结合枚举和计算属性名称的对象字面量

Question

我有一个枚举

enum Action {
  action1 = 'action1',
  action2 = 'action2'
};

我将其值用作对象中的计算属性名称：

/*
  ...to be able to use these as values, not just types
*/
const ActionState = {
  [Action.action1]: 'actionState1' as const,
  [Action.action2]: 'actionState2' as const,
};

我想定义一个类型/接口，其键是模板文字，将 Action 映射到相应的 ActionState:

/*
  {
    actionState1: boolean;
    actionState2: boolean;
  }
*/
type ActionModelState {
  [key in keyof Action as `${typeof ActionState[key]}`]: boolean // all booleans but I need the keys to be restricted 
}

// throwing:
// Type 'key' cannot be used to index type '{ action1: "actionState1"; action2: "actionState2"; }'.

我将用它来扩展我的基本类型：

type BaseAction = {
  id: number;
  foo: string;
  bar: string;
};

形成：

type EnrichedAction = BaseAction & ActionModelState;

最后：

const enrichedAction: EnrichedAction = {
  id: 123,
  foo: 'foo',
  bar: 'bar',
  actionState1: true,
  actionState2: false,
}

当我清楚地知道Action和ActionState的成员时，如何定义ActionModelState？编译器抱怨：

Type 'key' cannot be used to index type '{ action1: "actionState1"; action2: "actionState2"; }'

但我认为 typeof key === keyof Action。我做错了什么？

Playground link

Answer 1

事实证明，我可以使用 -- Playground.

而不是试图驯服 keyof

type ValueOf<T> = T[keyof T];

type ActionModelState = {
  [key in ValueOf<typeof ActionState>]: boolean;
}

我仍然想知道我是否可以使用某种形状或形式的模板文字...

Answer 2

你根本不需要keyof Action；类型 Action 已经是这些字符串值的并集。所以你想要 key in Action，而不是 key in keyof Action。

type ActionModelState = {
  [key in Action as `${typeof ActionState[key]}`]: boolean
}

Playground Link

结合枚举和计算属性名称的对象字面量

Object literals combining enums and computed property names

typescript

typescript4.0

结合枚举和计算 属性 名称的对象字面量

Object literals combining enums and computed property names

typescript

typescript4.0

结合枚举和计算属性名称的对象字面量