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);
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.
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.
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!