Question

我如何拥有一个具有常数项目位置的listView？

假设您listview上searchbox actionbar position。现在您想要过滤结果，并希望即使在过滤后项目的listView号也是不变的。

当我在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); } }上搜索时，以下代码给出了不同的位置：

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Answer 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。

Answer 2

您可以将索引或位置作为数据模型的一部分。

我如何在listview上有一个具有常量项目位置的过滤项目？

2 个答案: