所有派生接口(interface)的 TypeScript 类型

Question

我有一个像这样的接口(interface)继承层次结构:

interface IBase { type: stirng; }
interface IDerived1 extends IBase { type: "Derived1"; }
interface IDerived2 extends IBase { type: "Derived2"; }
...

这些实际上是为 C# 类型生成的接口(interface)，这就是它们上的 type 鉴别器属性的原因。

是否可以在 TypeScript 中为所有可能的 type 字符串文字值编写类型，
仅基于现有类型层次结构，
那么不用手动一一指定它们吗？

我想要这样的smg:

type AllTypesOf<T extends {type: string}> = /* ? what comes here ? */;

...

const myVar: AllTypesOf<IBase>; // "Derived1" | "Derived2" | ...

我会这样使用:

function mySwitch(myArg: IBase) {
    const type = myArg.type as AllTypesOf<IBase> // "Derived1" | "Derived2" | ...
    switch(type) {
        case "Derived1":
        case "Derived2":
            // ok

        case "Derived3":
            // NOK, error, does not exist such
    }
}

Answer 1

TypeScript 遵循基本的继承规则: parent 对 child 一无所知。您可以对所有值进行枚举，然后在您的类型中引用它们:

enum AllTypesOf {
  DERIVED_1 = "Derived1",
  DERIVED_2 = "Derived2"
}

interface IBase { type: AllTypesOf }

interface IDerived1 extends IBase { type: AllTypesOf.DERIVED_1 }

function mySwitch(myArg: IBase) {
    const type = myArg.type // Automatically typed as AllTypesOf
    switch(type) {
        case AllTypesOf.DERIVED_1:
        case AllTypesOf.DERIVED_2:
            // ok

        case AllTypesOf.DERIVED_3: // error, does not exist in enum
            // NOK, error, does not exist such
    }
}