Persist Alertdialog Options

I'm creating an AlertDialog. Here's the snippet of code that I'm using: final CharSequence[] items = {'Red', 'Green', 'Blue'}; AlertDialog.Builder builder = new AlertDialog.Builde

Solution 1:

Use the code below :

privatestaticfinalStringSELECTED_ITEM="SelectedItem";
private SharedPreferences sharedPreference;
private Editor sharedPrefEditor;

privatevoidshowAlert() {
    final CharSequence[] items = {"Red", "Green", "Blue"};

    AlertDialog.Builderbuilder=newAlertDialog.Builder(this);
    builder.setTitle("Pick a color");               
    builder.setSingleChoiceItems(items, getSelectedItem(), newDialogInterface.OnClickListener() {
        publicvoidonClick(DialogInterface dialog, int item) {
            saveSelectedItem(item);
            Toast.makeText(getApplicationContext(), items[item], Toast.LENGTH_SHORT).show();
        }       
    });
    AlertDialogalert= builder.create();
    alert.show();

}

privateintgetSelectedItem() {
    if (sharedPreference == null) {
        sharedPreference = PreferenceManager
                .getDefaultSharedPreferences(this);
    }
    return sharedPreference.getInt(SELECTED_ITEM, -1);
}

privatevoidsaveSelectedItem(int item) {
    if (sharedPreference == null) {
        sharedPreference = PreferenceManager
                .getDefaultSharedPreferences(this);
    }
    sharedPrefEditor = sharedPreference.edit();
    sharedPrefEditor.putInt(SELECTED_ITEM, item);
    sharedPrefEditor.commit();
}

For the first time, nothing will be selected. From second time onwards the item which is previously selected will be selected by default.