Flex设置stage.mouseChildren为false可防止使用忙光标

Question

我试图禁用鼠标单击并从UIComponent外部显示忙光标。即时做到这一点：

protected function setBusyCursor() : void { 

     const stage:Stage = mx.core.FlexGlobals.topLevelApplication.stage; 
     if (stage){ 
      stage.mouseChildren = false; 
     } 
     CursorManager.setBusyCursor(); 
    } 

这确实禁用鼠标单击，但光标出现是正常指针（不忙）。任何想法，我做错了什么？

Answer 1

最后我管理这个做到这一点：

protected function setBusyCursor() : void 
    { 
     var i : int = 0; 
     var uiComponent : UIComponent = FlexGlobals.topLevelApplication.parent.getChildAt(i); 
     while (uiComponent != null){ 
      uiComponent.mouseChildren = false; 
      uiComponent.cursorManager.setBusyCursor(); 
      i+=1; 
      if (FlexGlobals.topLevelApplication.parent.getChildAt(i) is UIComponent) { 
       uiComponent = FlexGlobals.topLevelApplication.parent.getChildAt(i); 
      } 
      else { 
       uiComponent = null; 
      } 
     } 
    } 

    protected function removeBusyCursor() : void { 
     var i : int = 0; 
     var uiComponent : UIComponent = FlexGlobals.topLevelApplication.parent.getChildAt(i);   
     while (uiComponent != null){ 
      uiComponent.cursorManager.removeBusyCursor(); 
      uiComponent.mouseChildren = true; 
      i+=1; 
      if (FlexGlobals.topLevelApplication.parent.getChildAt(i) is UIComponent) { 
       uiComponent = FlexGlobals.topLevelApplication.parent.getChildAt(i); 
      } 
      else { 
       uiComponent = null; 
      } 
     } 

    } 

禁用它在屏幕上的所有鼠标点击，并把繁忙的光标。

Answer 2

所有代码需要，我测试了4.5.1 flex框架，工作完美。用途：

protected function setBusyCursor() : void 
{  
    FlexGlobals.topLevelApplication.mouseChildren = false; 
    CursorManager.removeAllCursors(); 
    CursorManager.setBusyCursor(); 
}