在 Typescript 中将一个对象拆分为两个免费对象的正确类型是什么？

Question

将对象拆分为两个子对象的正确 Typescript 类型是什么，每个子对象都包含互补键，例如，表单的对象

{x: "x", y: "y", z:"z"}

应该拆分成一个对象

{ a: {x: "x"}, b: {y: "y", z:"z"}}. 

一般记录的此对象的类型是什么？

我试过这个，但它给了我一个错误：

type Split<Union extends Record<string, string | number>>= <A extends Partial<Union>>() => {
    a: A,
    b: Omit<Union, keyof A>
};

type SpecificUnion = {
    x: string
    y: number
    z: string
}

const split: Split<SpecificUnion> = () => {
    return {
        a: { x: "x" },
        b: { y: "y", z: "z"}
}}

我得到错误

Type '() => { a: { x: string; }; b: { y: string; z: string; }; }' is not assignable to type 'Split<SpecificUnion>'.
  Call signature return types '{ a: { x: string; }; b: { y: string; z: string; }; }' and '{ a: A; b: Omit<SpecificUnion, keyof A>; }' are incompatible.
    The types of 'a' are incompatible between these types.
      Type '{ x: string; }' is not assignable to type 'A'.
        '{ x: string; }' is assignable to the constraint of type 'A', but 'A' could be instantiated with a different subtype of constraint 'Partial<SpecificUnion>'.

这个错误似乎自相矛盾，因为它说“'{ x: string; }' is not assignable to type 'A'”和“'{ x: string; }' is assignable to the constraint of type 'A'”。

Answer 1

错误信息说A extends Partial<Union>可以是{ x: string }，但也可以是{ y: number }。或者{ x: string; y: number }。

对于A的不同值，您会得到Split<SpecificUnion>的不同定义，但是Split<Union>没有被A参数化，因此它不应该依赖于A的定义。

解决方法是让A成为Split类型的参数：

type Split<Union extends Record<string, string | number>, A extends Partial<Union>> = () => {
    a: A,
    b: Omit<Union, keyof A>
};

type SpecificUnion = {
    x: string
    y: number
    z: string
}

const split1: Split<SpecificUnion, { x: string }> = () => {
    return {
        a: { x: "x" },
        b: { y: 3, z: "z" },
    }
}

const split2: Split<SpecificUnion, { x: string, y: number }> = () => {
    return {
        a: { x: "x", y: 3 },
        b: { z: "z" },
    }
}

检查它online。