我有一个简单的活动,它包含一个tabbar来切换两个片段.这两个片段都是listFragment并实现了searchview,以便在listfragment中进行搜索.搜索视图始终显示在标签栏上方的操作栏中.
我遇到的问题是,切换标签后,搜索视图的输入不会重置(转到其他片段).因此,第二个片段从searchview读取输入并相应地过滤listfragment,它实际上读取了我在片段1中时输入的输入.
我想要的是searchview是两个片段的单独搜索视图.有没有办法实现这个目标？
这是我的代码：
活动
public class ActivityMainApp extends Activity implements ActionBar.TabListener {
private FragmentManager fragmentManager = getFragmentManager();
@Override
public void onCreate(Bundle icicle) {
super.onCreate(icicle);
setContentView(R.layout.mainapp);
ActionBar actionBar = getActionBar();
actionBar.setNavigationMode(ActionBar.NAVIGATION_MODE_TABS);
// Add tabs
ActionBar.Tab relatieTab = actionBar.newTab().setText("Relaties");
ActionBar.Tab takenTab = actionBar.newTab().setText("Taken");
//ActionBar.Tab urenTab = actionBar.newTab().setText("Uren");
// Listeners
relatieTab.setTabListener(this);
takenTab.setTabListener(this);
// Tabs toevoegen aan actionbar
actionBar.addTab(relatieTab);
actionBar.addTab(takenTab);
// Create fragmentmanager to switch fragments
FragmentTransaction fragmentTransaction = this.fragmentManager.beginTransaction();
Fragment fragment = fragmentManager.findFragmentById(R.id.fragment_content);
if(fragment == null){
FragmentTransaction ft = fragmentManager.beginTransaction();
ft.add(R.id.fragment_content,new FragRelaties());
}
}
public void onTabSelected(Tab tab,FragmentTransaction ft) {
FragmentTransaction fragmentTransaction = this.fragmentManager.beginTransaction();
if(tab.getText().equals("Taken")){
FragTaken fragment = new FragTaken();
fragmentTransaction.replace(R.id.fragment_content,fragment);
fragmentTransaction.commit();
}
if(tab.getText().equals("Relaties")){
FragRelaties fragment = new FragRelaties();
fragmentTransaction.replace(R.id.fragment_content,fragment);
fragmentTransaction.commit();
}
}
public void onTabUnselected(Tab tab,FragmentTransaction ft) {
}
public void onTabReselected(Tab tab,FragmentTransaction ft) {
}
}
片段一
public class FragRelaties extends ListFragment implements SearchView.OnQueryTextListener {
private LayoutInflater inflater;
private ModelRelatie modelRelatie;
private AdapterRelatie relatieAdapter;
public ArrayList<Relatie> relaties;
public View onCreateView(final LayoutInflater inflater,ViewGroup container,Bundle savedInstanceState) {
this.inflater = inflater;
Activity activity = getActivity();
final Context context = activity.getApplicationContext();
setHasOptionsMenu(true);
new AsyncTask<Void,Void,ArrayList<Relatie>>(){
protected ArrayList<Relatie> doInBackground(Void... params) {
// Get all my contacts
modelRelatie = ModelRelatie.instantiate(context);
ArrayList<Relatie> relaties = modelRelatie.getRelaties();
return relaties;
}
protected void onPostExecute(ArrayList<Relatie> relaties) {
// Initial input of objects in the list
relatieAdapter = new AdapterRelatie(inflater.getContext(),R.layout.relatieslijstitem,relaties);
setListAdapter(relatieAdapter);
}
}.execute();
View view = inflater.inflate(R.layout.relaties,container,false);
return view;
}
@Override
public void onCreateOptionsMenu(final Menu menu,final MenuInflater inflater) {
inflater.inflate(R.menu.relatie_menu,menu);
// Get the searchview
MenuItem zoekveld = menu.findItem(R.id.zoekveld_fragrelatie);
SearchView zoekview = (SearchView) zoekveld.getActionView();
// Execute this when searching
zoekview.setOnQueryTextListener(this);
super.onCreateOptionsMenu(menu,inflater);
}
public void onListItemClick(ListView l,View v,int position,long id) {
// Things that happen when i click on an item in the list
}
public boolean onQueryTextSubmit(String query) {
return true;
}
public boolean onQueryTextChange(String zoekterm) {
Activity activity = getActivity();
Context context = activity.getApplicationContext();
// We start searching for the name entered
ModelRelatie modelRelatie = ModelRelatie.instantiate(context);
ArrayList<Relatie> relaties = modelRelatie.zoekRelatie(zoekterm);
// The returned objects are placed in the list
this.relatieAdapter = new AdapterRelatie(inflater.getContext(),relaties);
setListAdapter(relatieAdapter);
return true;
}
片段二
public class FragTaken extends ListFragment implements SearchView.OnQueryTextListener {
private LayoutInflater inflater;
private AdapterTaak adapterTaak;
private ModelTaken modelTaken;
public View onCreateView(LayoutInflater inflater,Bundle savedInstanceState) {
this.inflater = inflater;
Activity activity = getActivity();
final Context context = activity.getApplicationContext();
setHasOptionsMenu(true);
this.modelTaken = ModelTaken.instantiate(context);
ArrayList<Taak> taken = modelTaken.getTaken();
// Initial input for the list
adapterTaak = new AdapterTaak(inflater.getContext(),R.layout.takenlijstitem,taken);
setListAdapter(adapterTaak);
View view = inflater.inflate(R.layout.taken,false);
return view;
}
@Override
public void onCreateOptionsMenu(final Menu menu,final MenuInflater inflater) {
inflater.inflate(R.menu.taken_menu,menu);
// get the searview
MenuItem zoekveld = menu.findItem(R.id.zoekveld_fragtaken);
SearchView zoekview = (SearchView) zoekveld.getActionView();
// Execute this when searching
zoekview.setOnQueryTextListener(this);
super.onCreateOptionsMenu(menu,inflater);
}
public boolean onQueryTextSubmit(String query) {
return true;
}
public boolean onQueryTextChange(String zoekterm) {
Activity activity = getActivity();
Context context = activity.getApplicationContext();
// Search the task by the inputed value
ModelTaken modelTaken = ModelTaken.instantiate(context);
ArrayList<Taak> taken = modelTaken.zoekTaak(zoekterm);
// Return the found items to the list
this.adapterTaak = new AdapterTaak(inflater.getContext(),taken);
setListAdapter(adapterTaak);
return true;
}
}
除搜索部分外,两个片段几乎相同.
解决方法
已经有一段时间了,但如果有人发现同样的问题,
这是我如何设法做到这一点.
这是我如何设法做到这一点.
只需在每个片段中实现onPrepareOptionsMenu()即可在里面,打电话onQueryTextChange( “”);传递“”作为查询字符串.
