Create a copy of object, then deleted an element only from the copy, not the original

Question

How do I make obj2 the same as obj1 but not affected by the deletion?

Currently the name: 42 entry will be deleted from both objects.

Is it because of hoisting - that is, the deletion occurs before obj2 is created?

var obj1 = {
 'name': 'John',
 'age': 42,
}

var obj2 = obj1

delete obj1['name']

console.log(obj1)
console.log(obj2)

assigning doesn't make a copy. https://stackoverflow.com/questions/728360/how-do-i-correctly-clone-a-javascript-object — zmag, Feb 22 '19 at 15:48
If you simply assign an object to a variable JavaScript will copy it by `reference`. It means that both variables are pointing to the same object. You really need to create a new object with the same properties as your target object. Possible duplicate [How do I correctly clone a JavaScript Object](https://stackoverflow.com/questions/728360/how-do-i-correctly-clone-a-javascript-object) — Alberti Buonarroti, Feb 22 '19 at 15:49
[How much research effort is expected of Stack Overflow users?](https://meta.stackoverflow.com/a/261593/3082296) — adiga, Feb 22 '19 at 15:57

score 4 · Answer 1 · answered Feb 22 '19 at 15:49

4

Please use the following

var obj2 = Object.assign({}, obj1)

Documentation

Please note that this is not the deep copy or a copy of an object with all the possible properties (it is explicitely stated in the doc attached above)

answered Feb 22 '19 at 15:49

nesteant

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And for the complicated object. We can use `_.cloneDeep(obj)` from `lodash` or using `JSON.parse(JSON.stringify(obj))` to clone a new object. – bird Feb 22 '19 at 16:04

score 0 · Accepted Answer · answered Feb 22 '19 at 15:49

You can use spread operator of ES6 as:

let obj2 = {...obj1};

Create a copy of object, then deleted an element only from the copy, not the original

2 Answers2