How to get value instead of Promise from Firebase?

Question

I want to get a "value of Data" from Firebase Realtime Database. You can find my code below:

  async function getName() {
    const result = await firebase.database().ref('/users/' + user.user.uid)
    .once('value').
    then(async function(snapshot) {
      console.log("SNapshot =>");
      console.log(snapshot.val().displayName); // returns "John Doe"

      const result2 = await snapshot.val().displayName;

      return result2;
    });

    return result; // returns "Promise Object"
  };

It returns "Promise". But i want to get value not promise.

How should i fix my code?

Thanks.

`async` functions return a Promise by contract - it is the way they work. — Randy Casburn, Mar 02 '19 at 22:40
You can't get a value from the future. Return the promise, have the user wait for the value. — Bergi, Mar 02 '19 at 22:52
I only want to get a value of any data from my Firebase Realtime Database. It doesn't matter to use "async" function for me. How can i get value as `firebase.auth().currentUser.email`? — John Valdetine, Mar 02 '19 at 23:05

score 4 · Answer 1 · answered Mar 02 '19 at 22:50

4

You are mixing up use of async/await with then()/catch(). You should pick one syntax or the other, not both together in this case. You are also trying to use await on something that is not a promise. Just write your function like this:

async function getName() {
  const snapshot = await firebase.database().ref('/users/' + user.user.uid).once('value')
  return snapshot.val().displayName
}

And call it like this:

const name = await getName()

answered Mar 02 '19 at 22:50

Doug Stevenson

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2

It still returns "Promise". How can i get any value of data from Firebase with a function ? – John Valdetine Mar 02 '19 at 23:03
Async functions all return promises, no exceptions. That's why I'm showing how to call getName using await to unwrap the promise. – Doug Stevenson Mar 02 '19 at 23:05
How can i get value from Database as `firebase.auth().currentUser.email`? It returns `abc@gmail.com`. – John Valdetine Mar 02 '19 at 23:07

score 1 · Answer 2 · answered Mar 02 '19 at 22:37

1

The getname function is doing async operation so you can do:

try {
  await result = getName();
} catch (e) {
  throw e;
}

within an async function, or, if you prefer without async :

result.then((res) => console.log(res));

answered Mar 02 '19 at 22:37

Dimitri Kopriwa

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While this *might* work, the entire getName function is poorly written and has way too many lines of code for what's necessary. – Doug Stevenson Mar 02 '19 at 22:53
Thanks but it still returns "Promise". – John Valdetine Mar 02 '19 at 23:03

How to get value instead of Promise from Firebase?

2 Answers2