Actionscript 3 按钮只工作一次

我有一个问题，我的按钮只工作一次。该按钮触发一个自定义事件。自定义事件激发，DocumentClass上的侦听器第一次拾取该事件，但此后任何时候都不响应

复制 ，然后单击“开始游戏”，然后单击右上角的菜单按钮，然后单击主菜单，然后单击“开始游戏”，然后单击右上角的菜单按钮，请注意，您不能再次单击主菜单。这是我的问题。它应该继续发挥作用

DocumentClass添加子openScreen。openScreen为DocumentClass触发事件以添加子playScreen并删除openScreen。单击playscreen上的菜单按钮时，playscreen会添加子菜单屏幕**单击主菜单按钮时，会触发DocumentClass侦听的事件。触发的功能将删除播放屏幕并添加打开的屏幕。当**步骤执行时，此系列事件仅在挂断前完全工作一次

文档类

public class DocumentClass extends MovieClip 
{
    public var playScreen:PlayScreen = new PlayScreen();
    public var openScreen:OpenScreen = new OpenScreen();
    public var started:Boolean = false;     

    public function DocumentClass() 
    {
        openScreen.addEventListener( GameEvent.STAR, OnPlayClick);
        playScreen.addEventListener( GameEvent.NG, NewGame);
        playScreen.menuScreen.addEventListener( GameEvent.NG, NewGame, true,  0, true);
        playScreen.menuScreen.addEventListener( GameEvent.MM, onMain);
        addChild(openScreen)


    }

    public function OnPlayClick(gameEvent:GameEvent):void{
        trace("start")

        addChildAt(playScreen,0)
        var tweenX:Tween = new Tween(openScreen, "x", None.easeIn, 0, -480, .5, true);
        tweenX.addEventListener(TweenEvent.MOTION_FINISH, onTweenDone);


    }
    public function NewGame(gameEvent:GameEvent):void{
        removeChild(playScreen)
        playScreen = new PlayScreen();
        addChild(playScreen)
        playScreen.begin()
    }
    public function onMain(gameEvent:GameEvent):void{
        playScreen.removeChild(playScreen.menuScreen)
        this.removeChild(playScreen)
        //openScreen = new OpenScreen();
        openScreen.x = 0
        addChild(openScreen)
        playScreen = new PlayScreen();

菜单屏幕 公共类MenuScreen扩展了MovieClip {

露天屏幕

 public class OpenScreen extends MovieClip 
{

    public var hs:String

    public function OpenScreen() 
    {
        startButton.buttonMode = true
        startButton.addEventListener( MouseEvent.CLICK, OnStartClick );

        exitButton.buttonMode = true
        exitButton.addEventListener( MouseEvent.CLICK, OnExitClick );


        //NativeApplication.nativeApplication.addEventListener(KeyboardEvent.KEY_DOWN, onSystemKey);
        readObject()
        highScoreText.text = String(hs)
        trace(highScoreText.text)
        startButton.AnswerText.text = "Start"
        //restartButton.AnswerText.text = "Restart"
        //mainButton.AnswerText.text = "Main Menu"
        exitButton.AnswerText.text = "Exit"
    }

    public function OnStartClick(myEvent:MouseEvent):void{
        trace(this.hasEventListener( GameEvent.STAR ))
        this.dispatchEvent( new GameEvent( GameEvent.STAR ))


    }
    public function OnExitClick(myEvent:MouseEvent):void{
        //NativeApplication.nativeApplication.exit();

    }


    protected function onSystemKey(e:KeyboardEvent):void
    {
        if(e.keyCode == Keyboard.BACK)
        {

            //NativeApplication.nativeApplication.exit();
        }
        else if(e.keyCode == Keyboard.HOME)
        {
            //handle the button press here. 
        }
        else if(e.keyCode == Keyboard.MENU)
        {

        }
    }

---

原因似乎是，在你的

onMain

方法中，你从舞台上移除，然后创建一个全新的播放屏幕实例。因此，你的侦听器连接到旧的实例。这也会造成内存泄漏

public function onMain(gameEvent:GameEvent):void{
        playScreen.removeChild(playScreen.menuScreen)
        this.removeChild(playScreen)
        //openScreen = new OpenScreen();
        openScreen.x = 0
        addChild(openScreen)
        playScreen = new PlayScreen(); //RIGHT HERE - this is creating a whole new play screen, your listeners are attached to the old one.
}

应该是这样的：（取出构造函数中的play screen代码，在声明时不要实例化它）

---

这是一个合理的问题，这样做可以解决我的按钮不工作的问题。新的游戏屏幕是我制作新游戏的方式，因为所有的游戏机制都包含在游戏屏幕中。有没有更好的方式将孩子重置为调用前的状态？可能是一种更有效的方式，但这取决于如果应用程序的性能足够，则清除并重新实例化没有什么错——只需确保没有任何东西将上一个播放屏幕保留在内存中（如事件侦听器）

public function onMain(gameEvent:GameEvent):void{
        playScreen.removeChild(playScreen.menuScreen)
        this.removeChild(playScreen)
        //openScreen = new OpenScreen();
        openScreen.x = 0
        addChild(openScreen)
        playScreen = new PlayScreen(); //RIGHT HERE - this is creating a whole new play screen, your listeners are attached to the old one.
}

//Create a function that kill your current play screen
public function clearPlayScreen():void {
    if(!playScreen) return; //don't do anything if there isn't a play screen
    playScreen.removeChild(playScreen.menuScreen)
    this.removeChild(playScreen);

    //remove your listeners so the old play screen can be garbage collected - other they will all stay in memory causing a memory leak
    playScreen.removeEventListener( GameEvent.NG, NewGame);
    playScreen.menuScreen.removeEventListener( GameEvent.NG, NewGame, true);
    playScreen.menuScreen.removeEventListener( GameEvent.MM, onMain);

    playScreen = null;
}

public function NewGame(gameEvent:GameEvent):void{
    clearPlayScreen(); //kill the old play screen if it exists

    playScreen = new PlayScreen();
    addChild(playScreen);

    //add the listeners to the new play screen
    playScreen.addEventListener( GameEvent.NG, NewGame);
    playScreen.menuScreen.addEventListener( GameEvent.NG, NewGame, true,  0, true);
    playScreen.menuScreen.addEventListener( GameEvent.MM, onMain);

    playScreen.begin();
}

public function onMain(gameEvent:GameEvent):void{
    clearPlayScreen();

    //openScreen = new OpenScreen();
    openScreen.x = 0;
    addChild(openScreen);
}