returning promise to a variable without $$state

Question

I have the following promise

let oembedData = oembed(uuid)
    .then(res => res);

console.log(oembedData);

What I'm hoping to achieve is to return the res to oembedData so the value can be used.

I know I can do

let oembedData;
oembed(uuid)
    .then((res) => {
        oembedData = res;
        console.log(oembedData);
    });

but I feel this way isn't as clean as the former example.

When I use the former my console log returns Promise {$$state: {…}}

Your two examples aren't equivalent. `console.log` should just be inside the `.then` callback doing `console.log(res);` so you don't need `oembedData` at all. — loganfsmyth, Dec 23 '17 at 00:35
You don't do that. You do your thing inside the `.then(function(){/* here */}`. Promises are Asynchronous. — StackSlave, Dec 23 '17 at 00:39

Giacomo Cosimato · Accepted Answer · 2017-12-24T09:07:56.237

2

If you're free to use async/await, it looks almost like you'd want.

async function oembed(uuid) {
  return new Promise(resolve => setTimeout(() => {
    resolve('foo');
  }, 3000));
}

async function getOembed() {
  try {
      const oembedData = await oembed('1234');
      console.log(oembedData);
  } catch(err) {
      handleError(err);
  }
}

getOembed();

edited Dec 24 '17 at 09:07

answered Dec 23 '17 at 00:54

Giacomo Cosimato

635
5
9

I still need to use `getOembed().then()` here done i? – Dec 23 '17 at 14:20
No, you can wrap await oembed in a try/catch block instead of then...catch – Giacomo Cosimato Dec 23 '17 at 18:51
updated example – Giacomo Cosimato Dec 24 '17 at 09:08

returning promise to a variable without $$state

1 Answers1