actionscript-3 - AS3 开/关菜单不起作用

Question

我想做一个菜单，当它不存在时，按Esc打开它，当它存在时，按Esc关闭它。但这不起作用，它显示此错误：

1176：具有静态类型的值与flash.text:TextField可能不相关的类型之间的比较String。

这是我的代码

stage.addEventListener(KeyboardEvent.KEY_DOWN, down);
function down(keyEvent:KeyboardEvent):void
{
    var keyPressed:String = "";
    keyPressed = keyEvent.keyCode.toString();
    if (keyPressed == "27")
    {
        if (now == "0")
        {
            menu._x = 100;
            now.text = "1";
        }
        else if (now == "1")
        {
            menu._x = -400;
            now.text = "0";
        }
    }
}

Answer 1

如果now是 aTextField你需要比较它的text属性

stage.addEventListener(KeyboardEvent.KEY_DOWN, down);
function down(keyEvent:KeyboardEvent):void
{
    if (keyEvent.keyCode == Keyboard.ESCAPE)
    {
        if (now.text == "0")
        {
            menu._x = 100;
            now.text = "1";
        }
        else if (now.text == "1")
        {
            menu._x = -400;
            now.text = "0";
        }
    }
}

您还可以设置visible为false/true隐藏显示菜单而不是将其移出舞台。我还更改了 keyCode 以使用Keyboard该类。