Adding keys in one JavaScript Object to another only if the first object doesn't have those keys

Question

function extend(obj1, obj2) {

  var obj2Keys = Object.keys(obj2);
  var obj2Values = Object.values(obj2);

  var obj1Keys = Object.keys(obj1);
  var obj1Values = Object.keys(obj1);
  
  var newObj = {};

  for(var i=0; i<obj1Keys.length; i++) {
    if(obj1Keys[i] !== obj2Keys[i]) {
      

    }
  }
  
}

var obj1 = {
  a: 1,
  b: 2
};
var obj2 = {
  b: 4,
  c: 3
};

extend(obj1, obj2);

console.log(obj1); // --> {a: 1, b: 2, c: 3}
console.log(obj2); // --> {b: 4, c: 3}

/*

 1. Add any keys that are not in the 1st object.
 2. If the 1st object already has a given key, ignore it  (do not
    overwrite the property value).
 3. Do not modify the 2nd object at all.

*/

Hey guys, trying to figure this one out. I'm sure i'm doing this the most inefficient way. I'm struggling to compare indexes of these objects.. Any help?? I was using for in loops originally but I couldn't conceptually understand what I was doing lol. Kinda stuck at this point.

Object.assign({}, second, first) - overwriting second's keys by the first's. — georg
– georg, Commented Mar 13, 2021 at 20:27

Rashid Iqbal · Accepted Answer · 2021-03-13 20:52:36Z

3

const firstObject = {a:1,b:2,c:3}
const secondObject = {a:11,b:22,c:33,d:44}
var newObject = {...secondObject,...firstObject}

console.log(newObject)

result:

{"a":1,"b":2,"c":3,"d":44}

edited Mar 13, 2021 at 20:52

answered Mar 13, 2021 at 20:41

Rashid Iqbal

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6 Comments

symlink Over a year ago

This is identical to my answer

Rashid Iqbal Over a year ago

no, your answer is different. your output will be : {"a":11,"b":22,"c":33,"d":44}

Rashid Iqbal Over a year ago

@symlink "only if the first object doesn't have those keys".

symlink Over a year ago

Slightly different output, identical method.

Rashid Iqbal Over a year ago

but he is asking, not to replace the values of the first object

|

Majed Badawi · Accepted Answer · 2021-03-13 20:32:28Z

1

Get keys of obj2
Iterate over them using the for-loop
In each iteration, if obj1 does not have this key, and the record to it from obj2

function extend (obj1, obj2) {
  const obj2Keys = Object.keys(obj2);
  for(let i = 0; i < obj2Keys.length; i++) {
    const currentKey = obj2Keys[i]
    if(!obj1[currentKey]) {
      obj1[currentKey] = obj2[currentKey];
    }
  }
}

const obj1 = { a: 1, b: 2 }; const obj2 = { b: 4, c: 3 };

extend(obj1, obj2);

console.log(obj1); // --> {a: 1, b: 2, c: 3}
console.log(obj2); // --> {b: 4, c: 3}

Another way using .forEach:

function extend (obj1, obj2) {
  Object.keys(obj2).forEach(currentKey => {
    if(!obj1[currentKey]) {
      obj1[currentKey] = obj2[currentKey];
    }
  });
}

const obj1 = { a: 1, b: 2 }; const obj2 = { b: 4, c: 3 };

extend(obj1, obj2);

console.log(obj1); // --> {a: 1, b: 2, c: 3}
console.log(obj2); // --> {b: 4, c: 3}

answered Mar 13, 2021 at 20:32

Majed Badawi

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2 Comments

Brixsta Over a year ago

How is if(!obj1[currentKey]) different from if(obj1[currentKey] !== obj2[currentKey]) why do I get the right answer with the first statement and the wrong value with the second?

Majed Badawi Over a year ago

!obj1[currentKey] means that obj1 has no such key. obj1[currentKey] !== obj2[currentKey] means that for the key currentKey, thevalues from obj1 and obj2 are different. In your case, you want to only add a key-value pair from obj2 to obj1 if it doesn't exist in the first place.

symlink · Accepted Answer · 2021-03-13 20:57:23Z

1

Use the spread syntax to combine the objects. The spread syntax basically inserts the objects key/value pairs wherever you write it, so you're creating a new object with {...obj1, ...obj2}. If there are pairs with duplicate keys, the pair from the second object will be selected.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax

let obj1 = {a: 1, b: 2}
let obj2 = {b: 4, c: 3}

let obj3 = {...obj2, ...obj1}
console.log(obj3)

edited Mar 13, 2021 at 20:57

answered Mar 13, 2021 at 20:28

symlink

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3 Comments

Fabio Lopez Over a year ago

I agree this is the most readable and simple way. However I think you should put ...obj1 last so it grabs its values on the shared keys

symlink Over a year ago

@FabioLopez yeah, it depends on which object's values he wants to take precedence. He didn't say so I used the second one, but explained how he could use either one by putting it last.

Rashid Iqbal Over a year ago

@symlink your answer is different. your script output will be : {"a":11,"b":22,"c":33,"d":44}

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