首页 > 解决方案 > TypeScript:泛型检查枚举类型

问题描述

有一个过程枚举。它们有不同的步骤,也表示为枚举,如下所示。

enum Process {
  Simple = "simple",
  Advanced = "advanced"
}

enum SimpleStep {
  A = "A",
  B = "B"
}

enum AdvancedStep {
  A = "A",
  B = "B",
  C = "C"
}

通过以下语句,我创建了一系列步骤。

const SIMPLE_STEPS = Object.keys(SimpleStep).map(
  (k: string) => SimpleStep[k]
);

const ADVANCED_STEPS = Object.keys(AdvancedStep).map(
  k => AdvancedStep[k]
);

const ALL_STEPS = {
  [Process.Simple]: SIMPLE_STEPS,
  [Process.Advanced]: ADVANCED_STEPS
};

我编写了以下函数来获取步骤号。

// ???: Check if S is a step of Process
const getStepNumber = <P extends Process, S>(process: P, step: S) => {
  return ALL_STEPS[process].indexOf(step) + 1;
};

// returns 2, which is correct
console.log('step number of B', getStepNumber(Process.Advanced, AdvancedStep.B)); 

// returns 0. Is it possible to prevent at compile-time?
console.log('step number of C', getStepNumber(Process.Simple, AdvancedStep.C));

正如您在代码示例中看到的那样,是否可以在编译时使用泛型防止以错误的步骤调用函数?

这是游乐场,如果您想尝试整个示例:TS Playground

标签: typescriptgenericstypescript-typingstypescript-generics

解决方案


一种选择是引入一个条件类型,它允许您根据提供给函数的值推断所需的步骤枚举(即SimpleStep或)。这可以按如下方式完成:AdvancedStepProcess

type StepFromProcess<P extends Process> =
    P extends Process.Simple ? SimpleStep : AdvancedStep

然后您可以更改您的函数以使用该类型:

const getStepNumber = <P extends Process>(process: P, step: StepFromProcess<P>) => ...

编译器现在将阻止您进行此(无效)调用:

console.log('step number of C', getStepNumber(Process.Simple, AdvancedStep.C));

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