node.js async series function's arguments

Question

I need to do the code like following:

function taskFirst(k, v) {
    console.log(k, v);
}

function taskSecond(k, v) {
    console.log(k, v);
}

function run() {
    var g1 = "Something";
    var g2 = "Something";
    var g3 = "Something";
    var g4 = "Something";

    async.series(
        [
            taskFirst(g1, g2),
            taskSecond(g3, g4)
        ],
        function(error, result){

        }
    );
}

What is the right way to pass custom variables and async.js callback function?

score 20 · Accepted Answer · edited Dec 12 '14 at 10:17

You could do something like this:

function taskFirst(k, v, callback) {
    console.log(k, v);

    // Do some async operation
    if (error) {
        callback(error);
    } else {
        callback(null, result);
    }
}

function taskSecond(k, v, callback) {
    console.log(k, v);

    // Do some async operation
    if (error) {
        callback(error);
    } else {
        callback(null, result);
    }
}

function run() {
    var g1 = "Something";
    var g2 = "Something";
    var g3 = "Something";
    var g4 = "Something";

        async.series(
            [
                // Here we need to call next so that async can execute the next function.
                // if an error (first parameter is not null) is passed to next, it will directly go to the final callback
                function (next) {
                    taskFirst(g1, g2, next);
                },
                // runs this only if taskFirst finished without an error
                function (next) {
                    taskSecond(g3, g4, next);    
                }
            ],
            function(error, result){

            }
        );
}

score 2 · Answer 2 · answered Jan 22 '14 at 16:15

2

It can be as follows

function taskFirst(k, v) {
    console.log(k, v);
}

function taskSecond(k, v) {
    console.log(k, v);
}

async.series([
    function(callback) { 
        callback(null, taskFirst(g1, g2));
    },
    function(callback) { 
        callback(null, taskFirst(g3, g4));
    }
],function(error, result){

});

answered Jan 22 '14 at 16:15

Fizer Khan

71,869
26
133
149

This won't work. The callback is executed immediately so async will move on to the next function. You should pass the callback to your function and all execute it one your asynchronous work is complete. – Douglas Ferguson Feb 15 '15 at 23:49
Yes - this answer assumes the code inside taskFirst is blocking. – 64_ Feb 26 '15 at 18:17

score 2 · Answer 3 · answered Jan 20 '15 at 02:02

2

This answer to async's github issue has worked for me perfectly. https://github.com/caolan/async/issues/241#issuecomment-14013467

for you it would be something like:

var taskFirst = function (k, v) {
   return function(callback){
      console.log(k, v);
      callback();
   }
};

answered Jan 20 '15 at 02:02

Juan Solano

1,075
1
17
29

Thanks, I was missing that I needed to wrap the function and return it :) – nyarasha Jan 07 '16 at 17:11
I needed to create a dynamic array of functions with parameters that I needed to give to async.series and your answer helped me in the right direction, thanks – Daniel van Niekerk Jun 10 '16 at 17:03

score 1 · Answer 4 · answered Apr 06 '17 at 12:21

Better way.

const a1 = (a, callback) => {
    console.log(a, 'a1')
    callback()
}
const a2 = (a, callback) => {
    console.log(a, 'a2')
    callback()
}

const run = () => {
    async.series([
        a1.bind(null, 'asd'),
        a2.bind(null, 'asd2')
    ], () => {
        console.log('finish')
    })
}
run()

node.js async series function's arguments

4 Answers4

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