没有包装的Promise.All中的链式Promise

Question

Is it possible for Promise.all to return the last value of the chain without a wrapper promise?

Without using await, it doesn't work in my context

Without wrapper example :

function sum1(x){
  return new Promise(resolve => {
    setTimeout(t => resolve(x+1),3000)
  })
}
const p1 = sum1(1);

p1
.then(sum1)
.then(sum1)

Promise.all([p1])
.then(v => console.log(v[0]));

It logs 2 instead of the expected 4.

But if I use a wrapper it works :

function sum1(x){
  return new Promise(resolve => {
    setTimeout(t => resolve(x+1),3000)
  })
}

function sum3(x){
  return sum1(x)
  .then(sum1)
  .then(sum1)
}
const p2 = sum3(1);

Promise.all([p2])
.then(v => console.log(v[0]));

But in my context it gets complicated if I need to create and name a wrapper function for every chain of promises...

Is this possible?

Answer 1

You can store the value returned by p1.then(sum1).then(sum1) and call Promise.all on this value. It waits for the resolution of the promise chain not only the first one. Here is an example:

function sum1(x) {
  return new Promise(resolve => {
    setTimeout(t => resolve(x + 1), 10);
  });
}

const p1 = sum1(1);
const p2 = p1.then(sum1).then(sum1);

Promise.all([p1]).then(v => console.log('P1', v[0]));
Promise.all([p2]).then(v => console.log('P2', v[0]));

Answer 2

Explanation: the problem with your code was that you store in const p1 = sum1(1); only first part of chain, and in Promise.all([p1]) you get result only from this first part (one solution is just to store all chain in p1 like this: p1=sum1(1).then(sum1).then(sum1) . However in your case, you don't need to use Promie.all at all (because in your example there is only one promise p1/2 ):

 function sum1(x){ return new Promise(resolve => { setTimeout(t => resolve(x+1),300) }) } // Promise.all([sum1(1).then(sum1).then(sum1)]).then(r => console.log(r)); // this works too sum1(1).then(sum1).then(sum1).then(r => console.log(r)); 

Answer 3

Actually all I had to do was call the chain in the variable declaration, so it references the last called promise

function sum1(x){
  return new Promise(resolve => {
    setTimeout(t => resolve(x+1),3000)
  })
}

//The change is here
const p1 = sum1(1)
.then(sum1)
.then(sum1)

Promise.all([p1])
.then(v => console.log(v[0]));

Answer 4

How about create a new function to do your task, it seems like promise.all does not fit to your case

const runInWaterfall = (promises) => new Promise((resolve, reject) => {
    const result = promises.reduce((acc, curr, index) => {
        if(!acc) {
            return curr();
        }
        return acc.then(curr);
    }, null);
    result.then(resolve).catch(reject);
})

and your task can be rewritten as following

runInWaterfall([() => Promise.resolve(1), sum1, sum1, sum1]).then(result => console.log(result))