Question

TYPESCRIPT 3.4.3

我想做这样的功能

exportObjectUnit({ a: 1, b: 2, c: 3, d: 4 }, ['b', 'c'])

输出 { a: 1, d: 4 };

我不知道如何键入此函数的返回类型

export const exportObjectKey = <T, K extends keyof T>(value: T, exports: K[]) => {
  const returnValue = {};

  Object.keys(value)
    .filter(key => {
      if (exports.indexOf(key) !== -1) {
        return false;
      }
      return true;
    })
    .map(key => {
      returnValue[key] = value[key];
    });

  return returnValue as T;
};

如果我使用此函数，则返回值仍具有类型（第二个字符串数组参数除外）

---------- EDIT ----------------

export const exportObjectKey = <T>(value: T, exports: Array<keyof T>) => {
  const returnValue = {};

  Object.keys(value)
    .filter(key => {
      if (exports.indexOf(key as keyof T) !== -1) {
        return false;
      }
      return true;
    })
    .map(key => {
      returnValue[key] = value[key];
    });

  return returnValue as T;
};

我不知道如何退货。从第一个对象中删除秒参数数组属性

-----------编辑2 ----------------

type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>;

export const exportObjectKey = <T, K extends keyof T>(value: T, omit: K): Omit<T, K> => {
  delete value[omit];
  return value;
};
export const exportObjectKeys = <T, K extends Array<keyof T>>(value: T, removes: K) =>
  removes.reduce((object, key) => exportObjectKey(object, key), value);

// This is not perfect version

const a = { a: 1, b: 2, c: 3 };
const keyOmitOne = exportObjectKey(a, 'b');
// When I type keyOmitOne.
// Type definition available, It works (a, c)

// ------------------------------------------

// But, when I use exportObjectKeys
const b = { a: 1, b: 2, c: 3 };
const keyOmitArray = exportObjectKey(b, ['b', 'c']);
// I thought type definition works (a available)
// But there is no definition in b value)

Answer 1

It shouldn't be hard I guess?

type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>

function exportObjectUnit<T extends Object, K extends (keyof T)[]>(obj: T, ...keys: K): Omit<T, K[number]> {
    keys.forEach(k => delete obj[k])
    return obj
}

const t1 = exportObjectUnit({ a: 1, b: 2, c: 3, d: 4})
const t2 = exportObjectUnit({ a: 1, b: 2, c: 3, d: 4} , 'a')
const t3 = exportObjectUnit({ a: 1, b: 2, c: 3, d: 4} , 'a', 'b')
const t4 = exportObjectUnit({ a: 1, b: 2, c: 3, d: 4} , 'a', 'b', 'c')
const t5 = exportObjectUnit({ a: 1, b: 2, c: 3, d: 4} , 'a', 'b', 'c', 'd')

Answer 2

不确定这是否是您想要的，但是如果没有泛型，则可以使用index signature输入和输出具有未知键的对象。

type MyObject = { [key: string]: number }

function doSomething(obj: MyObject, arr: string[]) : MyObject {
    let a : MyObject = {a:3, b:4}
    return a
}

doSomething({a:4, b:4, c:4, d:2}, ["a", "d"])

我不知道打字稿中的返回类型

2 个答案: