如何从angularjs的$ q.defer（）promise对象中删除回调

Question

Angularjs $q.defer() promise object receives notify callback, and when we give more than one, it keeps and notify all of them, ie it does not overwrite to older one.

var def=$q.defer();
def.promise.then(null, null, callback1);
def.promise.then(null, null, callback2);

Then if I want to remove (unregister), for example, callback2 , what should I do?

Here is live example:

jsfiddle

Answer 1

A quick look at the source for $q shows us that:

then: function(onFulfilled, onRejected, progressBack) {
  var result = new Deferred();

  this.$$state.pending = this.$$state.pending || [];
  this.$$state.pending.push([result, onFulfilled, onRejected, progressBack]);
  if (this.$$state.status > 0) scheduleProcessQueue(this.$$state);

  return result.promise;
}

So there is no specific identifier that would point to your anonymous callback in the $$state.pending stack in order to splice it out.

I haven't personally tried this before but if you wanted to wipe the pending stack, maybe def.$$state.pending = []; would do the trick. Then you could simply reassign only the def.then() callbacks you want.

Answer 2

Promises are used to make a sequence of asynchronous processes. Once a step of this sequence is set, it's not possible (in an elegant way at least) to unset it.

Correct me if I'm wrong, but that's what you are trying to do here: you set a step with the callback2 and then you try to remove that step. Instead, I would recommend setting the callback2 only if a certain condition is fulfilled, for example:

var notifCallback;
if (true) {
    notifCallback = function(notif) {
        console.log('notify 1', notif);
    };
} else {
    notifCallback = function(notif) {
        console.log('notify 2', notif);
    };
}

def.promise.then(null, null, notifCallback);

Example on jsFidle .