Reputation: 135
Given I have an object like this:
const props = [
{ name: 'car', list: { brand: 'audi', model: 's4' }
{ name: 'motorcycle', list: { type: 'ktm', color: 'orange' }
] as constant;
I want to create a type that represents something like this:
type Props = {
car?: 'brand' | 'model',
motorcycle?: 'type' | 'color'
};
The closest I got was something like this:
type Props = Partial<Record<typeof props[number]['name'], typeof props[number]['list']>
But returns something like this:
type Props = {
car?: { brand: 'audi', model: 's4' } | { type: 'ktm' | color: 'orange' } | undefined,
motorcycle?: { brand: 'audi', model: 's4' } | { type: 'ktm' | color: 'orange' } | undefined
}
How can I achieve the desired result?
Upvotes: 1
Views: 480
Reputation: 15106
You can use key remapping that was introduced in TypeScript 4.1 (docs) to achieve this:
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;
// }
Note that just like when using Partial
, the optional property types get an extra but harmless | undefined
.
Upvotes: 1