CODEWITHSUNDEEP
javascriptfirebaseasynchronousgoogle-cloud-firestore
Notorious776
Notorious776

Reputation: 493

How do I return a value in a promise? (Firebase.get())

I've been struggling for the past four hours on this. I am writing a function to see if a property named "data" exists in my Firebase storage. If it does, I want to do one thing, if it doesn't I want to do something else. However, I cannot figure out how this asynchronous stuff works for the life of me. I simplified my code below. Basically I just want to wait for the data to be fetched before I reach the if/else. I've been playing around with different options but keep getting errors or some other issue. The code below is the closest I've gotten to working where the code doesn't crash but even if "data" does not exist in the Firestore, I'm always going through the else clause and I don't know why. Can someone help me figure out what I am doing wrong?

const fetchDataFromDB = async (user) => {
  let query = db
    .collection("users")
    .doc(user)
    .get()
    .then((doc) => {
      doc.data();
      console.log(doc.data().data);
    });

  return await query;
};

export const getSchedule = (miles, user) => {
  const firebaseData = fetchDataFromDB(user);
  console.log(firebaseData);

  // WAIT FOR FETCH BEFORE CONTINUING
  if (!firebaseData) {
    console.log("NOT getting data from FB");
    return;
  } else {
    console.log("getting data from FB");
    return;
  }
};

Upvotes: 0

Views: 752

Answers (1)

see sharper
see sharper

Reputation: 12055

Change up the code as follows:

const fetchDataFromDB = (user) => {
  return db
    .collection("users")
    .doc(user)
    .get()
    .then((doc) => {
      const data = doc.data();
      console.log(data);
      return data;
    });
};

export const getSchedule = async (miles, user) => {
  const firebaseData = await fetchDataFromDB(user);
  console.log(firebaseData);

  if (!firebaseData) {
    console.log("NOT getting data from FB");
    return;
  } else {
    console.log("getting data from FB");
    return;
  }
};

The point to remember about async await is that it doesn't really make asynchronous calls synchronous, it just makes them look that way so that your code is a bit less unwieldy with wrapped promises and the like. Every async function returns a promise, so if you want to deal with what it returns, you need to either deal with the promise directly (using .then...), or by using another await. In the latter case, you of course need to declare the consuming function as async as well.

With regards to the first function, there's no need for the async await there. Just return the promise. (Thanks @charlieftl for pointing out the problem in that code)

Upvotes: 2

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