5

我想使用 Map 而不是对象映射来声明一些键和值。但是 Typescript 似乎不支持 ES6 Map 的索引类型，是否正确，是否有任何解决方法？

此外，我还想让值类型安全，以便映射中的每个条目都具有与键对应的值的正确类型。

这是一些描述我要实现的目标的伪代码：

type Keys = 'key1' | 'key2';

type  Values = {
  'key1': string;
  'key2': number;
}

/** Should display missing entry error */
const myMap = new Map<K in Keys, Values[K]>([
  ['key1', 'error missing key'],
]);

/** Should display wrong value type error for 'key2' */
const myMap = new Map<K in Keys, Values[K]>([
  ['key1', 'okay'],
  ['key2', 'error: this value should be number'],
]);

/** Should pass */
const myMap = new Map<K in Keys, Values[K]>([
  ['key1', 'all good'],
  ['key2', 42],
]);

编辑：更多部分描述我的用例的代码

enum Types = {
  ADD = 'ADD',
  REMOVE = 'REMOVE',
};

/** I would like type-safety and autocompletion for the payload parameter */
const handleAdd = (state, payload) => ({...state, payload});

/** I would like to ensure that all types declared in Types are implemented */
export const reducers = new Map([
  [Types.ADD, handleAdd],
  [Types.REMOVE, handleRemove]
]);
4

1 回答 1

5

这是我能想象得到的最接近的，虽然我仍然不明白为什么我们不只使用普通对象开始：

type ObjectToEntries<O extends object> = { [K in keyof O]: [K, O[K]] }[keyof O]

interface ObjectMap<O extends object> {
  forEach(callbackfn: <K extends keyof O>(
    value: O[K], key: K, map: ObjectMap<O>
  ) => void, thisArg?: any): void;
  get<K extends keyof O>(key: K): O[K];
  set<K extends keyof O>(key: K, value: O[K]): this;
  readonly size: number;
  [Symbol.iterator](): IterableIterator<ObjectToEntries<O>>;
  entries(): IterableIterator<ObjectToEntries<O>>;
  keys(): IterableIterator<keyof O>;
  values(): IterableIterator<O[keyof O]>;
  readonly [Symbol.toStringTag]: string;
}

interface ObjectMapConstructor {
  new <E extends Array<[K, any]>, K extends keyof any>(
    entries: E
  ): ObjectMap<{ [P in E[0][0]]: Extract<E[number], [P, any]>[1] }>;
  new <T>(): ObjectMap<Partial<T>>;
  readonly prototype: ObjectMap<any>;
}

const ObjectMap = Map as ObjectMapConstructor;

这个想法是创建一个新的接口，ObjectMap它具体依赖于对象类型O来确定其键/值关系。然后你可以说Map构造函数可以充当ObjectMap构造函数。我还删除了任何可以更改实际存在哪些键的方法（并且该has()方法也是冗余的true）。

我可以通过解释每个方法和属性定义的麻烦，但这是很多类型的杂耍。简而言之，您想使用K extends keyof OandO[K]来表示通常由K和Vin表示的类型Map<K, V>。

构造函数有点烦人，因为类型推断不能按照您想要的方式工作，因此保证类型安全分为两个步骤：

// let the compiler infer the type returned by the constructor
const myMapInferredType = new ObjectMap([
  ['key1', 'v'], 
  ['key2', 1],  
]);

// make sure it's assignable to `ObjectMap<Values>`: 
const myMap: ObjectMap<Values> = myMapInferredType;

If your myMapInferredType doesn't match ObjectMap<Values> (e.g., you are missing keys or have the wrong value types) then myMap will give you errors.

Now you can use myMap as an ObjectMap<Values>, similarly to how you'd use a Map instance, with get() and set(), and it should be type safe.

Please note again... this seems like a lot of work for a more complex object with trickier typings and no more functionality than a plain object. I would seriously warn anyone using a Map whose keys are subtypes of keyof any (that is, string | number | symbol) to strongly consider using a plain object instead, and be sure that your use case really necessitates a Map.

Playground link to code

于 2019-02-27T14:44:18.747 回答