如何在列表视图项单击上显示弹出菜单？

Question

我想在每次单击列表视图项目时显示弹出菜单 。 但是它不起作用。 帮帮我。 但吐司很好。

public class MainActivity extends AppCompatActivity {
    ListView list;
    @Override
    protected void onCreate(final Bundle savedInstanceState) {
    String[] mobileArray = {"Android", "IPhone", "WindowsMobile", "Blackberry",
    "WebOS", "Ubuntu", "Windows7", "Max OS X"};

    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    ArrayAdapter adapter = new ArrayAdapter<string>(this,
    R.layout.activity_listview, mobileArray);

    list = (ListView) findViewById(R.id.mobile_list);
    list.setAdapter(adapter);

    list.setOnItemClickListener(new AdapterView.OnItemClickListener() {
    public void onItemClick(AdapterView<?> parent, View v, int position, long id) {
    Toast.makeText(getApplicationContext(), "Hello Javatpoint", Toast.LENGTH_SHORT).show();
    PopupMenu popup = new PopupMenu(MainActivity.this, list);
    popup.getMenuInflater()
    .inflate(R.menu.popup_menu, popup.getMenu());
    }
    });
    }

    }

Answer 1

试试看：在您的listview onitemclicklistener内部调用此函数：

      PopUp(); 

    private void PopUp() {
    final Dialog dialog = new Dialog(Activity.this, android.R.style.Theme_Black_NoTitleBar_Fullscreen);
    dialog.requestWindowFeature(Window.FEATURE_NO_TITLE);
    dialog.setContentView(R.layout.yourlayotu);


    dialog.getWindow().setBackgroundDrawable(new ColorDrawable(Color.TRANSPARENT));
    dialog.show();
}

Answer 2

要显示弹出菜单，您需要使用弹出菜单的show()