POSTing only changed fields on $save of a $resource

Question

When calling $save on an ngResource, is it possible to POST only the edited fields rather than POSTing the entire model each time?

var User = $resource('http://example.com/user/123/');

User.get(function(user) {
  user.name="John Smith";
  user.$save();
  // What I *want* -> POST: /user/123/ {name:'John Smith'}
  // What currently happens -> POST: /user/123/ {name:'John Smith', age: 72, location: 'New York', noOfChildren: 5}
});

Answer 1

No, it's not possible, at least not on instances, see http://docs.angularjs.org/api/ngResource.$resource

[...] The action methods on the class object or instance object can be invoked with the following parameters:

• HTTP GET "class" actions: Resource.action([parameters], [success], [error])
• non-GET "class" actions: Resource.action([parameters], postData, [success], [error])
• non-GET instance actions: instance.$action([parameters], [success], [error])

So, it's only possible by passing the data to save to the "static" save method, ie User.save. Something like this:

User.get(function(user)
{
    user.name = 'John Smith';
    User.save({name: user.name});
});

Whether this works out for you will probably depend on what you're going to do with the user instance.

Answer 2

When I want to save only one field, I use the static .save() method, with a callback that takes the response from that and updates the local object on success:

$scope.saveOneField = function(modelInstance) {
  ModelName.save({
    id: modelInstance.id,
    theField: <some value>
  }, function(response) {
    // If you want to update *all* the latest fields:
    angular.copy(response, modelInstance.data);
    // If you want to update just the one:
    modelInstance.theField = response.data.theField;
  });
};

This assumes that when a POST request is sent to the resource (ie, /modelnames/:id), your server responds with the latest updated version of the modelInstace.