How to call a JavaScript function sequentially in a for loop?

Question

I want to pass each item into a function that take times. But seems that the JS function is asynchronized. How can I call the function sequentially ? (Pass next item to function after the previous done)

function main() {
    for (var i = 0; i < n ; i++) {
        doSomething(myArray[i]);
    }
}

function doSomething(item) {
    // do something take time
}

My solution is call the function recusively. But I want to know is there a different way to solve this issue ? Thanks.

function main() {
    doSomething(myArray, 0);
}

function doSomething(item, i) {
    // do something take time
    doSomething(myArray, i + 1);
}

As far as I know, for-cycles and function calls in javascript are synchronous, so it must be something else that's making the function async. Maybe it's something inside the function itself. — Andros Rex
– Andros Rex, Commented Jul 7, 2016 at 2:19
Sorry for not mention that, my code is running on a node.js server. Is that why the function is async ? — Aaron
– Aaron, Commented Jul 7, 2016 at 2:22
Functions don't become async just because they're on node.js. What does your function actually do? Asynchronous functions usually have a callback facility or return a promise, so you can use that to coordinate the subsequent call. — nnnnnn
– nnnnnn, Commented Jul 7, 2016 at 2:26
Hi @nnnnnn, the function do something like http post request, and i have a callback function. My solution is put the doSomething(myArray, i + 1); inside the callback function. — Aaron
– Aaron, Commented Jul 7, 2016 at 2:31

gtzilla · Accepted Answer · 2020-01-17 22:58:34Z

6

In JavaScript, as of 2020, the for-loop is async/await aware. You can return a promise and then await that promise inside of the for loop. This causes the loop to execute in a series, even for long running operations.

function randomSleep(ms, seed=10) {
    ms = ms * Math.random() * seed;
    return new Promise((resolve, reject)=>setTimeout(resolve, ms));
} 
async function doSomething(idx) {
  // long running operations
  const outcome = await randomSleep(500);
  return idx;
}   
const arrItems = ['a','b','c','d'];
for(let i=0,len=arrItems.length;i<len;i++) {
  const result = await doSomething(i);
  console.log("result is", result)
}

Read more about async await in for loops and forEach loops here https://zellwk.com/blog/async-await-in-loops/

answered Jan 17, 2020 at 22:58

gtzilla

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Comments

chenkehxx · Accepted Answer · 2016-07-11 09:43:41Z

0

if you want to pass next item to function after the previous done, you can try to use promise, just like this

var Q = require('q'); 
var promise;
main();
function main() {
     promise = doSomethingPromise(0)
    for (var i = 1; i < 10 ; i++) {
        (function (i) {
            promise = promise.then(function (res) {
                return doSomethingPromise(res + ' ' + i)
            });
        })(i)

    }
}

function doSomethingPromise (item) {
   var d = Q.defer();
      d.resolve(doSomething(item));
   return d.promise;
}

function doSomething(item) {
    // do something take time
    console.log('doSomething', item);
    return item;
}

it can make you function to be called by order.

edited Jul 11, 2016 at 9:43

answered Jul 7, 2016 at 2:39

chenkehxx

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6 Comments

Aaron Over a year ago

Hi chenkehxx, I've tried your answer. But looks item[1] to item[item.length-1] are stuck in line Q.defer().

chenkehxx Over a year ago

can I see your all code?, I don't run my code neither. it is just pseudocode.

Aaron Over a year ago

HI chenkenhxx, my code is just like what you wrote here.

chenkehxx Over a year ago

@Aaron, I have update my code, you can try it again.

Aaron Over a year ago

Hi chenkehxx, I've tried your modified answer, but the log shows that only the first item is run in the doSomething function.

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