typescript - 如何推断函数对象的返回类型？

Question

const actions = {
  setF1: (a: number) => ({
    type: 'a',
    a
  }),
  setF2: (b: string) => ({
    type: 'b',
    b
  })
};

type ActionTypes = ?

export default reducer = (state = {}, action: ActionTypes) => {
  switch (action.type) {
    case 'a':
      console.log(`${action.a} is a number`);
      return {
        ...state,
        a
      }
    case 'b':
      console.log(`${action.b} is a string`);
      return {
        ...state,
        b
      }
    default:
      return state;
  }
};

目标是每次我向actions对象添加函数时，reducer 都会自动推断switch语句中的操作返回类型。我试图避免我需要做类似action: Action1 | Action2 | Action3.

标签： typescripttypescript-typings

Answer 1

一种方法是在您的定义中使用const 断言actions，以便编译器不会将您的type属性扩大到string，其余的相对简单。

const actions = {
  setF1: (a: number) => ({
    type: 'a' as const,
    a
  }),
  setF2: (b: string) => ({
    type: 'b' as const,
    b
  })
};

type ActionKeys = keyof typeof actions; // "setF1" | "setF2"
type ActionTypes = ReturnType<typeof actions[ActionKeys]> // { type: "a", a: number } | { type: "b", b: string }

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