Creating a promise chain in a for loop

Question

I would expect the code below to print one number on the console, then wait a second and then print another number. Instead, it prints all 10 numbers immediately and then wa

Answer 1

You have to assign the return value of .then back to chain:

chain = chain.then(()=>getProm(i))
         .then(Wait)

Now you will basically be doing

chain
  .then(()=>getProm(1))
  .then(Wait)
  .then(()=>getProm(2))
  .then(Wait)
  .then(()=>getProm(3))
  .then(Wait)
  // ...

instead of

chain
  .then(()=>getProm(1))
  .then(Wait)

chain
  .then(()=>getProm(2))
  .then(Wait)

chain
  .then(()=>getProm(3))
  .then(Wait)
// ...

You can see that the first one is actually a chain, while the second one is parallel.

Answer 2

Now that we have await/async, a better way to do this is:

function getProm(v) {
    return new Promise(resolve => {
        console.log(v);
        resolve();
    })
}

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

async function createChain() {
    let a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
    for (let i of a) {
        await getProm(i);
        await Wait();
    }
}


createChain();