How to make promises wait each other without using async/await pair?

Question

I am working on old project that doesn't supports ES6 standards.

For ease of understanding i will write the code i am trying to reach, Please translate the following code without using async/await pair.

async function doSomeCalls(arrayWithNeededCalls){
    for(let i = 0; i < arrayWithNeededCalls.length; i++){
      await makeSomeCall(arrayWithNeededCalls[i]);
      console.log("Call: ", i, " Completed going to next one");
    }
}

I tried to use

  Promise.all()

But it doesn't wait one promise to be completed before running second one.

Hi, could you add what you've tried so far ? This is not a coding service, try to solve your problem and if you are having issues, we can help you. — Nicolas, Oct 01 '19 at 17:56
arrayWithNeededCalls map to Promises and use `Promise.all` for parallel or reduce arrayWithNeededCalls to one promise — HMR, Oct 01 '19 at 17:56
Hello @Nicolas this code is pretty much everything i am trying to achieve but in older ways for legacy project. — Levan Lomia, Oct 01 '19 at 18:02
Aren't there tools out there that can translate newer features to older standards? Babel or something? — melpomene, Oct 01 '19 at 18:04

ChrisG · Accepted Answer · 2019-10-01T18:16:43.500

2

You want to wait for your promises one by one, in order, so you should use recursion:

const doSomeCalls = (arrayWithNeededCalls) => {
  const processCall = (index) => makeSomeCall(arrayWithNeededCalls[index]).then(res => {
    console.log(`Call ${index} Completed going to next one`);
    if (index < arrayWithNeededCalls.length) {
      return processCall(index+1);
    }
  });
  return processCall(0);
}

edited Oct 01 '19 at 18:16

answered Oct 01 '19 at 17:58

ChrisG

2,637
18
20

Will this solution wait first promise to complete before go to another one? – Levan Lomia Oct 01 '19 at 18:00
`Promise.all(...promiseArray)`; will execute them simultaneously, not one by one – Alex Vovchuk Oct 01 '19 at 18:00
This will run all promises concurrently. – ChrisG Oct 01 '19 at 18:00
@ChrisG exactly, therefore it is wrong solution. In the question one by one is required – Alex Vovchuk Oct 01 '19 at 18:03
@LevanLomia I updated my response to a recursive solution that runs all the promises one after the other. – ChrisG Oct 01 '19 at 18:08
Recursion is on option, you could also do this with reduce. – HMR Oct 01 '19 at 18:14
@HMR Yes, you have a nice solution as well. You should post it as an answer. – ChrisG Oct 01 '19 at 18:14
@LevanLomia Thanks for noticing. I have updated to fix syntax issues. – ChrisG Oct 01 '19 at 18:17

Alex Vovchuk · Answer 2 · 2019-10-01T18:56:18.610

0

UPDATE: HRM comment looks much better:

Since OP is ignoring the value(s) returned by the promises you can do:

arrayWithNeededCalls.reduce( (result, item) => result.then(() => makeSomeCall(item)), Promise.resolve() );

OLD: In case if you need to run them one by one:

makeSomeCall(arrayWithNeededCalls[0])
    .then(() => makeSomeCall(arrayWithNeededCalls[1]))
    .then(() => makeSomeCall(arrayWithNeededCalls[2]))
    .then(() => makeSomeCall(arrayWithNeededCalls[3]))

But if you don't know exact promises but use just array of them - await is the best solution

edited Oct 01 '19 at 18:56

answered Oct 01 '19 at 17:56

Alex Vovchuk

2,828
4
19
40

You should redue arrayWithNeededCalls to one promise if you want to not parallel process them, then he amount of items in arrayWithNeededCalls can be dynamic. – HMR Oct 01 '19 at 17:57
@HMR could you explain how to reduce it one by one? (not concurrently) – Alex Vovchuk Oct 01 '19 at 18:04
1

Since OP is ignoring the value(s) returned by the promises you can do: `arrayWithNeededCalls.reduce( (result, item) => result.then(() => makeSomeCall(item)), Promise.resolve() ); ` – HMR Oct 01 '19 at 18:07
@HMR yes, it looks nice – Alex Vovchuk Oct 01 '19 at 18:18
You can use it in your answer – HMR Oct 01 '19 at 18:19
@Bergi thanks, just missed – Alex Vovchuk Oct 01 '19 at 18:56

How to make promises wait each other without using async/await pair?

2 Answers2