TypeScript：具有自定義方法和可變id型別的工廠

我在一家工廠作業；我最終需要添加自定義方法，但多虧了這個答案和這個答案，我們才能讓它按預期作業。

現在我需要通過一個選項來控制 id 的型別，我再一次如此接近，但我仍然缺少一些東西。

我需要id通過str選項來控制屬性true的型別：如果型別id必須是string，而如果false不提供，則型別id必須是number，但我得到string | number.

type Base = { save: () => Promise<boolean> }; // Base and BaseId must be kept separated for next steps ;)
type BaseId<B extends boolean> = { id: B extends true ? string : number };
type Options<B extends boolean> = { opt1?: boolean; opt2?: number; opt3?: string; str?: B };
type Factory<E> = new () => E;

function factory<B extends boolean, E extends Base & BaseId<B>>(name: string, options?: Options<B>): Factory<E>;
function factory<B extends boolean, E extends Base & BaseId<B>, M extends Record<string, <S extends M>(this: E & S) => unknown>>(
  name: string, options: Options<B>, methods: M & Record<string, ((this: E & M) => void)>
): Factory<E & M>;
function factory<B extends boolean, E extends Base & BaseId<B>, M extends Record<string, <S extends M>(this: E & S) => unknown>>(
  name: string, options?: Options<B>, methods?: M & Record<string, ((this: E & M) => void)>
): Factory<E> {
  const ret = function (this: E) {};

  Object.defineProperty(ret, "name", { value: name });
  if(methods) Object.assign(ret.prototype, methods);

  return ret as unknown as Factory<E>;
}

const T1 = factory("T1");
const t1 = new T1();
t1.id = "0"; // Error: string | number but number is expected
console.log(t1, t1.id);

const T2 = factory(
  "T2", { str: true }, {
    test: function (repeat = true) {
      if (repeat) this.test(false);
      return "test";
    }
  }
);
const t2 = new T2();
t2.id = "0"; // Ok: string
console.log(t2, t2.id, t2.test());

這里是一個操場進行試驗。

uj5u.com熱心網友回復：

您string|number在第一種情況下有，因為options未提供第二個引數。您需要更改上/最后一個多載簽名并擺脫options. 您需要提供多載簽名而不是options因為options是可選引數。您提供了帶有 optional 的簽名options，帶有 requiredoptions但您沒有沒有沒有簽名options。由于B泛型引數與options引數系結，因此您需要使用一些默認值而不是B.

考慮這個例子：

type Base = { save: () => Promise<boolean> }; // Base and BaseId must be kept separated for next steps ;)
type BaseId<B extends boolean> = { id: B extends true ? string : number };
type Options<B extends boolean> = { opt1?: boolean; opt2?: number; opt3?: string; str?: B };
type Factory<E> = new () => E;

function factory<E extends Base & BaseId<false>>(name: string): Factory<E>;
function factory<B extends boolean, E extends Base & BaseId<B>, M extends Record<string, <S extends M>(this: E & S) => unknown>>(
  name: string, options: Options<B>, methods: M & Record<string, ((this: E & M) => void)>
): Factory<E & M>;
function factory<B extends boolean, E extends Base & BaseId<B>, M extends Record<string, <S extends M>(this: E & S) => unknown>>(
  name: string, options?: Options<B>, methods?: M & Record<string, ((this: E & M) => void)>
): Factory<E> {
  const ret = function (this: E) { };

  Object.defineProperty(ret, "name", { value: name });
  if (methods) Object.assign(ret.prototype, methods);

  return ret as unknown as Factory<E>;
}

const T1 = factory("T1");
const t1 = new T1();
t1.id = "0"; // Error: string | number but number is expected
t1.id = 1; // Ok number

我已用作false默認型別的游樂場B

更新

boolean 是一個聯合 true | false

如果你不想接觸多載簽名，你可以檢查是否B是聯合。如果它是一個聯合 ( boolean) 則意味著它B不是推斷出來的，我們需要使用一些默認值：

// credits goes to https://stackoverflow.com/a/50375286
type UnionToIntersection<U> = (U extends any ? (k: U) => void : never) extends ((k: infer I) => void) ? I : never

// credits goes to https://stackoverflow.com/questions/53953814/typescript-check-if-a-type-is-a-union
type IsUnion<T> = [T] extends [UnionToIntersection<T>] ? false : true

type Base = { save: () => Promise<boolean> };
type BaseId<B extends boolean> = { id: IsUnion<B> extends true ? number : B extends true ? string : number }; // <------ CHANGE IS HERE
type Options<B extends boolean> = { opt1?: boolean; opt2?: number; opt3?: string; str?: B };
type Factory<E> = new () => E;

function factory<B extends boolean, E extends Base & BaseId<B>>(name: string, options?: Options<B>): Factory<E>;
function factory<B extends boolean, E extends Base & BaseId<B>, M extends Record<string, <S extends M>(this: E & S) => unknown>>(
  name: string, options: Options<B>, methods: M & Record<string, ((this: E & M) => void)>
): Factory<E & M>;
function factory<B extends boolean, E extends Base & BaseId<B>, M extends Record<string, <S extends M>(this: E & S) => unknown>>(
  name: string, options?: Options<B>, methods?: M & Record<string, ((this: E & M) => void)>
): Factory<E> {
  const ret = function (this: E) { };

  Object.defineProperty(ret, "name", { value: name });
  if (methods) Object.assign(ret.prototype, methods);

  return ret as unknown as Factory<E>;
}

const T1 = factory("T1");
const t1 = new T1();
t1.id = "0"; // Error: id should be number

操場

一般來說，既然id是必需的屬性而options不是，id不應該依賴options.

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