Check if a value is an object in JavaScript

Question

How do you check if a value is an object in JavaScript?

Answer 1

Try this

if (objectName instanceof Object == false) {
  alert('Not an object');
}
else {
  alert('An object');
}

Answer 2

Oh My God! I think this could be more shorter than ever, let see this:

Short and Final code

function isObject(obj)
{
    return obj != null && obj.constructor.name === "Object"
}

console.log(isObject({})) // returns true
console.log(isObject([])) // returns false
console.log(isObject(null)) // returns false

Explained

Return Types

typeof JavaScript objects (including null) returns "object"

console.log(typeof null, typeof [], typeof {})

Checking on Their constructors

Checking on their constructor property returns function with their names.

console.log(({}).constructor) // returns a function with name "Object"
console.log(([]).constructor) // returns a function with name "Array"
console.log((null).constructor) //throws an error because null does not actually have a property

Introducing Function.name

Function.name returns a readonly name of a function or "anonymous" for closures.

console.log(({}).constructor.name) // returns "Object"
console.log(([]).constructor.name) // returns "Array"
console.log((null).constructor.name) //throws an error because null does not actually have a property

Note: As of 2018, Function.name might not work in IE https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/name#Browser_compatibility

Answer 3

var a = [1]
typeof a //"object"
a instanceof Object //true
a instanceof Array //true

var b ={a: 1}
b instanceof Object //true
b instanceof Array //false

var c = null
c instanceof Object //false
c instanceof Array //false

I was asked to provide more details. Most clean and understandable way of checking if our variable is an object is typeof myVar. It returns a string with a type (e.g. "object", "undefined").

Unfortunately either Array and null also have a type object. To take only real objects there is a need to check inheritance chain using instanceof operator. It will eliminate null, but Array has Object in inheritance chain.

So the solution is:

if (myVar instanceof Object && !(myVar instanceof Array)) {
  // code for objects
}

Answer 4

If you would like to check if the prototype for an object solely comes from Object. Filters out String, Number, Array, Arguments, etc.

function isObject (n) {
  return Object.prototype.toString.call(n) === '[object Object]';
}

Or as a single-expression arrow function (ES6+)

const isObject = n => Object.prototype.toString.call(n) === '[object Object]'

Answer 5

Little late... for "plain objects" (i mean, like {'x': 5, 'y': 7}) i have this little snippet:

function isPlainObject(o) {
   return (o === null || Array.isArray(o) || typeof o == 'function' || o.constructor === Date ) ?
           false
          :(typeof o == 'object');
}

It generates the next output:

console.debug(isPlainObject(isPlainObject)); //function, false
console.debug(isPlainObject({'x': 6, 'y': 16})); //literal object, true
console.debug(isPlainObject(5)); //number, false
console.debug(isPlainObject(undefined)); //undefined, false
console.debug(isPlainObject(null)); //null, false
console.debug(isPlainObject('a')); //string, false
console.debug(isPlainObject([])); //array?, false
console.debug(isPlainObject(true)); //bool, false
console.debug(isPlainObject(false)); //bool, false

It always works for me. If will return "true" only if the type of "o" is "object", but no null, or array, or function. :)

Answer 6

lodash has isPlainObject, which might be what many who come to this page are looking for. It returns false when give a function or array.