Cannot get view from preferencefragment

Question

I have the following preference fragment xml:

<?xml version="1.0" encoding="utf-8"?>
<PreferenceScreen xmlns:android="http://schemas.android.com/apk/res/android" >

    <PreferenceScreen
        android:order="1"
        android:summary="Sound settings"
        android:title="Sound"
        android:widgetLayout="@layout/preference_switch_sound" >
        <intent
            android:action="android.intent.action.VIEW"
            android:targetClass="com.test.SoundPreferencesActivity"
            android:targetPackage="com.test" />
    </PreferenceScreen>

</PreferenceScreen>

And my sound switch widgetLayout that I point to above:

<?xml version="1.0" encoding="utf-8"?>
<Switch xmlns:android="http://schemas.android.com/apk/res/android"
    android:id="@+id/soundSwitchWidget"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:layout_gravity="center"
    android:padding="16dip"
    android:focusable="false"
    android:clickable="true" />

My problem is that I need to grab the switch in code such that I can attach the onClickListener.

I have tried finding the view in my Preference Activity:

public class PublicPreferencesActivity extends PreferenceActivity {

... 

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        getFragmentManager().beginTransaction().replace(android.R.id.content, preference).commit();
            Switch s = (Switch) findViewById(R.id.soundSwitchWidget);
                // I always get null here
    }

}

I have also tried in my fragment:

public static class PublicPreferenceFragment extends PreferenceFragmentSummary  {       
    @Override
    public void onCreate(final Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        addPreferencesFromResource(R.xml.publicpreferences);
                    Switch s = (Switch) findViewById(R.id.soundSwitchWidget);
                    // switch is null here too

    }
}

How do I get access programmatically to the switch I put in the preference screen?

EDIT: There was a question about why I am not using a SwitchPreference. Basically I want to be able to turn sound on/off with the switch but also be able to switch to a new PreferenceScreen if they click in the preference (not on the switch, but the title). I tried to use a SwitchPreference however when I click on the text it switches the switch and sends me to the the PreferenceScreen, rather than just send my to the PreferenceScreen.

I am trying to mimic the wifi behavior in main setting. I.E. you can turn on/off the switch from the top level, but also click the text to visit a more detailed screen.

Answer 1

Any particular reason you aren't using a SwitchPreference?

<SwitchPreference
    android:order="1"
    android:summary="Sound settings"
    android:title="Sound" >
   <intent
        android:action="android.intent.action.VIEW"
        android:targetClass="com.test.SoundPreferencesActivity"
        android:targetPackage="com.test" />
</SwitchPreference>

But as far as calling View.findViewById for your Switch, you'd need to subclass Preference, as you can't subclass PreferenceScreen, then override Preference.bindView and initialize it from there.

Answer 2

onCreate() function of Fragment should not create any View (this correct for all types of fragments).

Regardless of needs, to access views of PreferenceFragment it is necessary to override onCreateView(). See example of preferences ListView padding removal:

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                         Bundle savedInstanceState) {
    View view = super.onCreateView(inflater, container, savedInstanceState);

    if (null != view) {
        ListView lv = (ListView) view.findViewById(android.R.id.list);
        if (null != lv)
            lv.setPadding(0, 0, 0, 0);
    }

    return view;
}