How to run concurrent promises in JS and then wait for them to finish without Promise.all()?

Question

So I need to do something like:

promiseFunction1().then((result) => {

}).catch((err) => {
    // handle err
});

promiseFunction2().then((result) => {

}).catch((err) => {
    // handle err
});

....

promiseFunctionN().then((result) => {

}).catch((err) => {
    // handle err
});

// WAIT FOR BOTH PROMISES TO FINISH
functionWhenAllPromisesFinished();

I cannot use Promise.all, as I DO NOT CARE IF ONE or ALL OF THEM FAIL. I need to be sure that ALL promises have finished. Also, the callback functions in then() are quite unique to each of the promiseFunctionX().

I am sure this is somewhat trivial, but I cannot figure it out. My initial idea was to keep a counter in the upper scope of running promises and incrementing it when one is run and decrementing it in finally(). Then I would need some async function checkIfRunningPromisesAre0() , but I am not sure how to implement this either, as it looked like recursion hell.

Here is my sample, but consider it just a material to laugh at poor implementation:

async function RunningPromisesFinished(){
    if(RunningPromises > 0){
        await sleep(2000);
        return await RunningPromisesFinished();
    }else{
        return true;
    }
}

on top of that Id have to implement async function sleep(N) and in few seconds the recursion level would be high, which im sure is not good for RAM.

Answer 1

Collect all the promises:

  const promise1 = promiseFunction1().then((result) => {
   }).catch((err) => {
     // handle err
   });

Then you can use Promise.all on them:

  await Promise.all([promise1, promise2, /*...*/ ]);

I cannot use Promise.all, as I DO NOT CARE IF ONE or ALL OF THEM FAIL

For sure you can. As you added a .catch to each promise that gets the promise chain back into the resolution branch, promise1 will never reject, therefore Promise.all will never reject too.