How do I test if a Javascript object has only empty objects at the last nested level?

Question

I have an object that has other objects inside it up to n levels. I want to check if the object only contains other objects. Like this:

var emptyObj = {
    obj1 : {
        obj1a : {},
        obj1b : {}
    },
    obj2 : {},
    obj3 : {
        obj3a : {
            obj3aa : {}
        }
    }
};

Changing the data structure is not an option. No jQuery.

-- edit--

If there is anything apart from an empty object {} at the last level, the test should fail. Failure examples:

var notEmpty1 = {
    obj1 : []
};
var notEmpty2 = {
    obj1: {
        obj1a: ""
    },
    obj2: {}
};

please clarify, I can write recursive function but I don't understand the question — WhiteHat
– WhiteHat, Commented Sep 12, 2015 at 2:37
@WhiteHat I think he wants to check that the properties of the object are: not arrays, functions etc. Just objects. — hammus
– hammus, Commented Sep 12, 2015 at 2:39
Sorry, after reading the post again I realized it was not clear. There should be only empty objects at the leaf level. — Douglas Henrique
– Douglas Henrique, Commented Sep 12, 2015 at 2:48

score 3 · Accepted Answer · 2015-09-12 03:20:43Z

3

An object is empty if it's, well, an object, and every one of its keys is in turn an empty object.

function empty(o) {

  function isObject()      { return o && o.constructor === Object; }
  function keyEmpty(key)   { return empty(o[key]; }
  function everyKeyEmpty() { return Object.keys(o) . every(keyEmpty); }

  return isObject() && everyKeyEmpty();
}

Note that if an object has zero keys then "every" one of the zero keys passes the condition.

edited Sep 12, 2015 at 3:20

answered Sep 12, 2015 at 3:00

user663031

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3 Comments

Douglas Henrique Over a year ago

Very good solution. I changed the second line to return true when !o, so {key: undefined} is understood as {}

user663031 Over a year ago

Do really want {key: undefined} to be treated as empty? Actually, there's a subtle difference.

Douglas Henrique Over a year ago

Yes, it is different, you are right. Your solution fits the answer as it was asked perfectly.

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