java - Libgdx 将一个 Actor 的 Action 传递给另一个 Actor

Question

好吧，基本上我正在尝试用其他 Action 装饰 Libgdx Actor 类

 public Shake(Button buttonToBeDecorated) extends ButtonDecorator {
      super(buttonToBeDecorated);
      Array<Action> actions = buttonToBeDecorated.getActions();

      for (Action action : actions)
            addAction(action);

      addAction(Actions.forever(new SequenceAction(
             Actions.moveBy(10, 0, 0.5f),
             Actions.moveBy(-10, 0, 0.5f)))
      );

 }

但是来自 toBeDecorated 类的 Action (也包含在 SequenceAction 中)不适用于 Shake 的实例。我确定操作已正确通过，因为我能够将它们打印出来。但我没有得到综合效果，也许你们中的一些人会知道为什么？谢谢

编辑:(基于@DHa 的新回复)

我相信我已经理解了您提出的这个 Group-Workaround。但是我仍然无法让它发挥作用。对于这个实例，假设我们用 Shake Action 装饰按钮对象，然后用 FadeOut Action 装饰按钮对象(这两个类都有从父类 ButtonDecorator 扩展的“Group”变量)。所以创建这种类型的对象看起来像这样:

Button button = new Decorators.FadeOut(new Decorators.Shake(new Buttons.PlayButton()));

和类:

//Shake class - we just simply add Shake actor to group and then add a specific action
//this works perfectly fine by itself - new Decorators.Shake(new Buttons.PlayButton())
public static class Shake extends ButtonDecorator {
    public Shake(Button buttonToBeDecorated) {
        super(buttonToBeDecorated);
        group.addActor(this);
        group.addAction(Actions.forever(new SequenceAction(
                Actions.moveBy(10, 0, 0.5f),
                Actions.moveBy(-10, 0, 0.5f))));

    }
}

//In FadeOut we are trying to decorate Shake object with another Action
public static class FadeOut extends ButtonDecorator {
    public FadeOut(Button buttonToBeDecorated) {
        super(buttonToBeDecorated);
        Array<Action> actions = buttonToBeDecorated.group.getActions(); //getting actions from Shake
        group.addActor(buttonToBeDecorated);
       /* I'm guessing that the whole workaround is in this line. We are adding
          Shake-actor to FadeOut group so Shake-actions should no longer apply
          to Shake-object and can be applied to our new FadeOut button */

        group.addActor(this); //Adding FadeOut to it's own group
        for (Action action : actions) 
            group.addAction(Actions.parallel(action,new SequenceAction(Actions.fadeOut(3), Actions.fadeIn(3)))) 
            //besides adding shake actions to FadeOut object we are also adding parallel fadeout action



    }
}

我不知道为什么，但仍然只有一个 Action (淡出)应用于创建的对象

Answer 1

使用 Group actor 组合多个 actor。应用于该组的操作将应用于该组中的所有参与者。

Group group = new Group();

group.addActor(actor1);
group.addActor(actor2);

group.addAction(...);

但是，一个 Actor 只能是一个组的一部分，因此您不能混合和匹配不同 Actor 之间的 Action 。

例如:

Group group1 = new Group();

group1.addActor(actor1);
group1.addActor(actor2);

Group group = new Group();

group2.addActor(actor2);
group2.addActor(actor3);

group1.addAction(...); // will only apply to actor1 since actor2 left group1 when joining group2

我想出一个比我以前的答案更好的答案，另一个答案在技术上仍然是正确的，所以我也会保留它。