Android: Passing AppWidgetId to a Service

Question

Hi there

I got a Widget on my HomeScreen on which i add an click to a Button. I pass the widget id from the Widget to a Service, but when im reading the WidgetId at the Service it's always 3.

Widget:

@Override
public void onUpdate(Context context, AppWidgetManager appWidgetManager,
int[] appWidgetIds) {

    for (int appWidgetId : appWidgetIds) {
        RemoteViews remoteView = new RemoteViews(context.getPackageName(), R.layout.widget);

        Intent intent = new Intent(context, WidgetService.class);


        // prints the correct appWidgetId
        System.out.println(appWidgetId);


        intent.putExtra(AppWidgetManager.EXTRA_APPWIDGET_ID, appWidgetId);
        PendingIntent pendingIntent = PendingIntent.getService(context, 0, intent, 0);

        remoteView.setOnClickPendingIntent(R.id.Button01, pendingIntent);

        appWidgetManager.updateAppWidget(appWidgetId, remoteView);
    }

    super.onUpdate(context, appWidgetManager, appWidgetIds);
}

The Service looks like this:

public int onStartCommand(Intent intent, int flags, int startId){
    if (intent != null && intent.getExtras() != null) {

        int widgetId = intent.getExtras().getInt(AppWidgetManager.EXTRA_APPWIDGET_ID);

        // 3 is always printend
        System.out.println(widgetId);

        // More stuff here

    }
    return super.onStartCommand(intent, flags, startId);
}

So what am I doing wrong?

Answer 1

What you're trying to do will not work, AFAIK, without some trickery.

There can be only one outstanding PendingIntent per Intent. So, if you create two Intents that are equivalent (match on non-extra data), and you try to create two PendingIntents, you really wind up with a single PendingIntent in the end. Which Intent is used depends a bit on the flags you supply to getService() (or getActivity() or getBroadcast()). In your case, the first Intent will always be used.

If you only had one outstanding PendingIntent but just wanted to update its extras, FLAG_UPDATE_CURRENT would work. But, you want N outstanding PendingIntents.

The only way I know of to make that work in your case is to put a unique action string in the Intent. Since you are specifying the component (WidgetService.class), the action string will not be used for routing. However, it will suffice to give you multiple PendingIntent objects.