Display ListPopupWindow auf der linken Seite des Bildschirms

Question

Ich könnte ListPopupWindow auf der linken Seite des Bildschirms mit dem folgenden Code anzeigen.

@Override 
public View onCreateActionView() { 
    LayoutInflater layoutInflater = LayoutInflater.from(mContext); 

    final View actionItem = layoutInflater.inflate(R.layout.list_table_view_action_provider, null); 

    final ImageButton button = (ImageButton) actionItem.findViewById(R.id.button); 
    button.setOnClickListener(new View.OnClickListener() { 
     @Override 
     public void onClick(View v) { 
      mListPopupWindow = new ListPopupWindow(mContext); 
      mListPopupWindow.setAnchorView(actionItem); 
      mListPopupWindow.setAdapter(mAdapter); 
      mListPopupWindow.setModal(true); 
      mListPopupWindow.setContentWidth(150); 

      // Display mListPopupWindow on most left of the screen 
      mListPopupWindow.setHorizontalOffset(-1000); 


      mListPopupWindow.setOnItemClickListener(ListTableViewActionProvider.this); 

      mListPopupWindow.show(); 
      mListPopupWindow.setOnDismissListener(ListTableViewActionProvider.this); 
     } 
    }); 

    return actionItem; 
} 

"mlistPopupWindow.setHorizontalOffset (-1000);" ist schade. Gibt es andere Lösungen? Hier

Answer 1

ist das ListPopupWindow Beispiel

 View menuItemView = getActivity().findViewById(R.id.menu_filter); 
     ListPopupWindow popup  = new ListPopupWindow(getActivity()); 

     popup.setHorizontalOffset(-200); 
     popup.setVerticalOffset(-100); 

     popup.setAnchorView(menuItemView); 
     popup.setWidth(400); 
     popup.setModal(true); 
     popup.setHeight(ListPopupWindow.WRAP_CONTENT); 

     ListAdapter adapter = new MyAdapter(getActivity()); 
     popup.setAdapter(adapter); 

     popup.show();