0

我怎样才能有一个listView恒定的项目位置？

假设您有 alistview和searchboxon actionbar。现在您要过滤结果，并且position即使在过滤后也希望项目的数量保持不变。

下面的代码在我搜索时给了我不同的位置listView：

public class About extends ActionBarActivity {
      private ListView mainListView ;  
      private ArrayAdapter<String> listAdapter ;
      SearchView searchview;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.about);
        getSupportActionBar().setDisplayHomeAsUpEnabled(true);
        getSupportActionBar().setHomeButtonEnabled(true);
        getSupportActionBar().setDisplayShowTitleEnabled(false);

        //searchview = new SearchView(getSupportActionBar().getThemedContext());

        mainListView = (ListView) findViewById( R.id.mainListView );
        String[] mTitles = getResources().getStringArray(R.array.lstview_Content);
        String[] values= new String[5];
        for(int i=0;i<values.length;i++){
            values[i] = mTitles[i];
        }
        ArrayList<String> planetList = new ArrayList<String>();
        planetList.addAll( Arrays.asList(values) );
        listAdapter = new ArrayAdapter<String>(About.this, R.layout.lstview_layout,R.id.rowTextView, planetList);
         mainListView.setAdapter( listAdapter );
         mainListView.setOnItemClickListener(new OnItemClickListener() {
             @Override
            public void onItemClick(AdapterView<?> arg0, View arg1, int position,long arg3) {
                int itemposition = position;
                String itemvalue = (String) mainListView.getItemAtPosition(position);
                Toast.makeText(getApplicationContext(),
                        "Position :"+itemposition+"  ListItem : " +itemvalue , Toast.LENGTH_LONG).show();
            }
        });
         mainListView.setTextFilterEnabled(true);
    }
    @Override
    public boolean onOptionsItemSelected(MenuItem item) {
        switch (item.getItemId()) {
        case android.R.id.home:
            finish();
            break;
        default:
            break;
        }
        return super.onOptionsItemSelected(item);
    }



    @Override
    public boolean onCreateOptionsMenu (Menu menu) {
        MenuInflater inflater = getMenuInflater();
        inflater.inflate(R.menu.about, menu);
        MenuItem searcehItem = menu.findItem(R.id.action_search);
        searchview = (SearchView) MenuItemCompat.getActionView(searcehItem);
        searchview.setQueryHint("ُُSearch");
        searchview.setOnQueryTextListener(new OnQueryTextListener() {
            @Override
            public boolean onQueryTextSubmit(String arg0) {

                return false;
            }
            @Override
            public boolean onQueryTextChange(String arg0) {
                mainListView.setFilterText(arg0);
                return false;
            }
        });
        return super.onCreateOptionsMenu(menu);
    }
}
4

2 回答 2

1

在ActionBarActivity写：

int ID = 0;

在onItemClick写：

for (int i = 0; i < values.length; i++) {
    if(values[i] == itemvalue)
    {
            ID = i;
        }
}
Toast.makeText(getApplicationContext(),"Position :"+ID+"  ListItem : " +itemvalue , Toast.LENGTH_LONG).show();

祝你好运。

于 2014-06-16T06:06:11.313 回答
0

您可以将索引或位置作为数据模型的一部分。

于 2014-06-14T12:59:04.110 回答