How to get a JavaScript object's class?

Question

I created a JavaScript object, but how I can determine the class of that object?

I want something similar to Java's .getClass() method.

Answer 1

There's no exact counterpart to Java's getClass() in JavaScript. Mostly that's due to JavaScript being a prototype-based language, as opposed to Java being a class-based one.

Depending on what you need getClass() for, there are several options in JavaScript:

• typeof
• instanceof
• obj.constructor
• func.prototype, proto.isPrototypeOf

A few examples:

function Foo() {}
var foo = new Foo();

typeof Foo;             // == "function"
typeof foo;             // == "object"

foo instanceof Foo;     // == true
foo.constructor.name;   // == "Foo"
Foo.name                // == "Foo"    

Foo.prototype.isPrototypeOf(foo);   // == true

Foo.prototype.bar = function (x) {return x+x;};
foo.bar(21);            // == 42

Note: if you are compiling your code with Uglify it will change non-global class names. To prevent this, Uglify has a --mangle param that you can set to false is using gulp or grunt.

Answer 2

obj.constructor.name

is a reliable method in modern browsers. Function.name was officially added to the standard in ES6, making this a standards-compliant means of getting the "class" of a JavaScript object as a string. If the object is instantiated with var obj = new MyClass(), it will return "MyClass".

It will return "Number" for numbers, "Array" for arrays and "Function" for functions, etc. It generally behaves as expected. The only cases where it fails are if an object is created without a prototype, via Object.create( null ), or the object was instantiated from an anonymously-defined (unnamed) function.

Also note that if you are minifying your code, it's not safe to compare against hard-coded type strings. For example instead of checking if obj.constructor.name == "MyType", instead check obj.constructor.name == MyType.name. Or just compare the constructors themselves, however this won't work across DOM boundaries as there are different instances of the constructor function on each DOM, thus doing an object comparison on their constructors won't work.

Answer 3

This getNativeClass() function returns "undefined" for undefined values and "null" for null.
For all other values, the CLASSNAME-part is extracted from [object CLASSNAME], which is the result of using Object.prototype.toString.call(value).

getAnyClass() behaves the same as getNativeClass(), but also supports custom constructors

function getNativeClass(obj) {
  if (typeof obj === "undefined") return "undefined";
  if (obj === null) return "null";
  return Object.prototype.toString.call(obj).match(/^\[object\s(.*)\]$/)[1];
}

function getAnyClass(obj) {
  if (typeof obj === "undefined") return "undefined";
  if (obj === null) return "null";
  return obj.constructor.name;
}

getClass("")   === "String";
getClass(true) === "Boolean";
getClass(0)    === "Number";
getClass([])   === "Array";
getClass({})   === "Object";
getClass(null) === "null";

getAnyClass(new (function Foo(){})) === "Foo";
getAnyClass(new class Foo{}) === "Foo";

// etc...

Answer 4

To get the "pseudo class", you can get the constructor function, by

obj.constructor

assuming the constructor is set correctly when you do the inheritance -- which is by something like:

Dog.prototype = new Animal();
Dog.prototype.constructor = Dog;

and these two lines, together with:

var woofie = new Dog()

will make woofie.constructor point to Dog. Note that Dog is a constructor function, and is a Function object. But you can do if (woofie.constructor === Dog) { ... }.

If you want to get the class name as a string, I found the following working well:

http://blog.magnetiq.com/post/514962277/finding-out-class-names-of-javascript-objects

function getObjectClass(obj) {
    if (obj && obj.constructor && obj.constructor.toString) {
        var arr = obj.constructor.toString().match(
            /function\s*(\w+)/);

        if (arr && arr.length == 2) {
            return arr[1];
        }
    }

    return undefined;
}

It gets to the constructor function, converts it to string, and extracts the name of the constructor function.

Note that obj.constructor.name could have worked well, but it is not standard. It is on Chrome and Firefox, but not on IE, including IE 9 or IE 10 RTM.

Answer 5

You can get a reference to the constructor function which created the object by using the constructor property:

function MyObject(){
}

var obj = new MyObject();
obj.constructor; // MyObject

If you need to confirm the type of an object at runtime you can use the instanceof operator:

obj instanceof MyObject // true

Answer 6

i had a situation to work generic now and used this:

class Test {
  // your class definition
}

nameByType = function(type){
  return type.prototype["constructor"]["name"];
};

console.log(nameByType(Test));

thats the only way i found to get the class name by type input if you don't have a instance of an object.

(written in ES2017)

dot notation also works fine

console.log(Test.prototype.constructor.name); // returns "Test" 

Answer 7

In keeping with its unbroken record of backwards-compatibility, ECMAScript 6, JavaScript still doesn't have a class type (though not everyone understands this). It does have a class keyword as part of its class syntax for creating prototypes—but still no thing called class. JavaScript is not now and has never been a classical OOP language. Speaking of JS in terms of class is only either misleading or a sign of not yet grokking prototypical inheritance (just keeping it real).

That means this.constructor is still a great way to get a reference to the constructor function. And this.constructor.prototype is the way to access the prototype itself. Since this isn't Java, it's not a class. It's the prototype object your instance was instantiated from. Here is an example using the ES6 syntactic sugar for creating a prototype chain:

class Foo {
  get foo () {
    console.info(this.constructor, this.constructor.name)
    return 'foo'
  }
}

class Bar extends Foo {
  get foo () {
    console.info('[THIS]', this.constructor, this.constructor.name, Object.getOwnPropertyNames(this.constructor.prototype))
    console.info('[SUPER]', super.constructor, super.constructor.name, Object.getOwnPropertyNames(super.constructor.prototype))

    return `${super.foo} + bar`
  }
}

const bar = new Bar()
console.dir(bar.foo)

This is what that outputs using babel-node:

> $ babel-node ./foo.js                                                                                                                   ⬡ 6.2.0 [±master ●]
[THIS] [Function: Bar] 'Bar' [ 'constructor', 'foo' ]
[SUPER] [Function: Foo] 'Foo' [ 'constructor', 'foo' ]
[Function: Bar] 'Bar'
'foo + bar'

There you have it! In 2016, there's a class keyword in JavaScript, but still no class type. this.constructor is the best way to get the constructor function, this.constructor.prototype the best way to get access to the prototype itself.

Answer 8

For Javascript Classes in ES6 you can use object.constructor. In the example class below the getClass() method returns the ES6 class as you would expect:

var Cat = class {

    meow() {

        console.log("meow!");

    }

    getClass() {

        return this.constructor;

    }

}

var fluffy = new Cat();

...

var AlsoCat = fluffy.getClass();
var ruffles = new AlsoCat();

ruffles.meow();    // "meow!"

If you instantiate the class from the getClass method make sure you wrap it in brackets e.g. ruffles = new ( fluffy.getClass() )( args... );

Answer 9

In javascript, there are no classes, but I think that you want the constructor name and obj.constructor.toString() will tell you what you need.

Answer 10

I find object.constructor.toString() return [object objectClass] in IE ,rather than function objectClass () {} returned in chome. So,I think the code in http://blog.magnetiq.com/post/514962277/finding-out-class-names-of-javascript-objects may not work well in IE.And I fixed the code as follows:

code:

var getObjectClass = function (obj) {
        if (obj && obj.constructor && obj.constructor.toString()) {
            
                /*
                 *  for browsers which have name property in the constructor
                 *  of the object,such as chrome 
                 */
                if(obj.constructor.name) {
                    return obj.constructor.name;
                }
                var str = obj.constructor.toString();
                /*
                 * executed if the return of object.constructor.toString() is 
                 * "[object objectClass]"
                 */
                 
                if(str.charAt(0) == '[')
                {
                        var arr = str.match(/\[\w+\s*(\w+)\]/);
                } else {
                        /*
                         * executed if the return of object.constructor.toString() is 
                         * "function objectClass () {}"
                         * for IE Firefox
                         */
                        var arr = str.match(/function\s*(\w+)/);
                }
                if (arr && arr.length == 2) {
                            return arr[1];
                        }
          }
          return undefined; 
    };
    

Answer 11

Agree with dfa, that's why i consider the prototye as the class when no named class found

Here is an upgraded function of the one posted by Eli Grey, to match my way of mind

function what(obj){
    if(typeof(obj)==="undefined")return "undefined";
    if(obj===null)return "Null";
    var res = Object.prototype.toString.call(obj).match(/^\[object\s(.*)\]$/)[1];
    if(res==="Object"){
        res = obj.constructor.name;
        if(typeof(res)!='string' || res.length==0){
            if(obj instanceof jQuery)return "jQuery";// jQuery build stranges Objects
            if(obj instanceof Array)return "Array";// Array prototype is very sneaky
            return "Object";
        }
    }
    return res;
}

Answer 12

I suggest using Object.prototype.constructor.name:

Object.defineProperty(Object.prototype, "getClass", {
    value: function() {
      return this.constructor.name;
    }
});

var x = new DOMParser();
console.log(x.getClass()); // `DOMParser'

var y = new Error("");
console.log(y.getClass()); // `Error'

Answer 13

Here's a implementation of getClass() and getInstance()

You are able to get a reference for an Object's class using this.constructor.

From an instance context:

function A() {
  this.getClass = function() {
    return this.constructor;
  }

  this.getNewInstance = function() {
    return new this.constructor;
  }
}

var a = new A();
console.log(a.getClass());  //  function A { // etc... }

// you can even:
var b = new (a.getClass());
console.log(b instanceof A); // true
var c = a.getNewInstance();
console.log(c instanceof A); // true

From static context:

function A() {};

A.getClass = function() {
  return this;
}

A.getInstance() {
  return new this;
}

Answer 14

We can read Class's name of an instance by just doing 'instance.constructor.name' like in this example:

class Person {
  type = "developer";
}
let p = new Person();

p.constructor.name // Person

Answer 15

If you need to not only GET class but also EXTEND it from having just an instance, write:

let's have

 class A{ 
   constructor(name){ 
     this.name = name
   }
 };

 const a1 = new A('hello a1');

so to extend A having the instance only use:

const a2 = new (Object.getPrototypeOf(a1)).constructor('hello from a2')
// the analog of const a2 = new A()

console.log(a2.name)//'hello from a2'

Answer 16

You can also do something like this

 class Hello {
     constructor(){
     }
    }
    
      function isClass (func) {
        return typeof func === 'function' && /^class\s/.test(Function.prototype.toString.call(func))
    }
    
   console.log(isClass(Hello))

This will tell you if the input is class or not

Answer 17

getClass() function using constructor.prototype.name

I found a way to access the class that is much cleaner than some of the solutions above; here it is.

function getClass(obj) {

   // if the type is not an object return the type
   if((let type = typeof obj) !== 'object') return type; 
    
   //otherwise, access the class using obj.constructor.name
   else return obj.constructor.name;   
}

How it works

the constructor has a property called name accessing that will give you the class name.

cleaner version of the code:

function getClass(obj) {

   // if the type is not an object return the type
   let type = typeof obj
   if((type !== 'object')) { 
      return type; 
   } else { //otherwise, access the class using obj.constructor.name
      return obj.constructor.name; 
   }   
}

Answer 18

There is one another technique to identify your class You can store ref to your class in instance like below.

class MyClass {
    static myStaticProperty = 'default';
    constructor() {
        this.__class__ = new.target;
        this.showStaticProperty = function() {
            console.log(this.__class__.myStaticProperty);
        }
    }
}

class MyChildClass extends MyClass {
    static myStaticProperty = 'custom';
}

let myClass = new MyClass();
let child = new MyChildClass();

myClass.showStaticProperty(); // default
child.showStaticProperty(); // custom

myClass.__class__ === MyClass; // true
child.__class__ === MyClass; // false
child.__class__ === MyChildClass; // true

Answer 19

Javascript is a class-less languages: there are no classes that defines the behaviour of a class statically as in Java. JavaScript uses prototypes instead of classes for defining object properties, including methods, and inheritance. It is possible to simulate many class-based features with prototypes in JavaScript.

Answer 20

Question seems already answered but the OP wants to access the class of and object, just like we do in Java and the selected answer is not enough (imho).

With the following explanation, we can get a class of an object(it's actually called prototype in javascript).

var arr = new Array('red', 'green', 'blue');
var arr2 = new Array('white', 'black', 'orange');

You can add a property like this:

Object.defineProperty(arr,'last', {
  get: function(){
    return this[this.length -1];
  }
});
console.log(arr.last) // blue

But .last property will only be available to 'arr' object which is instantiated from Array prototype. So, in order to have the .last property to be available for all objects instantiated from Array prototype, we have to define the .last property for Array prototype:

Object.defineProperty(Array.prototype,'last', {
  get: function(){
    return this[this.length -1];
  }
});
console.log(arr.last) // blue
console.log(arr2.last) // orange

The problem here is, you have to know which object type (prototype) the 'arr' and 'arr2' variables belongs to! In other words, if you don't know class type (prototype) of the 'arr' object, then you won't be able to define a property for them. In the above example, we know arr is instance of the Array object, that's why we used Array.prototype to define a property for Array. But what if we didn't know the class(prototype) of the 'arr'?

Object.defineProperty(arr.__proto__,'last2', {
  get: function(){
    return this[this.length -1];
  }
});
console.log(arr.last) // blue
console.log(arr2.last) // orange

As you can see, without knowing that 'arr' is an Array, we can add a new property just bu referring the class of the 'arr' by using 'arr.__proto__'.

We accessed the prototype of the 'arr' without knowing that it's an instance of Array and I think that's what OP asked.