I would like to know how to change the value in a nested object array using javascript.

how to change the "load": "undefined" to "load":1 in the obj

var obj=[{
  "id": "service",
  "country": "AR",
  "load": "undefined"
},{
  "id": "fund",
  "country": "CA",
  "load": "undefined"
}]


var result = obj.forEach(e=>e.load=1);

Expected Output:
[{
  "id": "service",
  "country": "AR",
  "load": 1
},{
  "id": "fund",
  "country": "CA",
  "load": 1
}]

I would like to know how to change the value in a nested object array using javascript.

how to change the "load": "undefined" to "load":1 in the obj

var obj=[{
  "id": "service",
  "country": "AR",
  "load": "undefined"
},{
  "id": "fund",
  "country": "CA",
  "load": "undefined"
}]


var result = obj.forEach(e=>e.load=1);

Expected Output:
[{
  "id": "service",
  "country": "AR",
  "load": 1
},{
  "id": "fund",
  "country": "CA",
  "load": 1
}]

• javascript
• arrays
• object

• Why the two key names are different load and loading – Maheer Ali Commented Aug 3, 2019 at 14:34
• 2 Your code works. The only issue is that forEach doesn't return anything - you're instead changing obj in-place, so you can just return that, if needed. – VLAZ Commented Aug 3, 2019 at 14:37

2 Answers 2

You can do this easily with new ES2015+ syntax. Using the spread operator:

var result = obj.map(e => ({ ...e, load: 1 }));

This will keep all the other props and only change load to what you want.

Also, FYI - [].forEach() does not work this way.

forEach does not return anything,your code is working fine so in your code the original object that is obj will be changed and if you log it show load:1. If you don't want to change original array use map which returns a new array

var obj = [{
  "id": "service",
  "country": "AR",
  "load": "undefined"
}, {
  "id": "fund",
  "country": "CA",
  "loading": "undefined"
}]


var result = obj.map(function(elem) {
  return Object.assign({}, elem, {
    // only change to 1 if value is undefined
    load: elem.load === 'undefined' ? 1 : elem.load 
  })

});
console.log(result)