javascript - $.Deferred: How to detect when every promise has been executed

Question

I have a number of async tasks that need to be completed, so I'm using promises.

I need to detect when each one of the promises has been executed (both resolved and rejected). I must not continue execution until that point.

I was using something like this:

$.when(promise1, promise2, ...).always();

But this code is wrong, because the when method has lazy evaluation, and it returns as soon as one of the promises fails. So the always callback also runs as soon as one of the promises fail.

I was thinking in coding a workaround, but this use case is so common that maybe somebody has done it already, or maybe there's even a way of doing this using just jQuery (if not, it would be nice to add a Promise.whenNonLazy or a Promise.when(promise1, promise2, ..., false) in the future.

Is this possible?

Answer 1

More sophisticated promise libraries have an allSettled() function like Q or Promise.settle like Bluebird.

In jQuery, you could implement such a function yourself as well and extend the $ namespace with it, but that will only be necessary if you need it often and performance-optimized.

A simpler solution would be to create a new promise for each of the ones you are waiting for, and fulfilling them even when the underlying one is rejected. Then you can use $.when() on them without problems. In short:

// using Underscore's .invoke() method:
$.when.apply(null, _.invoke(promises, "then", null, $.when)).done(…)

More stable:

$.when.apply($, $.map(promises, function(p) {
    return p.then(null, function() {
        return $.Deferred().resolveWith(this, arguments);
    });
})).then(…);

You might change the then callbacks a bit to distinguish between fulfilled and rejected results in the final done.