I have made a simple punch in / punch out time clock application. I want to add the user the option of making a shortcut on the homescreen that will toggle the state of the app(time out / time in) but I don't want this shortcut to open up the app on the screen at all.

here is my setupShortcut()

private void setupShortcut() {
    Intent shortcutIntent = new Intent(this, Toggle.class);
   // shortcutIntent.setClassName(this, Toggle.class.getName());  
    shortcutIntent.putExtra(EXTRA_KEY, "ToggleShortcut");  

    shortcutIntent.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);  


    Intent intent = new Intent();
    intent.putExtra(Intent.EXTRA_SHORTCUT_INTENT, shortcutIntent); 
    intent.putExtra(Intent.EXTRA_SHORTCUT_NAME, "ToggleShortcut");
    Parcelable iconResource = Intent.ShortcutIconResource.fromContext(this,  R.drawable.app_sample_code);
    intent.putExtra(Intent.EXTRA_SHORTCUT_ICON_RESOURCE, iconResource);
    setResult(RESULT_OK, intent);

}

Toggle.class is an activity that toggles the state for me. In the manifest I have these settings on it

<activity android:name=".Toggle" android:exported="true" android:theme="@android:style/Theme.Translucent.NoTitleBar">

As it is now I can create a shortcut on the home screen then press it. The first time I press it it starts the Toggle activity and completes it fine, but it also opens up the TimeClock activity on the screen. If I then hit the back button I go back to the home. I can now press this shortcut and it will start the Toggle activity and not change the screen. Before I added: shortcutIntent.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP); It would open up the TimeClock activity every time. Now it is just the first time. But I want it to never show anything on the screen when the Toggle shortcut is pressed. Does anyone have any idea of how to get rid of that?