Question 210578 in Javascript Question posted in
A very good W3school tutorial can be found here.

Javascript – Run multiple n promises, when done, then another n promises

In JavaScript, in the array of promises (can have n promises), run first n promises, when finished, run another n;

Example:

const promises = [Promise_1, Promise_2, Promise_3, Promise_4, Promise_5, Promise_6];

Run Promise_1 and Promise_2, when both finish, run Promise_3, Promise_4, then Promise_5, Promise_6.

axios.post(‘/api’)
axios.post(‘/api’)
WAIT
axios.post(‘/api’)
axios.post(‘/api’)
WAIT
axios.post(‘/api’)
axios.post(‘/api’)

Tried with Promise.all, for of loop, async/await…

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2

Answers


  1. Chosen as BEST ANSWER
    0 
    const promises = [];
const apiPromises = [promise_1, promise_2, promise_3, ....etc]; 
let counter = 1;

const resolvePromisesSeq = async () => {
  for (const prom of apiPromises) {
    if (counter % 3 !== 0) {
      promises.push(axios.post('/fakeApi/' + prom));
    } else {
      await axios.post('/fakeApi/' + prom);
    }
    counter++;
  }
  return Promise.all(promises);
};
resolvePromisesSeq().then(whatEverYouWant)

  2. 0 
    const array = [wait, wait, wait, wait, wait, wait]

run(array)

async function run(array, batch = 2) {
  let i = 0
  while (i < array.length) {
    await Promise.all(array.slice(i, i + batch).map(fn => fn()))
    console.log(`finished [${i} - ${i + batch}]`)
    i += batch
  }
}

function wait() {
  return new Promise(r => setTimeout(r, 1000))
}
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