Object Destructuring ({ X, Y, ...rest }) For Whitelisting Properties Of An Object

Using Object rest destructuring is straightforward to blacklist properties of an object, like in the following example: const original = { a: 1, b: 2, c: 3, evil: 'evil',

Solution 1:

For this purpose I use 2 helper functions

exportconstpickProps = (object, ...props) => (
  props.reduce((a, x) => {
    if (object.hasOwnProperty(x)) a[x] = object[x];
    return a;
  }, {})
);

exportconstomitProps = (object, ...props) => {
  const no = {...object};
  props.forEach(p =>delete no[p]);
  return no;
};

You can also do

const original = {
  a: 1,
  b: 2,
  c: 3,
  evil: "evil",
  ugly: "ugly",
};

const { a, b, c } = original;
const filtered = { a, b, c };

Solution 2:

I don't think your way to "blacklist" is good, because it unnecessarily assigns original.evil to evil, and original.ugly to ugly.

You can try this approach:

constblacklistFilter = (obj, blacklist) => Object.entries(obj)
  .filter(([key, value]) => !blacklist.includes(key))
  .reduce((obj, [key, value]) => (obj[key] = value, obj), {})

constwhitelistFilter = (obj, whitelist) => Object.entries(obj)
  .filter(([key, value]) => whitelist.includes(key))
  .reduce((obj, [key, value]) => (obj[key] = value, obj), {})

const original = {
  a: 1
 ,b: 2
 ,c: 3
 ,evil: 'evil'
 ,ugly: 'ugly'
}

console.log(blacklistFilter(original, ['evil', 'ugly']))
console.log(whitelistFilter(original, ['a', 'b', 'c']))

Object.entries() returns an array of object's keys and corresponding values in format [key, value], the filter() method filters keys based on whether they aren't blacklisted or whether they are whitelisted, and reduce() method converts the [key, value] array back to an object (similar method as in this answer).