How to copy JavaScript object to new variable NOT by reference? [duplicate]

Question

I wrote a quick jsfiddle here, where I pass a small JSON object to a new variable and modify the data from the original variable (not the new variable), but the new variable's data gets updated as well. This must mean that the JSON object was passed by reference, right?

Here is my quick code:

var json_original = {one:'one', two:'two'}

var json_new = json_original;

console.log(json_original); //one, two
console.log(json_new); //one, two

json_original.one = 'two';
json_original.two = 'one';

console.log(json_original); //two, one
console.log(json_new); //two, one

Is there a way to make a deep copy of a JSON object so that modifying the original variable won't modify the new variable?

Answer 1

Important edit: this answer was made in 2014 and there are now better ways of deep-cloning an object.

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I've found that the following works if you're not using jQuery and only interested in cloning simple objects (see comments).

JSON.parse(JSON.stringify(json_original));

Documentation

• JSON.parse()
• JSON.stringify()

Answer 2

Your only option is to somehow clone the object.

See this stackoverflow question on how you can achieve this.

For simple JSON objects, the simplest way would be:

var newObject = JSON.parse(JSON.stringify(oldObject));

if you use jQuery, you can use:

// Shallow copy
var newObject = jQuery.extend({}, oldObject);

// Deep copy
var newObject = jQuery.extend(true, {}, oldObject);

UPDATE 2017: I should mention, since this is a popular answer, that there are now better ways to achieve this using newer versions of javascript:

In ES6 or TypeScript (2.1+):

var shallowCopy = { ...oldObject };

var shallowCopyWithExtraProp = { ...oldObject, extraProp: "abc" };

Note that if extraProp is also a property on oldObject, its value will not be used because the extraProp : "abc" is specified later in the expression, which essentially overrides it. Of course, oldObject will not be modified.