Javascript: how to test an object is an object (and not an array or a string)?

Question

I know about (a instanceof Array) but how would I test for an object?

var c = {};

if ( c instanceof XXXX) {
   // should get thru
}

var s = "abscdef";

if ( s instanceof XXXX) {
   // should not get thru
}

var a = [];

if ( a instanceof XXXX) {
   // should not get thru
}

what do you mean exactly? {} is an object, not a collection. — Yoeri
– Yoeri, Commented Jun 10, 2014 at 10:25
Your terminology is confused - c is a Object, not a "collection". — joews
– joews, Commented Jun 10, 2014 at 10:26
Ok, sorry then. How to detect an object of any type then? (I edit the question) — Stéphane de Luca
– Stéphane de Luca, Commented Jun 10, 2014 at 10:26

joews · Accepted Answer · 2014-06-10 10:34:56Z

1

function isObject(c) {
  return c instanceof Object 
    && !(c instanceof Array) 
    && !(c instanceof Function)
}

The Array and Function checks are necessary because JavaScript Arrays are also Objects (assuming that you don't want the function to return true for an Array or Function arguments)

Example output:

isObject([])
> false

isObject({})
> true

isObject(1)
> false

isObject('something')
> false

isObject(isObject)
> false

edited Jun 10, 2014 at 10:34

answered Jun 10, 2014 at 10:29

joews

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2 Comments

hsz Over a year ago

You can also add && !(c instanceof Function) because c = function(){} is an instance of Object AND Function

Stéphane de Luca Over a year ago

Perfect, thanks @joews!: a fiddle to see it in action jsfiddle.net/stephanedeluca/43Jza

Yoeri · Accepted Answer · 2014-06-10 10:51:55Z

0

I think what you mean is something like this? I'm sorry if I don't really understand the question.

function MyOwnType(name){
    this.name = name
}
var myInstance = new MyOwnType("StackOverflow");
console.log(myInstance instanceof(MyOwnType)) //this evaluates to true
console.log(myInstance instanceof(Array)) //this evaluates to false

answered Jun 10, 2014 at 10:51

Yoeri

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