How to create and clone a JSON object?

Question

I was wondering how can I create a JSON (JS) object and then clone it.

Answer 1

This is what I do and it works like a charm

if (typeof JSON.clone !== "function") {
    JSON.clone = function(obj) {
        return JSON.parse(JSON.stringify(obj));
    };
}

Answer 2

Let's suppose, We have a JSONOBJECT EmailData which have some Data in it. Now if you want the same data in another object (i.e clone the data) then you can do something like this:

JSONOBJECT clone_EmailData=new JSONOBJECT(EmailData.toString());

The above statement will give you a new object with the same data and the new object is not a reference to the EmailData object.

Answer 3

How to create a JSON object in javascript/jquery?

There is nothing like a JSON object. JSON stands for JavaScript Object Notation and is basically a string that encodes information similar to JavaScript's object literals.

You can however create such an encoding (which would result in a string) with JSON.stringify(object), see JSON in JavaScript. You could also create such a string manually, but it is very error prone and I don't recommend it.

How do I clone a JSON object in javascript/jquery?

As it is just a string:

var jsonString2 = jsonString;

I can`t work anymore with javascript arrays

JSON is a format to exchange data, it is not a data structure you can use in an application.

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Maybe you want to read more about JSON, objects in JS and arrays in JS.

Answer 4

We can clone JSON object as below.

EmployeeDetails = 
{
  Name:"John Deer",
  Age:29,
  Company:"ABC Limited."

}

Now create one clone function

function clonning(Employee)
{
  // conversion from Object to String
  var EmployeeString = JSON.stringify(Employee);

  // conversion from String to Object type

  var EmployeeConvertedObject = JSON.parse(EmployeeString);

  // printing before changing prperty value.

  console.log(EmployeeConvertedObject);

  // modifying EmployeeConvertedObject property value 

  EmployeeConvertedObject.Name="Kelvin Bob";

   // printing After changing prperty value.

  console.log(EmployeeConvertedObject);

  // Now printing original json object.

  console.log(Employee);

  // Here original JSON object is not affecting. Only Cloned object affecting.


}

Now Call function.

clonning(EmployeeDetails);

Result:

clonning(EmployeeDetails)
VM212:22 {Name: "John Deer", Age: 29, Company: "ABC Limited."}
VM212:30 {Name: "Kelvin Bob", Age: 29, Company: "ABC Limited."}
VM212:34 {Name: "John Deer", Age: 29, Company: "ABC Limited."}

Answer 5

Q1: How to create a JSON object in javascript/jquery?

Creating a Javascript object is so simple:

var user = {}; // creates an empty user object
var user = {firstName:"John", lastName:"Doe"}; // creates a user by initializing 
// its firstName and lastName properties.

After the creation you can add extra fields to your object like user.age = 30;.

If you have the object as a JSON string, you can convert it to a JSON object using built-in JSON.parse(yourJsonString) function or jQuery's $.parseJSON(yourJsonString) function.

Q2: How do I clone a JSON object in javascript/jquery?

My way to clone JSON objects is extend function of jQuery. For example, you can generate a clone of your user object like below:

var cloneUser = $.extend(true, {}, {firstName:"John", lastName:"Doe"});

The first parameter designates whether the clone object will be a shallow or deep copy of the original (see Object copy on wiki).

To see other JSON cloning alternatives you can read this article.

Answer 6

As of ES6. Object.assign is a good way to do this.

newjsonobj = Object.assign({}, jsonobj, {})

The items in the first argument mutate the existing object, and the third argument are changes in the new object returned.

In ES7 it is proposed that the spread operator be used.

newjsonobj = {...jsonobj}