如何在 TypeScript 中键入对象聚合函数

根据一些业务规则，我创建了一个聚合函数，需要正确输入.

函数(JS):

它需要:

• 初始对象.
• n件物品.

export function aggregateObjects(initialObject, ...newObjects) {
  const aggregate = (objectA, objectB) => {
    return Object.keys(objectB).reduce(
      (acc, key) => ({
        ...acc,
        [key]: [...(acc[key] ?? []), ...(objectB[key] ?? [])],
      }),
      objectA
    );
  };

  return newObjects.reduce(
    (aggregatedObjects, newObject) => aggregate(aggregatedObjects, newObject),
    initialObject
  );
}

如何使用它(Typescript ):

下面是传递给aggregateObjects的两个新对象的示例，但它应该适用于5、10、...

const initialObject = {
  a: [objectOfType1],
};

const newObject1 = {
  a: [objectOfType2],
  b: [objectOfType1],
};

const newObject2 = {
  c: [objectOfType1, objectOfType2],
};

const result = aggregateObjects(initialObject, [newObject1, newObject2]);

我们所期望的(Typescript ):

根据前面的示例，我们预期结果如下:

• a的类型应为ObjectType1或ObjectType2的array.
• b的类型应为ObjectType1的array.
• c的类型应为ObjectType1或ObjectType2的array.

type Result = {
  a: (ObjectType1 | ObjectType2)[];
  b: ObjectType1[];
  c: (ObjectType1 | ObjectType2)[];
};

我try 了什么(Typescript ):

当然，它不起作用:

• 内部aggregate看起来没问题(即使有一些与传播相关的问题).
• 在外部，我没有找到正确输入返回值的解决方案.

请注意，CombineObjs是受https://dev.to/svehla/typescript-how-to-deep-merge-170c启发的TS帮助器

export function aggregateObjects<T extends object, U extends any[]>(
  initialObjects: T,
  ...newObjects: U
) {
  const aggregate = <A extends object, B extends object>(
    objectA: A,
    objectB: B
  ): CombineObjs<A, B> => {
    return (Object.keys(objectB) as Array<keyof B>).reduce(
      (acc, key) => ({
        ...acc,
        [key]: [
          // @ts-ignore
          ...(acc[key as keyof CombineObjs<A, B>] ?? []),
          // @ts-ignore
          ...(objectB[key] ?? []),
        ],
      }),
      objectA as unknown as CombineObjs<A, B>
    );
  };

  return newObjects.reduce(
    (aggregatedObjects, newObject) => aggregate(aggregatedObjects, newObject),
    initialObjects
  );
}

感谢您的解决方案.