Can I initialize Firebase without using google-services.json?
<p>EDIT: I should emphasize, I have flavors, for which I don't want to use any of these Google services, and attempting to apply the google-services plugin in such a case, without having an applicable google-services.json, would result in a failed build.</p>

<p>I would like to be able to initialize the Firebase SDK, specifically to use the Remote Config, WITHOUT using the google-services.json.</p>

<p>I see that FirebaseApp has an initialize method, which receives a FirebaseOptions object. </p>

<p>I've built the FirebaseOptions with values provided in the google-services.json, and after a call to FirebaseApp.initialize with these options, I always get </p>

<pre><code>FirebaseInitProvider: FirebaseApp initialization unsuccessful
</code></pre>

<p>I know it's the recommended way to use the google-services.json file, but I need to be able to make the app call different Firebase projects, depending on debug/release builds, while keeping the package name the same.</p>

<p>The way I would like to do this is to have a debug/release pairs for all values necessary to initialize Firebase, and then dynamically do the initialization. </p>

<pre><code>        FirebaseOptions options = new FirebaseOptions.Builder()
                .setApplicationId(appId)
                .setApiKey(apiKey)
                .setGcmSenderId(appContext.getString(
                        isDebug ? R.string.firebase_testing_gcm_sender_id :
                                R.string.firebase_production_gcm_sender_id))
                .setDatabaseUrl(appContext.getString(
                        isDebug ? R.string.firebase_testing_database_url :
                                R.string.firebase_production_database_url))
                .setStorageBucket(
                        appContext.getString(
                                isDebug ? R.string.firebase_testing_storage_bucket :
                                        R.string.firebase_production_storage_bucket))
                .build();
        FirebaseApp.initializeApp(appContext, options);
</code></pre>

<p>So far it doesn't seem to work.</p>

<p>All help is greatly appreciated!</p>