You cannot simply add promises to the array afterwards and expect it to affect the previous .all() call. This is not how .all() functions. One alternative approach is:
var promises = [promise1, promise2, promise3];
var allDone = $q.all(promises);
$q.all([allDone, promise4]).then(function() {
// All four promises are completed at this point
});
Alternatively, you can abandon the initial allDone promise and create a new .all() like so:
var promises = [promise1, promise2, promise3];
var allDone = $q.all(promises);
promises.push(promise4);
var all4Done = $q.all(promises);
Considering your feedback, here is another option. Say the final handler you want to call is in a function named allDoneNow(). You can try something like this:
var promises = [promise1, promise2, promise3];
var allDone = $q.all(promises);
allDone.then(function(results) {
// If no new promises have been added
if (promises.length === results.length) {
// Call the final handler
return allDoneNow(results);
} else {
// Wait for any new promises
return $q.all(promises).then(allDoneNow);
}
});
promises.push(promise4);
For a recursive approach to keep checking for more promises, you can use a function like this:
function waitAll(promiseArray) {
return $q.all(promiseArray).then(function(results) {
if (promiseArray.length <= results.length) {
// All promises in the array are completed
return results;
} else {
// There are new promises in the array, so wait for them
return waitAll(promiseArray);
}
});
}
var promises = [promise1, promise2, promise3];
waitAll(promises).then(function(results) {
// All promises in the array are done
});
promises.push(promise4);