Cloning javascript object omitting one property

Question

What's the best way in JavaScript for returning an object omitting just one or more properties?

I can assign a key to undefined and that works for sure, but what if want to completely get rid of that key?

function removeCKey() {
  const obj = {a: 'a', b: 'b', c: 'c'}
  return {
    ...obj,
    c: undefined,
  };
}

const myObj = removeCKey();

Also, I want to avoid creating an intermediate object where I use the spread operator like this

function removeCKey() {
  const obj = {a: 'a', b: 'b', c: 'c'}
  const {c, ...rest} = newObj

  return rest;
}

const myObj = removeCKey();

That works fine @SergiuParaschiv but what if I need to remove a lot of keys? — Andrea Zornada, Mar 19 '19 at 15:00
FYI, [operator delete](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/delete) — crashmstr, Mar 19 '19 at 15:00
Destructuring: `const { c, ...result } = obj; return result;` — Tyler Roper, Mar 19 '19 at 15:00
You make a list of the keys and `delete obj[k]` where `k` is each value in the list. — Sergiu Paraschiv, Mar 19 '19 at 15:01

Pranav C Balan · Answer 1 · 2019-03-19T17:03:07.317

2

You can use ES6 object destructuring assignment.

function removeKeys() {
  const obj = {
    a: 'a',
    b: 'b',
    c: 'c'
  };

  // extract property c in variable c
  // and rest of the value into res 
  let { c, ...res } = obj;

  return res;
}

console.log(removeKeys())

edited Mar 19 '19 at 17:03

answered Mar 19 '19 at 15:00

Pranav C Balan

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score 0 · Answer 2 · answered Mar 19 '19 at 14:59

Just delete what you don't need:

function omit(obj, keys) {
  const newObj = { ...obj };  // shallow copy...
  keys.forEach(key => {
    delete newObj[key];  // ... and `delete`.
  });
  return newObj;
}

omit({a: 'a', b: 'b', c: 'c'}, ['c']);

outputs

{ a: 'a', b: 'b' }

Cloning javascript object omitting one property

2 Answers2