Slobodan,
You can use a counter for this:
function
ifYouNeedToGetThreeNumbersAsynchronouslyAndThenDivideTheSumOfTheFirstTwoNumbersByTheThirdNumberAndPassTheResultOrErrorToTheCallbackThenCallThisFunctionBecauseItDoesThat(cb)
{
var num1, num2, num3;
retrieveNumOne(function(er, n) {
num1 = n
then(er)
})
retrieveNumTwo(function(er, n) {
num2 = n
then(er)
})
retrieveNumThree(function(er, n) {
num3 = n
then(er)
})
var i = 0, errState = null
function then(er) {
if (errState) return;
if (er) return cb(errState = er)
if (++i === 3) cb(null, (num1 + num2) / num3);
}
}
Of course, yes, this is more lines of code, and a somewhat more
verbose function name. But it's very explicit, and there's no
question about what it's doing.
"Call these functions and get the result, or error" is a pretty common
pattern, so there are some libraries that abstract it away. Check out
async.
Personally, I prefer to not use control flow libraries like async.
There's nothing wrong with them, per se, but I find that leaving that
out requires me to have a lower tolerance for complexity in my
programs, which is on the net a good thing. I definitely think that
you should *write* a control flow library, even if you then throw it
away, because there are few ways to get as familiar with these
concepts than writing a library that implements them.
On Mon, Apr 22, 2013 at 12:54 PM, Slobodan Blazeski
<slobodan.blaze...@gmail.com> wrote:
> Hi All
>
> I'm looking for suggestions of how to retrieve values asynchronously:
>
> In the synchronous world I have
>
> var num1 = Object1.retrieveNum1();
> var num2 = Object2.retrieveNum2();
> var num3 = Object3.retrieveNum3();
>
> var result = (num1 + num2) / num3;
>
>
> but since functions Object1.retrieveNum1,Object2.retrieveNum2 &
> Object3.retrieveNum3
> retrieve data from the database or represent long calculation I need to pass
> callbacks,
> that leads me to below
>
> Object1.retrieveNum1Async(function(num1){
> Object2.retrieveNum2Async(function(num2){
> Object3.retrieveNum3Async(function(num3){
> var result = (num1 + num2) / num3;
> });
> });
> });
>
> is this the idiomatic way of doing this or there is something better
>
>
> thanks
> Bobi
>
>
>
>
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