Basic question.

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Kevin Burton

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Apr 20, 2014, 12:11:59 AM4/20/14
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I have about 20 standard node type methods ( function(args, callback)). Each of these functions require the previous to be completed before the next can correctly proceed. For sake of discussion I will name these functions a,b,c,d,e,f..... So because these functions do not block I have "callback hell", that looks like:

a(args, function(err) {
    if(err) {
    } else {
        b(args, function(err) {
            if(err) {
            } else {
                c(args, function(err) {
                    if(err) {
                    } else {
                        d(args, function(err) {
                            if(err) {
                            } else {
                                e(args, function(err) {
                                     if(err) {
                                     } else {
                                        f(args, function(err) {
                                             if(err) {
                                             } else {

Looking at the documentation I am not sure if I should use 'defer', 'denodeify', or 'ninvoke'. Like I said I am new to 'Q' and am just trying to glean what I can form the docs. Can it be as simple as:

a(args)
.then(b(args))
.then(c(args))
.then(d(args))
.then(e(args))
.then(f(args))
.error(err) {
    //handle error
}
.done();

Any experience with this simple scenario would be greatly appreciated and would jumpstart my understanding of Q and promises.

Thank you.

Stuart Knightley

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Apr 21, 2014, 2:55:37 PM4/21/14
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If you adapt a,b,c,d,e,f... to return promises, then it can indeed be as simple as .then().then()...

If you control a, b, c... you will use Q.defer inside them to create a new promise to return. If you don't control them (e.g. an existing callback based package) then you can use either denodify or ninvoke to create a promise returning function or invoke the function immediately and return a promise respectively.

You may be interested in dropping some of your code into http://stuk.github.io/promise-me/ to see what it looks like (note this just converts the *usage* from callback to promise, it doesn't make the changes to the functions themselves). 

Stuart

// ravi

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Apr 21, 2014, 2:55:49 PM4/21/14
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Hello Kevin,

note that when you do .then(b(args)) you are actually invoking b(args), not passing a reference to it. What might work in your case:

Q.nfcall(a, arg1, arg2, …)
.then(Q.denodify(b, arg1, arg2, ...))
.then(Q.denodify(c, arg1, arg2, …))
…
…
.fail
(
    function(err)
    {
        // handle error
    }
)
.done();

—ravi


Stuart Knightley

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Apr 21, 2014, 2:58:50 PM4/21/14
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On Saturday, 19 April 2014 21:11:59 UTC-7, Kevin Burton wrote:
a(args)
.then(b(args))
.then(c(args))
.then(d(args))
.then(e(args))
.then(f(args))
.error(err) {
    //handle error
}
.done();

Just rereading this, your code would look like this:

a(args)
.then(function (aResult) {
   return b(args);
})
.then(function (bResult) {
   return c(args);
})
// etc.
.catch(function (error) {
   // handle error
})
.done();

As you wrote it a, b, c... would all be called immediately, rather then when the previous promise was resolved.

// ravi

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Apr 21, 2014, 3:04:46 PM4/21/14
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On Apr 21, 2014, at 2:55 PM, // ravi <ravi-...@g8o.net> wrote:

note that when you do .then(b(args)) you are actually invoking b(args), not passing a reference to it. What might work in your case:

Q.nfcall(a, arg1, arg2, …)
.then(Q.denodify(b, arg1, arg2, ...))
.then(Q.denodify(c, arg1, arg2, …))
…


Note that in the above, if a() returns a value, that will be passed to b() as its last parameter/argument; similarly for b()’s return value passed to c(), etc.

—ravi


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