json - How can I get a list of the differences between two JavaScript object graphs?

Question

I want to be able to get a list of all differences between two JavaScript object graphs, with the property names and values where the deltas occur.

For what it is worth, these objects are usually retrieved from the server as JSON and typically are no more than a handful of layers deep (i.e. it may be an array of objects that themselves have data and then arrays with other data objects).

I want to not only see the changes to basic properties, but differences in the number of members of an array, etc. etc.

If I don't get an answer, I will probably end up writing this myself, but hope someone has already done this work or know of someone who has.

---

EDIT: These objects will typically be very close in structure to one another, so we are not talking about objects that are utterly different from one another, but may have 3 or 4 deltas.

Answer 1

Here is a partial, na?ve solution to my problem - I will update this as I further develop it.

function findDifferences(objectA, objectB) {
   var propertyChanges = [];
   var objectGraphPath = ["this"];
   (function(a, b) {
      if(a.constructor == Array) {
         // BIG assumptions here: That both arrays are same length, that
         // the members of those arrays are _essentially_ the same, and 
         // that those array members are in the same order...
         for(var i = 0; i < a.length; i++) {
            objectGraphPath.push("[" + i.toString() + "]");
            arguments.callee(a[i], b[i]);
            objectGraphPath.pop();
         }
      } else if(a.constructor == Object || (a.constructor != Number && 
                a.constructor != String && a.constructor != Date && 
                a.constructor != RegExp && a.constructor != Function &&
                a.constructor != Boolean)) {
         // we can safely assume that the objects have the 
         // same property lists, else why compare them?
         for(var property in a) {
            objectGraphPath.push(("." + property));
            if(a[property].constructor != Function) {
               arguments.callee(a[property], b[property]);
            }
            objectGraphPath.pop();
         }
      } else if(a.constructor != Function) { // filter out functions
         if(a != b) {
            propertyChanges.push({ "Property": objectGraphPath.join(""), "ObjectA": a, "ObjectB": b });
         }
      }
   })(objectA, objectB);
   return propertyChanges;
}

And here is a sample of how it would be used and the data it would provide (please excuse the long example, but I want to use something relatively non-trivial):

var person1 = { 
   FirstName : "John", 
   LastName : "Doh", 
   Age : 30, 
   EMailAddresses : [
      "john.doe@gmail.com", 
      "jd@initials.com"
   ], 
   Children : [ 
      { 
         FirstName : "Sara", 
         LastName : "Doe", 
         Age : 2 
      }, { 
         FirstName : "Beth", 
         LastName : "Doe", 
         Age : 5 
      } 
   ] 
};

var person2 = { 
   FirstName : "John", 
   LastName : "Doe", 
   Age : 33, 
   EMailAddresses : [
      "john.doe@gmail.com", 
      "jdoe@hotmail.com"
   ], 
   Children : [ 
      { 
         FirstName : "Sara", 
         LastName : "Doe", 
         Age : 3 
      }, { 
         FirstName : "Bethany", 
         LastName : "Doe", 
         Age : 5 
      } 
   ] 
};

var differences = findDifferences(person1, person2);

At this point, here is what the differences array would look like if you serialized it to JSON:

[
   {
      "Property":"this.LastName", 
      "ObjectA":"Doh", 
      "ObjectB":"Doe"
   }, {
      "Property":"this.Age", 
      "ObjectA":30, 
      "ObjectB":33
   }, {
      "Property":"this.EMailAddresses[1]", 
      "ObjectA":"jd@initials.com", 
      "ObjectB":"jdoe@hotmail.com"
   }, {
      "Property":"this.Children[0].Age", 
      "ObjectA":2, 
      "ObjectB":3
   }, {
      "Property":"this.Children[1].FirstName", 
      "ObjectA":"Beth", 
      "ObjectB":"Bethany"
   }
]

The this in the Property value refers to the root of the object that was compared. So, this solution is not yet exactly what I need, but it is pretty darn close.

Hope this is useful to someone out there, and if you have any suggestions for improvement, I am all-ears; I wrote this very late last night (i.e. early this morning) and there may be things I am completely overlooking.

Thanks.