Is it possible to get the non-enumerable inherited property names of an object?

Question

In JavaScript we have a few ways of getting the properties of an object, depending on what we want to get.

1) Object.keys(), which returns all own, enu

Answer 1

Here is the solution that I came up with while studying the subject. To get all non-enumerable non-own properties of the obj object do getProperties(obj, "nonown", "nonenum");

function getProperties(obj, type, enumerability) {
/**
 * Return array of object properties
 * @param {String} type - Property type. Can be "own", "nonown" or "both"
 * @param {String} enumerability - Property enumerability. Can be "enum", 
 * "nonenum" or "both"
 * @returns {String|Array} Array of properties
 */
    var props = Object.create(null);  // Dictionary

    var firstIteration = true;

    do {
        var allProps = Object.getOwnPropertyNames(obj);
        var enumProps = Object.keys(obj);
        var nonenumProps = allProps.filter(x => !(new Set(enumProps)).has(x));

        enumProps.forEach(function(prop) {
            if (!(prop in props)) {
                props[prop] = { own: firstIteration, enum_: true };
            }           
        });

        nonenumProps.forEach(function(prop) {
            if (!(prop in props)) {
                props[prop] = { own: firstIteration, enum_: false };
            }           
        });

        firstIteration = false;
    } while (obj = Object.getPrototypeOf(obj));

    for (prop in props) {
        if (type == "own" && props[prop]["own"] == false) {
            delete props[prop];
            continue;
        }
        if (type == "nonown" && props[prop]["own"] == true) {
            delete props[prop];
            continue;
        }

        if (enumerability == "enum" && props[prop]["enum_"] == false) {
            delete props[prop];
            continue;
        }
        if (enumerability == "nonenum" && props[prop]["enum_"] == true) {
            delete props[prop];
        }
    }

    return Object.keys(props);
}