Javascript: Checking if an object has no properties or if a map/associative-array is empty [duplicate]

Question

Possible Duplicate:
How do I test for an empty Javascript object from JSON?

Is there an easy way to check if an object has no properties, in Javascript? Or in other words, an easy way to check if a map/associative array is empty? For example, let's say you had the following:

var nothingHere = {};
var somethingHere = {foo: "bar"};

Is there an easy way to tell which one is "empty"? The only thing I can think of is something like this:

function isEmpty(map) {
   var empty = true;

   for(var key in map) {
      empty = false;
      break;
   }

   return empty;
}

Is there a better way (like a native property/function or something)?

@Daniel - thanks for the link to that question. I tried searching on SO but I didn't find anything. Mods - please close this question. Thanks! — Vivin Paliath
– Vivin Paliath, Commented Aug 6, 2010 at 19:22
I would go with chryss's solution over yours because of the hasOwnProperty call. If anything extends the Object prototype (something many libraries do), your method will no longer return the correct results as it will read inherited properties. — Cristian Sanchez
– Cristian Sanchez, Commented Aug 6, 2010 at 19:36
@Daniel yeah, I like it for that reason as well. Prototype seems to pollute the namespace that way. — Vivin Paliath
– Vivin Paliath, Commented Aug 6, 2010 at 19:39

Frazer Kirkman · Accepted Answer · 2018-03-29 13:29:06Z

51

Try this:

function isEmpty(map) {
   for(var key in map) {
     if (map.hasOwnProperty(key)) {
        return false;
     }
   }
   return true;
}

Your solution works, too, but only if there is no library extending the Object prototype. It may or may not be good enough.

edited Mar 29, 2018 at 13:29

Frazer Kirkman

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answered Aug 6, 2010 at 19:19

chryss

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3 Comments

Cristian Sanchez Over a year ago

The hasOwnProperty call is pretty vital here if any libraries are messing around with the Object prototype. +1

chryss Over a year ago

Thanks. I put the remark in the solution -- I had adopted the hasOwnPrototype call for a while and wasn't even thinking about it any longer.

Zo72 Over a year ago

you ever not even thinking about it any longer and in fact you called it hasOwnPrototype. lol

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