这适用于我在Chrome控制台。有什么问题吗?

var { color, height } = elmo
var subelmo = { color, height }
console.log(subelmo) // {color: "red", height: "unknown"}

这适用于我在Chrome控制台。有什么问题吗?

var { color, height } = elmo
var subelmo = { color, height }
console.log(subelmo) // {color: "red", height: "unknown"}

function splice()
{
    var ret = new Object();

    for(i = 1; i < arguments.length; i++)
        ret[arguments[i]] = arguments[0][arguments[i]];

    return ret;
}

var answer = splice(elmo, "color", "height");

打印稿的解决方案:

function pick<T extends object, U extends keyof T>(
  obj: T,
  paths: Array<U>
): Pick<T, U> {
  const ret = Object.create(null);
  for (const k of paths) {
    ret[k] = obj[k];
  }
  return ret;
}

输入信息甚至允许自动补全:

Credit to definitelytyping for U extends keyof T trick!

打印稿操场

我知道它不是最干净的，但它简单易懂。

function obj_multi_select(obj, keys){
    let return_obj = {};
    for (let k = 0; k < keys.length; k++){
        return_obj[keys[k]] = obj[keys[k]];
    };
    return return_obj;
};

两种Array.prototype.reduce:

const selectable = {a: null, b: null};
const v = {a: true, b: 'yes', c: 4};

const r = Object.keys(selectable).reduce((a, b) => {
  return (a[b] = v[b]), a;
}, {});

console.log(r);

这个答案使用了神奇的逗号运算符: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comma_Operator

如果你想要更花哨，这个更紧凑:

const r = Object.keys(selectable).reduce((a, b) => (a[b] = v[b], a), {});

把所有这些放到一个可重用的函数中:

const getSelectable = function (selectable, original) {
  return Object.keys(selectable).reduce((a, b) => (a[b] = original[b], a), {})
};

const r = getSelectable(selectable, v);
console.log(r);