Determining if all attributes on a javascript object are null or an empty string

Question

What is the most elegant way to determine if all attributes in a javascript object are either null or the empty string? It should work for an arbitrary number of attributes.

{'a':null, 'b':''} //should return true for this object
{'a':1, 'b':''} //should return false for this object
{'a':0, 'b':1} //should return false
{'a':'', 'b':''} //should return true

Answer 1

Check all values with Object.values. It returns an array with the values, which you can check with Array.prototype.every or Array.prototype.some:

const isEmpty = Object.values(object).every(x => x === null || x === '');

const isEmpty = !Object.values(object).some(x => x !== null && x !== '');

Answer 2

Create a function to loop and check:

function checkProperties(obj) {
    for (var key in obj) {
        if (obj[key] !== null && obj[key] != "")
            return false;
    }
    return true;
}

var obj = {
    x: null,
    y: "",
    z: 1
}

checkProperties(obj) //returns false

Answer 3

Here's my version, specifically checking for null and empty strings (would be easier to just check for falsy)

function isEmptyObject(o) {
    return Object.keys(o).every(function(x) {
        return o[x]===''||o[x]===null;  // or just "return o[x];" for falsy values
    });
}

Answer 4

let obj = { x: null, y: "hello", z: 1 };
let obj1 = { x: null, y: "", z: 0 };

!Object.values(obj).some(v => v);
// false

!Object.values(obj1).some(v => v);
// true

Answer 5

Using Array.some() and check if the values are not null and not empty is more efficient than using Array.every and check it the other way around.

const isEmpty = !Object.values(object).some(x => (x !== null && x !== ''));

This answer should just make the excellent comment of user abd995 more visible.