How can I extend the constructor of an AngularJS resource ($resource)?

Alan H. picture Alan H. · May 9, 2013 · Viewed 9.4k times · Source

I have a model, defined using $resource, that I am successfully loading.

Each loaded instance is, as promised, an instance of the class I defined.

(The example below is from the Angular docs. In it, User.get results in an object that is an instanceof User.)

var User = $resource('/user/:userId', {userId:'@id'});

However, imagine each User comes over the wire like this:

{
  "username": "Bob",
  "preferences": [
    {
      "id": 1,
      "title": "foo",
      "value": false
    }
  ] 
}

I defined a Preference factory that adds valuable methods to Preference objects. But when a User loads, those preferences aren’t Preferences, naturally.

I attempted this:

User.prototype.constructor = function(obj) {
  _.extend(this, obj);
  this.items = _.map(this.preferences, function(pref) {
    return new Preference(pref);
  });
  console.log('Our constructor ran'); // never logs anything
}

But it has no effect and never logs anything.

How can I make each item in my Users’ preferences array an instance of Preference?

Answer

Andrew Joslin picture Andrew Joslin · May 9, 2013

$resource is a simple implementation, and lacks in things like this.

User.prototype.constructor won't do anything; angular doesn't try to act like it's object oriented, unlike other libraries. It's just javascript.

..But luckily, you have promises and javascript :-). Here's a way you could do it:

function wrapPreferences(user) {
  user.preferences = _.map(user.preferences, function(p) {
    return new Preference(p);
  });
  return user;
}

var get = User.get;
User.get = function() {
  return get.apply(User, arguments).$then(wrapPreferences);
};
var $get = User.prototype.$get;
User.prototype.$get = function() {
  return $get.apply(this, arguments).$then(wrapPreferences);
};

You could abstract this into a method which decorates any of a resource's methods: It takes an object, an array of method names, and a decorator function.

function decorateResource(Resource, methodNames, decorator) {
  _.forEach(methodNames, function(methodName) {
    var method = Resource[methodName];
    Resource[methodName] = function() {
      return method.apply(Resource, arguments).$then(decorator);
    };
    var $method = Resource.prototype[methodName];
    Resource.prototype[methodName] = function() {
      return $method.apply(this, arguments).$then(decorator);
    };
  });
}
decorateResource(User, ['get', 'query'], wrapPreferences);