Question

41

async/await always returns promise

rated 0 times [ 47] [ 6] / answers: 1 / hits: 57365 / 7 Years ago, sat, april 15, 2017, 12:00:00

I'm trying async/await functionality. I have such code imitating a request:

const getJSON = async () => {

  const request = () => new Promise((resolve, reject) => (

    setTimeout(() => resolve({ foo: 'bar'}), 2000)

  ));



  const json = await request();

  return json;

}

When I use the code in this way

console.log(getJSON()); // returns Promise

it returns a Promise

but when I call this line of code

getJSON().then(json => console.log(json)); // prints { foo: 'bar' }

it prints json as expected

Is it possible to use just code like console.log(getJSON())? What don't I understand?

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everardo

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4 Years ago

answered 7 Years ago jarettajb · Accepted Answer

Every async function returns a Promise object. The await statement operates on a Promise, waiting until the Promise resolves or rejects.

So no, you can't do console.log on the result of an async function directly, even if you use await. Using await will make your function wait and then return a Promise which resolves immediately, but it won't unwrap the Promise for you. You still need to unwrap the Promise returned by the async function, either using await or using .then().

When you use .then() instead of console.logging directly, the .then() method makes the result of the Promise available to you. But you can't get the result of the Promise from outside the Promise. That's part of the model of working with Promises.