TypeScript 映射类型值取决于键

Question

我可以根据项目的键来限制地图条目的值类型吗？

type Thing<T extends string = any> = {
    type: T
}

type ThingMap<T extends Thing> = {
    [K in T["type"]]: T
}

interface A {
    type: "A",
    foo: boolean,
}

interface B {
    type: "B",
}


// This compiles.
const map: ThingMap<A | B> = {
    A: { type: "A", foo: true },
    B: { type: "B" },
}

// But this also compiles, when it should not.
const map: ThingMap<A | B> = {
    A: { type: "A", foo: true },
    B: { type: "A", foo: true },
}

Answer 1

您要确保对于特定 K，键的类型不是整个联合（即 T），而只是类型为 [=11 的联合成员=].您可以使用 Extract 获取具有类型 K:

的属性 type 的联合成员

type Thing<T extends string = any> = {
    type: T
}

type ThingMap<T extends Thing> = {
    [K in T["type"]]:  Extract<T, { type : K }>  // Extract the union member that has { type: K }
}

interface A {
    type: "A",
    foo: boolean,
}

interface B {
    type: "B",
}


// This compiles.
const map: ThingMap<A | B> = {
    A: { type: "A", foo: true },
    B: { type: "B" },
}

// Err now
const map2: ThingMap<A | B> = {
    A: { type: "A", foo: true },
    B: { type: "A", foo: true },
}

TypeScript 映射类型值取决于键

TypeScript mapped type value depending on key

typescript

typescript-generics

typescript-typings