I created a JavaScript object, but how I can determine the class of that object?
I want something similar to Java's .getClass()
method.
I created a JavaScript object, but how I can determine the class of that object?
I want something similar to Java's .getClass()
method.
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.
func.prototype
(yes, functions are objects, but the prototype
property is only relevant on function objects). –
Cuboid instanceof
/isPrototypeOf()
and the non-standard __proto__
–
Eudocia Object.getPrototypeOf()
–
Eudocia Function Foo() {}
–
Considered Foo
pretty-printed. The comments don't indicate the return values, but equalities that hold for the return values. So the comment means that foo.constructor == Foo
holds, which will also be the case for you. –
Horan class
of an object is with Object.prototype.toString.call(myObject)
. What this question is really about is getting the constructor and/or prototype, not the class, but it may be worth adding this for completeness. See this MDN article –
Photometer new this.constructor()
to instantiate new instance of the current Object. #8454387 –
Fondle constructor.name
if your code is being minified. The function name is going to change arbitrarily. –
Besom var Foo = () => true;
It's returning function
whenever you call console.log(typeof Foo)
. Foo
is a function but a constructor. –
Unlearned construction.name
as a token to be ignored / not minimize. Besides most (if not all) minifier software provide exception rules. –
Frith 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.
Function.name
is not (yet) part of the JavaScript standard. It is currently supported in Chrome and Firefox, but not in IE(10). –
Camphor Object.create(something).constructor === something.constructor
, which is not quite correct too. So obj.constructor is unreliable for all objects made with Object.create, no matter with or without a prototype. –
Loculus constructor.name
if your code is being minified. The function name is going to change arbitrarily. –
Besom class A { }; class B extends A { }; b = new B(); b.constructor.name; "B"
–
Amphitryon constructor.name
behaves as expected with the new class support in ES6. –
Lundgren 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;
}
getNativeClass("") === "String";
getNativeClass(true) === "Boolean";
getNativeClass(0) === "Number";
getNativeClass([]) === "Array";
getNativeClass({}) === "Object";
getNativeClass(null) === "null";
getAnyClass(new (function Foo(){})) === "Foo";
getAnyClass(new class Foo{}) === "Foo";
// etc...
"Object"
. –
Camphor return obj.constructor.name
. That gives the same results, plus also handles non native objects. –
Sosthenna 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
n
JS really blows for this sort of thing. –
Lookthrough 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.
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
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"
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.
Don't use o.constructor
because it can be changed by the object content. Instead, use Object.getPrototypeOf()?.constructor
.
const fakedArray = JSON.parse('{ "constructor": { "name": "Array" } }');
// returns 'Array', which is faked.
fakedArray.constructor.name;
// returns 'Object' as expected
Object.getPrototypeOf(fakedArray)?.constructor?.name;
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... );
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'
getClass()
function using constructor.prototype.nameI 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;
}
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;
}
}
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.
.name
. –
Lundgren .name
is not defined even on IE 9 –
Lip 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:
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;
};
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;
}
getClass()
and getInstance()
You are able to get a reference for an Object's class using this.constructor
.
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
function A() {};
A.getClass = function() {
return this;
}
A.getInstance() {
return new this;
}
this.constructor
? –
Iconostasis 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
isClass
implementation fails for e.g. isClass(Array)
, isClass(Object)
, ... Besides that, the OP didn't want to know whether a function is a class constructor but instead wants to get/access the class (which is the constructor function) reference of any given type. –
Rovner 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'
class ICT{
//your logic
}
const student= new ICT();
if (student instanceof ICT) {
console.log('student is an instance of ICT');
}
If you have access to an instance of the class Foo (say foo = new Foo()) then there is exactly one way to get access to the the class from the instance: foo.Contructor in Javascript = foo.getClass() in Java.
eval() is another way, but since eval() is never recommended and works for everything (analogous to Java reflection), that answer is not recommended. foo.Constructor = Foo
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.
class
type. It does have a class
keyword and class
syntax for creating prototypes in which the methods can more easily access super
. –
Cohleen 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.
__proto__
property is deprecated, and has almost no advantage over the prototype
property. –
Lytic 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
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class
type. It does have aclass
keyword andclass
syntax for creating prototypes in which the methods can more easily accesssuper
. – Cohleen