Combine async function result in separate function in more efficient way

Question

I have two async functions, that separately work on their own - they each return data if I call them singularly, but I can't get a result from the second when I try and combine them. Basically I'm pinging an API and if the first function returns a result, use that data. If it returns empty, run the second function and use that data. I'm getting the data through a node proxy (again these work fine separately).

Function 1 that checks for a live event:

async function getLive(){
  const response = await fetch("video-live");
  const getLiveData = await response.json();
  return getLiveData;
}

Function 2 that should run if Function 1 returns empty:

async function getArchived() {
  const response = await fetch("video-archived");
  const getArchivedData = await response.json();
  return getArchivedData;
}

The final function that applies the logic:

function showVideo(getLiveData, getArchivedData) {
  if( (getLiveData) == "" ) {
    console.log('live'+getLiveData);
  } else {
    console.log('no live'+getArchivedData);
  }
}

Then I call them like this:

getLive().then(showVideo);

The above returns 'no live' but not any data from the getArchivedData. How do I combine these in an elegant, efficient way? Can I combine both getLive and getArchived into one function with .then()?

Per @MichaelM's code:

async function getLive(){
  const response = await fetch("video-live");
  const getLiveData = await response.json();
  return getLiveData;
}

async function getArchived() {
  const response = await fetch("video-archived");
  const getArchivedData = await response.json();
  return getArchivedData;
}
async function showVideo() {
  const liveData = await getLive();
  if(liveData == "") {
    console.log('live' + getLiveData);
  } else {
    const archivedData = await getArchived();
    console.log('no live' + archivedData);
  }
}

"Uncaught (in promise) ReferenceError: getLiveData is not defined"

Are the functions getLive() and getArchivedData only used in showVideo? — Michael M.
– Michael M., Commented Dec 22, 2022 at 23:29
Yes, that is my intent. To run the two functions, and use the returned data in showVideo(). Basically if there's anything returned in getLiveData, use that. If nothing is returned, use the returned data from getArchivedData — Dirty Bird Design
– Dirty Bird Design, Commented Dec 22, 2022 at 23:31

Michael M. · Accepted Answer · 2022-12-22 23:51:44Z

2

Try rewriting your showVideo() function to use getLive() and getArchived() directly, instead of passing the results into showVideo(). Like this:

async function showVideo() {
  const liveData = await getLive();
  if(liveData == "") {
    console.log('live' + liveData);
  } else {
    const archivedData = await getArchived();
    console.log('no live' + archivedData);
  }
}

Then you just call showVideo() without the .then() statement.

edited Dec 22, 2022 at 23:51

answered Dec 22, 2022 at 23:33

Michael M.

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3 Comments

Dirty Bird Design Over a year ago

I updated my question with your code, formatting in comments is non existent. Still getting undefined error

Michael M. Over a year ago

@DirtyBirdDesign Oops, I had a typo. It should be fixed now.

Dirty Bird Design Over a year ago

Thanks Michael - took a little tweaking but you got me on the right path. Here is working code async function showVideo() { const liveData = await getLive(); const archivedData = await getArchived(); if(liveData == "") { console.log('no live data' + JSON.stringify(archivedData)); } else { console.log('live' + JSON.stringify(liveData)); } }

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