题:我有一个枚举变量难度。在函数中，我想根据 Difficulty 的值设置配置 DifficultyConfig 。这是我能想到的一种不优雅的方式:

export interface DifficultyConfig {
    healthModifier: number,
    deathIsPermanent: boolean,
}

export interface AppProps {
    difficultyConfig: DifficultyConfig
}

export const NormalDifficultyConfig: DifficultyConfig = {
    enemyHealthModifier: 1,
    deathIsPermanent: false,
}

export const HigherDifficultyConfig: DifficultyConfig = {
    enemyHealthModifier: 1.3,
    deathIsPermanent: true, 
}

export enum Difficulty {
    NORMAL = 'Normal',
    HARD = 'Hard',
    ADVANCED = 'Advanced',
}

function createApp(difficulty: Difficulty) {
    let difficultyConfig: DifficultyConfig;
    
    switch(difficulty) {
        case Difficulty.NORMAL: 
            difficultyConfig = NormalDifficultyConfig;
            break;
        // Both HARD and ADVANCED get HigherDifficultyConfig
        case Difficulty.HARD:
            difficultyConfig = HigherDifficultyConfig;
            break;
        case Difficulty.ADVANCED: 
            difficultyConfig = HigherDifficultyConfig;
            break;
        default: 
            difficultyConfig = NormalDifficultyConfig;
            break;
    }
    
    return new App({
        difficultyConfig
    });
}

我不喜欢这么简单的 Switch case 语法。我的理想在 Scala 中是这样的:

val difficultyConfig = difficulty match {
    case Difficulty.NORMAL => NormalDifficultyConfig
    case Difficulty.HARD | Difficulty.ADVANCED => HigherDifficultyConfig
    case _ => NormalDifficultyConfig
}

JavaScript 中有类似的东西吗？

Answer 1

如果逻辑存在差异，则 switch或if/else if/else (稍微不太冗长)可能是可行的方法(下面有更多介绍)。如果它像您的示例中那样纯粹是数据，那么您可以使用配置难度映射对象:

const difficultyConfigs = {
    [Difficulty.NORMAL]: NormalDifficultyConfig,
    [Difficulty.HARD]: HigherDifficultyConfig,
    [Difficulty.ADVANCED]: HigherDifficultyConfig,
} as const;

您甚至可以将其声明为 Record<Difficult, DifficultyConfig>像这样，如caTS在评论中指出:

const difficultyConfigs: Record<Difficult, DifficultyConfig> = {
//                     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    [Difficulty.NORMAL]: NormalDifficultyConfig,
    [Difficulty.HARD]: HigherDifficultyConfig,
    [Difficulty.ADVANCED]: HigherDifficultyConfig,
} as const;

稍后会详细介绍，但无论哪种方式，该函数都只是:

function createApp(difficulty: Difficulty) {
    let difficultyConfig = difficultyConfigs[difficulty];
    
    return new App({
        difficultyConfig
    });
}

Playground 链接:Without Record | With Record (我还更新了问题代码中的一些看似拼写错误/编辑错误。)

这还有一个优点，如果您添加 Difficulty但忘记在 difficultyConfigs 中包含一个，当您使用它时，您会遇到一个方便的编译错误。我模拟过这样的错误here通过添加 MEDIUM困难却“忘记”更新功能。

但更好的是，如果我们包含类型 Record<Difficulty, DifficultyConfig>输入 caTS建议和difficultyConfigs没有每个 Difficulty 的条目值，您会得到更早的编译时错误，like this .

---

即使对于逻辑，如果您愿意，也可以使用相同的概念来创建调度对象:

const difficultyConfigs: Record<Difficulty, () => DifficultyConfig> = {
    [Difficulty.NORMAL]: () => { /*...build and return NORMAL config...*/ },
    [Difficulty.HARD]: () => { /*...build and return HARD config...*/ },,
    [Difficulty.ADVANCED]: () => { /*...build and return ADVANCED config...*/ },,
} as const;
// ...
const difficultyConfig = difficultyConfigs[difficulty]();