类型记录问题--类型记录不能推断对象属性

Question

我对类型记录有困难，我试图动态地推断一个对象的属性。

看看下面的代码

const Actions = {
    globalUI: {
        openModal: 'openModal'
    }
}

type Param = keyof typeof Actions;
type T = <T, K extends keyof T>(action: T, key: K) => T[K];
const fn = (param: Param) => {

    const action = mapper(Actions, param);

    return action
}
const mapper:T = (action, name) => {
    return action[name]
}

const action = fn('globalUI');

​

​

如您所见，类型记录可以推断出全局UI属性。

但是，当我在Actions对象中放置附加属性时

​

类型记录不能推断对象属性

const Actions = {
    globalUI: {
        openModal: 'openModal'
    },
    LP: {
        addDirectory: 'addDirectory'
    }
}

type Param = keyof typeof Actions;
type T = <T, K extends keyof T>(action: T, key: K) => T[K];
const fn = (param: Param) => {

    const action = mapper(Actions, param);

    return action
}
const mapper:T = (action, name) => {
    return action[name]
}

const action = fn('globalUI');

​

​

我怎样才能做到这一点？请给出任何想法

Answer 1

您的mapper函数按预期工作：

type T = <T, K extends keyof T>(action: T, key: K) => T[K];

const mapper:T = (action, name) => {
    return action[name]
}

const action = mapper(Actions, 'globalUI');

action.openModel // works greak 

问题在于fn函数，它将联合推断为返回类型。在您的示例中使用{ openModal: string } | { addDirectory: string }。

因此，您需要使用泛型来显化返回类型：

const fn = <S extends Param>(param: S): typeof Actions[S]  => {
    const action = mapper(Actions, param);
    return action
}

const action = fn('globalUI'); 

action.globalUI; // Ok now

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