How can I return a promise using the syntax of async/await?

Question

So I am writing a function to upload a file and I am using a class Upload to do so. This class has a method called start that basically uploads the file to where I need it and return an object with the info of the file uploaded.

What I have done is that:

upload = async (param, onFinish) => {
  const upload = new Upload(param)
  let response = await upload.start()

  response = doSomeWork(response)

  onFinish(response)
}

And I use it like that:

upload(param, (response) => {console.log(response)} )

As you can see I am little confused with async function in javascript, I can see how that looks weird and one of the reasons is because I read everywhere that an async function should always return a promise.

But my question is how in this case could I return a promise if I need to do some work in the response first? what is the best way of implementing this so my async function is solid?

I have searched on Stack overflow and did not find a clear answer for my question I don't know if because i did not understand the answers correctly or because there isn't actually one but I hope that is not a duplicated question.

An async function does always return a promise. If you don't return anything explicitly, it's a promise of undefined. If you return a promise, it's a promise of whatever that promise resolves to. Otherwise it's a promise of whatever you return. — jonrsharpe
– jonrsharpe, Commented Apr 24, 2020 at 15:10
Don't pass a callback, just return doSomeWork(response); then when you call it, do upload(param).then(response => { console.log(response); }); instead. — Patrick Roberts
– Patrick Roberts, Commented Apr 24, 2020 at 15:11

David · Accepted Answer · 2020-04-24 15:11:40Z

4

If the function is async then it's already returning a Promise. Just return your value and it will be passed to that Promise:

upload = async (param) => {
  const upload = new Upload(param);
  let response = await upload.start();

  response = doSomeWork(response);

  return response;
}

Then you can await that result:

let response = await upload(param);
console.log(response);

Or use .then() on the returned Promise:

upload(param).then(response => console.log(response));

answered Apr 24, 2020 at 15:11

David

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