8

I am having a hard time running promises sequentially.

var getDelayedString = function(string) {
    var deferred = Q.defer();

    setTimeout(function() {
        document.write(string+" ");
        deferred.resolve();
    }, 500);

    return deferred.promise;
};

var onceUponATime = function() {
    var strings = ["Once", "upon", "a", "time"];

    var promiseFuncs = [];

    strings.forEach(function(str) {
        promiseFuncs.push(getDelayedString(str));
    });

    //return promiseFuncs.reduce(Q.when, Q());
    return promiseFuncs.reduce(function (soFar, f) {
        return soFar.then(f);
    }, Q());    
};

getDelayedString("Hello")
.then(function() {
    return getDelayedString("world!")
})
.then(function() {
    return onceUponATime();
})
.then(function() {
    return getDelayedString("there was a guy and then he fell.")
})
.then(function() {
    return getDelayedString("The End!")
})

onceUponATime() should sequentially output ["Once", "upon", "a", "time"] but instead they are being output immediately for some reason.

jsFiddle here: http://jsfiddle.net/6Du42/2/

Any idea what I am doing wrong?

Bergi
  • 630,263
  • 148
  • 957
  • 1,375
Nick
  • 11,483
  • 8
  • 41
  • 44

1 Answers1

15

but instead they are being output immediately for some reason.

You are calling them already here:

promiseFuncs.push(getDelayedString(str));
//                                ^^^^^

You would need to push function(){ return getDelayedString(str); }. Btw, instead of using pushing to an array in an each loop you rather should use map. And actually you don't really need that anyway but can reduce over the strings array directly:

function onceUponATime() {
    var strings = ["Once", "upon", "a", "time"];

    return strings.reduce(function (soFar, s) {
        return soFar.then(function() {
            return getDelayedString(s);
        });
    }, Q());    
}

Oh, and don't use document.write.

Community
  • 1
  • 1
Bergi
  • 630,263
  • 148
  • 957
  • 1,375