Prevent user from clicking twice on Facebook and Google sign in button

Question

I added Facebook and Google sign in in my app using:

private LoginButton facebook; //package com.facebook.login.widget;
private SignInButton googleSignIn; //package com.google.android.gms.common;

Both buttons work, I would like to prevent the users from clicking the buttons multiple times (every time it's being clicked another request is being sent):

Both cases are handled in the Google and Facebook API as follows:

1. In the Google sign in, there is a list of Google accounts of the device and once pressed I can not catch that event to disable the button once the user chose his account

2. In the Facebook sign in, it is handled by a callback registered in the Facebook package (which I don't access to) and I also can not disable the button once clicked.

Edit: Facebook code:

private void ActivateFacebook() {
        FacebookSdk.sdkInitialize(getApplicationContext());
        callbackManager = CallbackManager.Factory.create();
        facebook = (LoginButton) rootView.findViewById(R.id.fb);
        ((LoginButton) facebook).setReadPermissions("email");
        ((LoginButton) facebook).setFragment(this);

        ((LoginButton) facebook).registerCallback(callbackManager, new FacebookCallback<LoginResult>() {
            @Override
            public void onSuccess(LoginResult loginResult) {
                onFacebookLogin(loginResult);
            }

            @Override
            public void onCancel() {}

            @Override
            public void onError(FacebookException error) {
                Log.d(TAG, error.getMessage());
            }
        });
    }

What is the best practice in achieving this? Thanks!

Answer 1

If you're starting an intent for the result, you can wrap the call in a conditional block, such as:

boolean requestSent = false;

if(!requestSent) {
    Intent signInIntent = mGoogleSignInClient.getSignInIntent();     
    startActivityForResult(signInIntent, 101 );
    requestSent = true;
} else {
    //show a message
}