TypeScript 遍历对象并从值创建一个类型

Question

鉴于我有这样的对象：

const props = [
    { name: 'car', list: { brand: 'audi', model: 's4' }
    { name: 'motorcycle', list: { type: 'ktm', color: 'orange' }
] as constant;

我想创建一个代表这样的类型：

type Props = {
    car?: 'brand' | 'model',
    motorcycle?: 'type' | 'color'
};

我得到的最接近的是这样的：

type Props = Partial<Record<typeof props[number]['name'], typeof props[number]['list']>

但返回如下内容：

type Props = {
    car?: { brand: 'audi', model: 's4' } | { type: 'ktm' | color: 'orange' } | undefined,
    motorcycle?: { brand: 'audi', model: 's4' } | { type: 'ktm' | color: 'orange' } | undefined
}

我怎样才能达到预期的效果？

Answer 1

您可以使用 TypeScript 4.1 (docs) 中引入的 key remapping 来实现此目的：

type IndexKeys<T> = Exclude<keyof T, keyof []>
type ListProp = { name: string, list: object }
type GetName<P> = P extends ListProp ? P['name'] : never
type GetListKeys<P> = P extends ListProp ? keyof P['list'] : never

type PropsFromList<PropList extends ReadonlyArray<ListProp>> = {
  [i in IndexKeys<PropList> as GetName<PropList[i]>]?: GetListKeys<PropList[i]>
}

type Props = PropsFromList<typeof props>
// Inferred type:
// Props: {
//     car?: "brand" | "model" | undefined;
//     motorcycle?: "type" | "color" | undefined;
// }

请注意，就像使用Partial 时一样，可选属性类型会获得一个额外但无害的| undefined。

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