typescript :记录<>，其中值取决于键

Question

我想写这样的东西:

const structs: Record<T extends StructureConstant,Array<ConcreteStructure<T>>> = Object.create(null);

其中 StructureConstant 是常量的联合，而 ConcreteStructure 基本上是一个 type lookup .

这在 TypeScript 中可行吗？我不知道如何使值依赖于键。

---

这是一个完整的例子:

type StructureConstant = "road" | "tower";

interface Road {
    a: "i am a road",
}

interface Tower {
    b: "i be towerin"
}

type ConcreteStructure<T extends StructureConstant> = T extends "road"
    ? Road
    : T extends "tower"
    ? Tower
    : never;

type StuctureRecord<T extends StructureConstant> = Record<T, Array<ConcreteStructure<T>>>

const structs: StuctureRecord<StructureConstant> = Object.create(null); 

const roads = structs["road"]; // should be Array<Road>|undefined

console.log(roads[0].a); // should work (if structs was populated)

TS playground

Answer 1

在 typescript 中，只有函数值可以具有泛型类型。其他值需要完全指定。

所以也许你想要这样的东西:

type StuctureRecord<T extends StructureConstant> = Record<T, Array<ConcreteStructure<T>>>

// structs doesn't have a generic type.
const structs: StuctureRecord<StructureConstant> = Object.create(null); 

或

// function can have generic type.
const structsCreator = <T extends StructureConstant>(): Record<T, Array<ConcreteStructure<T>>> => Object.create(null) 

const structs = structsCreator<StructureConstant>()

Playground

---

要获得描述的效果，您不能使用 Record。记录将所有键映射到同一类型。如果每个条目需要不同的类型，则需要自定义映射类型

type StuctureMappedType<T extends StructureConstant> = 
    { [K in T]: Array<ConcreteStructure<K>>}

Playground