I am copying objA
to objB
const objA = { prop: 1 }, const objB = objA; objB.prop = 2; console.log(objA.prop); // logs 2 instead of 1
same problem for Arrays
const arrA = [1, 2, 3], const arrB = arrA; arrB.push(4); console.log(arrA.length); // `arrA` has 4 elements instead of 3.
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Answer
It is clear that you have some misconceptions of what the statement var tempMyObj = myObj;
does.
In JavaScript objects are passed and assigned by reference (more accurately the value of a reference), so tempMyObj
and myObj
are both references to the same object.
Here is a simplified illustration that may help you visualize what is happening
// [Object1]<--------- myObj var tempMyObj = myObj; // [Object1]<--------- myObj // ^ // | // ----------- tempMyObj
As you can see after the assignment, both references are pointing to the same object.
You need to create a copy if you need to modify one and not the other.
// [Object1]<--------- myObj const tempMyObj = Object.assign({}, myObj); // [Object1]<--------- myObj // [Object2]<--------- tempMyObj
Old Answer:
Here are a couple of other ways of creating a copy of an object
Since you are already using jQuery:
var newObject = jQuery.extend(true, {}, myObj);
With vanilla JavaScript
function clone(obj) { if (null == obj || "object" != typeof obj) return obj; var copy = obj.constructor(); for (var attr in obj) { if (obj.hasOwnProperty(attr)) copy[attr] = obj[attr]; } return copy; } var newObject = clone(myObj);