How to check if a Firebase App is already initialized on Android

Question

With the following, the first time it's called it works, but then fails on subsequent calls with "FirebaseApp name [DEFAULT] already exists!"

public FirebaseDatabase conn(Context c) {
         FirebaseOptions options = new FirebaseOptions.Builder()
                .setApiKey("key")
                .setDatabaseUrl("url")
                .setApplicationId("ID")
                .build();


        /////I tried Try and Catch with no success//////
        FirebaseApp app = FirebaseApp.initializeApp(c, options);

        /// for this : FirebaseApp app = FirebaseApp.initializeApp(c, options, "some_app");
        //// will fail with "FirebaseApp name some_app already exists!"
        return FirebaseDatabase.getInstance(app);
}

All of the above is an attempt to connect to a second Firebase App.

Solution 1

On firebase web, you check if already initialized with:

if (firebase.apps.length === 0) {
    firebase.initializeApp({});
}

Solution 2

In v9, Firebase has been modularized for better tree shaking. So we can no longer import entire app and check the apps property AFAIK. The below approach can be used instead.

import { initializeApp, getApps, getApp } from "firebase/app";
getApps().length === 0 ? initializeApp(firebaseConfig) : getApp();

https://firebase.google.com/docs/reference/js/v9/app.md#getapps for documentation

Solution 3

For those wondering how to do the same as the accepted answer, in Android:

if (FirebaseApp.getApps(context).isEmpty()) {
    FirebaseApp.initializeApp(context);
}

and in an instrumented test environment, use this context:

InstrumentationRegistry.getContext()

Solution 4

Firebase Version 9

import { initializeApp, getApp } from "firebase/app";

const createFirebaseApp = (config = {}) => {
  try {
    return getApp();
  } catch () {
    return initializeApp(config);
  }
};

const firebaseApp = createFirebaseApp({/* your config */})

Solution 5

You can try to get the Firebase app instance, in it's code firebase checks if it's initialized, if not it throws an IllegalStateException.

    try{
        FirebaseApp.getInstance();
    }
    catch (IllegalStateException e)
    {
        //Firebase not initialized automatically, do it manually
        FirebaseApp.initializeApp(this);
    }

View more solutions

Comments

[Vlad]

Cool but in android this impossible

[Anand Rockzz]

doesn't work anymore - github.com/zeit/next.js/issues/1999#issuecomment-302244429

[eoinmullan]

I had been using the App returned from initializeApp() but with the above approach this is no longer possible, so I'm using now firebase.app('[DEFAULT]'), but I'm not sure if that's recommended.

[Pumuckelo]

@eoinmullan this worked for me githubmemory.com/repo/firebase/firebase-js-sdk/issues/5037

[cjmling]

and in case if you want to get instance in else you can get by firebase.app();

[Ayush Kumar]

Can you update it to web 9?

[Levi Wilkerson]

Thank you. One of the few explanations that make sense for being able to export the firebaseApp variable.

[user3399180]

I tried this but I get this error - stackoverflow.com/questions/72028182/…