gwt中的弹出面板

Question

当鼠标输入时，弹出式面板的工作方式将显示并在离开时将隐藏，当光标在弹出窗口内移动时，弹出式面板显示将保持结束。

Answer 1

我这样做但是它有问题弹出窗口没有停留....任何想法？

  Event.addNativePreviewHandler(new NativePreviewHandler() {
      public void onPreviewNativeEvent(final NativePreviewEvent event) {
        final int eventType = event.getTypeInt();
        switch (eventType) {

          case Event.ONMOUSEOUT:
          {
                        System.out.println("hello");
                        popup.setVisible(false);
           break;
          }
          case Event.ONMOUSEOVER:
          {
              l1.addMouseOverHandler(new MouseOverHandler()
              {
                  public void onMouseOver(MouseOverEvent event)
                  {
              popup.show();
              popup.setPopupPositionAndShow(new PopupPanel.PositionCallback() 
                {
                    public void setPosition(int offsetWidth, int offsetHeight) 
                        {
                            int left = (Window.getClientWidth() - offsetWidth) / 24;
                            int top = (Window.getClientHeight() - offsetHeight) / 5;
                            popup.setPopupPosition(left, top);
                        }
                });
              MouseOutHandler handler1=new MouseOutHandler()
                {
                    public void onMouseOut(MouseOutEvent event)
                    {
                        System.out.println("welcome2");
                        popup.setVisible(false);

                    }
                };
                popup.addHandler(handler1,MouseOutEvent.getType());
               popup.sinkEvents(Event.ONMOUSEOUT); 
               event.preventDefault();
                  }
              });
            break;
          }
          default: 

        }
      }
    });