typescript - 根据 typescript 中的通用属性选择类型

Question

如何根据泛型属性选择类型？

enum AnimalType {
  DOG,
  CAT,
  FROG,
  SNAKE,
}

const dog = {
  type: AnimalType.DOG, // type property exists for all animals. other properties are vary for different animals
  legs: 4,
}

const cat = {
  type: AnimalType.CAT,
  tail: 1,
}

const frog = {
  type: AnimalType.FROG,
  canSwim: true,
  tail: 0,
}

const snake = {
  type: AnimalType.SNAKE,
  tail: 'looong',
  age: 2,
}

type Animal<AnimalType> = typeof dog | typeof cat | typeof frog | typeof snake // how to rewrite this type definition?

如何根据AnimalType推断类型？例如Animal<AnimalType.DOG>应等于 typeof dog

可以使用条件类型:

type Animal<T extends AnimalType> = T extends AnimalType.DOG ? typeof dog : T extends AnimalType.CAT ? typeof cat : T extends AnimalType.FROG ? typeof frog : T extends AnimalType.SNAKE ? typeof snake : unknown

但是它看起来太长并且完全无法阅读。希望有更好的解决办法

Answer 1

你可能打算这样做

const dog = {
  type: AnimalType.DOG,
  legs: 4,
}

类型为 {type: AnimalType.DOG, legs: number} 。但 TypeScript 推断出 AnimalType.DOG 的类型变得更宽enum类型AnimalType而不是 literal type AnimalType.DOG .

这意味着有人可以写 dog.type = AnimalType.CAT ，这不好。这也意味着typeof dog没有可用于计算的信息 Animal<T> .

要解决此问题，您需要告诉编译器 type应该给出更具体的文字类型。通常的方法是使用 const assertion关于有问题的字面意思。您可以对完整对象文字 {type: AnimalType.DOG, legs: 4} as const 执行此操作，如果您希望编译器强制执行 dog.legs总是恰好 4 。但至少我们应该为 type 做到这一点属性(property)。产生这个:

const dog = { type: AnimalType.DOG as const, legs: 4 };
const cat = { type: AnimalType.CAT as const, tail: 1 };
const frog = { type: AnimalType.FROG as const, canSwim: true, tail: 0 };
const snake = { type: AnimalType.SNAKE as const, tail: 'looong', age: 2 };

现在是union type typeof dog | typeof cat | typeof frog | typeof snake看起来像:

type SomeAnimal = typeof dog | typeof cat | typeof frog | typeof snake;
/* type SomeAnimal = {
    type: AnimalType.DOG;
    legs: number;
} | {
    type: AnimalType.CAT;
    tail: number;
} | {
    type: AnimalType.FROG;
    canSwim: boolean;
    tail: number;
} | {
    type: AnimalType.SNAKE;
    tail: string;
    age: number;
} */

这让我们可以实现 Animal通过联合过滤 Extract utility type :

type Animal<T extends AnimalType> =
  Extract<SomeAnimal, { type: T }>;

让我们测试一下:

type Dog = Animal<AnimalType.DOG>;
/* type Dog = {
    type: AnimalType.DOG;
    legs: number;
} */

看起来不错。

Playground link to code