It's possible to use a recursive function that iterates by the object keys. Then use the Object.is to test for NaN and null. Then test if the second object is the type that cast to false like 0, NaN, or null. List the keys of both objects and concatenate them to test of missing keys in the obj1 and then iterate it.

When there is a difference between the same key values, it stores the value of object2 and proceeds. If both key values are object means that can be recursively compared and so it does.

function diff(obj1, obj2) {
    const result = {};
    if (Object.is(obj1, obj2)) {
        return undefined;
    }
    if (!obj2 || typeof obj2 !== 'object') {
        return obj2;
    }
    Object.keys(obj1 || {}).concat(Object.keys(obj2 || {})).forEach(key => {
        if(obj2[key] !== obj1[key] && !Object.is(obj1[key], obj2[key])) {
            result[key] = obj2[key];
        }
        if(typeof obj2[key] === 'object' && typeof obj1[key] === 'object') {
            const value = diff(obj1[key], obj2[key]);
            if (value !== undefined) {
                result[key] = value;
            }
        }
    });
    return result;
}

The code above is BSD licensed and can be used anywhere.

Test link: https://jsfiddle.net/gartz/vy9zaof2/54/

An important observation, this will convert arrays to objects and compare the values in the same index position. There are many other ways to compare arrays not covered by this function due to the required extra complexity.

EDIT 2/15/2019: This answer was changed to add the new ES2017 syntax and fix use-cases from comments.

This is just a kickoff, I haven't tested it, but I began with a filter or comparator function, that is recursive, change it however you need to get priority results.

function filter(obj1, obj2) {
    var result = {};
    for(key in obj1) {
        if(obj2[key] != obj1[key]) result[key] = obj2[key];
        if(typeof obj2[key] == 'array' && typeof obj1[key] == 'array') 
            result[key] = arguments.callee(obj1[key], obj2[key]);
        if(typeof obj2[key] == 'object' && typeof obj1[key] == 'object') 
            result[key] = arguments.callee(obj1[key], obj2[key]);
    }
    return result;
}

Tests: http://jsfiddle.net/gartz/Q3BtG/2/

Lodash _.isEqual allows you to do that:

var
remoteJSON = {"allowExternalMembers": "false", "whoCanJoin": "CAN_REQUEST_TO_JOIN"},
    localJSON = {"whoCanJoin": "CAN_REQUEST_TO_JOIN", "allowExternalMembers": "false"};
    
console.log( _.isEqual(remoteJSON, localJSON) );

<script src="https://cdn.jsdelivr.net/npm/[email protected]/lodash.min.js"></script>

You can try following as a simple comparison:

Note: This answer assumes the structure to be exactly same. You can take this as reference and make it more generic for cases where structures are different.

Points to consider:

• Number of items are different. o1.details have 2 entries and o2.details have say 5 entries.
• Object structure is different. Properties in both object are different.

function getDifference(o1, o2) {
  var diff = {};
  var tmp = null;
  if (JSON.stringify(o1) === JSON.stringify(o2)) return;

  for (var k in o1) {
    if (Array.isArray(o1[k]) && Array.isArray(o2[k])) {
      tmp = o1[k].reduce(function(p, c, i) {
        var _t = getDifference(c, o2[k][i]);
        if (_t)
          p.push(_t);
        return p;
      }, []);
      if (Object.keys(tmp).length > 0)
        diff[k] = tmp;
    } else if (typeof(o1[k]) === "object" && typeof(o2[k]) === "object") {
      tmp = getDifference(o1[k], o2[k]);
      if (tmp && Object.keys(tmp) > 0)
        diff[k] = tmp;
    } else if (o1[k] !== o2[k]) {
      diff[k] = o2[k]
    }
  }
  return diff;
}

var o1={id:"1",details:[{name:"Peter",address:"Arizona",phone:9900998899},{name:"Jam",address:"Kentucky",phone:56034033343}],profession:"Business"},
o2={id:"2",details:[{name:"Peter",address:"Arizona",phone:9900998899},{name:"David",address:"Boston",phone:434323434}],profession:"Business"};

var d = getDifference(o1, o2)
console.log(d)

I had written an answer to detect changes in objects. You can even take reference from this as well.