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up vote 29 down vote favorite
3

The below first logs 0, and then logs 1. How do I store a copy of the object, rather than a reference to it?

debug.log(vi.details.segment);
vi.nextSegment = vi.details;
vi.nextSegment.segment++;
debug.log(vi.details.segment);
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up vote 63 down vote accepted

To clone an object in jQuery:

var vi.nextSegment = jQuery.extend({}, vi.details);

NOTE: The above is a shallow copy: any nested objects or arrays will be copied by reference - meaning any changes you make to vi.nextSegment.obj[prop] will be reflected in vi.details.obj[prop]. If you want a completely new object which is completely separate from the original, you will need to do a deep copy (pass true as the first parameter):

var vi.nextSegment = jQuery.extend(true, {}, vi.details);

To read up more on extend, see here.

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8  
That isn't a deep copy. var a = {b: [1, 2]}; $.extend({}, a).b[0] = 'test'; and a.b[0] will be changed to 'test'. To do a deep copy, put true as the first argument: $.extend(true, {}, a); – Reid Mar 19 '11 at 20:25
    
I forgot to include deep copy, I just edited my answer. – Mike Lewis Mar 19 '11 at 20:26
3  
One gothcha that just got me: jQuery.extend(true, {}, obj); will create a copy. jQuery.extend(true, obj, {}); will not. – worldsayshi Apr 21 '13 at 21:41
    
@worldsayshi : I wish I saw your comment sooner. Having headaches on this for weeks. Thanks a lot. – nicolas Oct 9 '14 at 21:58
up vote 19 down vote

Take a look at the post: What is the most efficient way to clone a javascript object

As per John Resig's answer:

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

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

More information can be found in the jQuery documentation.

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up vote 4 down vote

This worked better for me cloning an object using jQuery "parseJSON()" and "JSON.stringify()"

$.ajax({
  url: 'ajax/test.html',
  dataType: 'json',
  success: function(data) {
    var objY = $.parseJSON(JSON.stringify(data));
    var objX = $.parseJSON(JSON.stringify(data));
  }
});

Cloning data object in objX & objY are two different object, you do no have to mess up with the "by reference" problem

Gracias!

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I like this :0) – kiranvj May 11 '15 at 20:06
up vote 2 down vote

Another way to clone object is

newObj = JSON.parse(JSON.stringify(oldObj));

But be careful if it's contains dates. JSON.parse will return date.toString() instead of date in that case.

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up vote 0 down vote

This is how I copy elements several times:

First I have a template:

<div class="forms-container">
    <div class="form-template">
        First Name <input>
         .. a lot of other data ...
        Last Name <input>
     <div>
     <button onclick="add_another();">Add another!</button>
<div>

Now, the JavaScript:

function add_another(){
    jQuery(".form-template").clone().appendTo(".forms-container");
}
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