Typescript 怎样把嵌套的属性全部变为可选类型

Question

type A = {
    name:string
}

type B = {
    age:number
}

type C = {
    aa:A,
    bb:B,
}

type c = Partial<C>

通过Partial<C>只能转换成这样

{
    aa?: { name: string },
    bb?: { age: number }
}

我想把嵌套的属性全部转为可选类型，应该怎么以做？

{
    aa?: { name?: string },
    bb?: { age?: number }
}

Answer 1

type DeepPartial<T> = {
    [P in keyof T]?: DeepPartial<T[P]>;
};

type c = DeepPartial<C>;