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?
