Question

我已经实现了拖放操作，但问题出在用户释放拖动之后。被拖走的演员从屏幕上消失了。如何将actor移动到新的位置，或者如果它放在不适当的位置，则返回到它的位置？

感谢您的回答。

此致

阿尔法 “

final DragAndDrop dragAndDrop = new DragAndDrop();
  final Box sourceBox =  boxList.get(0);
  final Box targetBox =  boxHolderList.get(0);

  Source source = new Source(sourceBox) {

     @Override
     public void dragStop(InputEvent event, float x, float y,
           int pointer, Payload payload, Target target) {

        Box sourceBox = (Box) payload.getDragActor();
        if (target == null){
           sourceBox.setPosition(sourceBox.getX(), sourceBox.getY());
        }
        else{
           Box targetBox = (Box) target.getActor();
           sourceBox.setPosition(targetBox.getX(), targetBox.getY());
        }
     }

     @Override
     public Payload dragStart(InputEvent event, float x, float y, int pointer) {

        dragAndDrop.setDragActorPosition(-1*sourceBox.getWidth()/2, 1*sourceBox.getHeight()/2);
        Payload payload = new Payload();
        payload.setDragActor(sourceBox);

        Label validLabel = new Label("Valid!", skin);
        validLabel.setColor(0, 1, 0, 1);
        payload.setValidDragActor(validLabel);

        Label invalidLabel = new Label("Invalid!", skin);
        invalidLabel.setColor(1, 0, 0, 1);
        payload.setInvalidDragActor(invalidLabel);

        return payload;
     }


  };

Answer 1

这种拖放方法很网状，你可以试试这个

 yourActor.addListener(new DragListener(){

        public void touchDragged(InputEvent event, float x, float y, int pointer) {
            yourActor.moveBy(x-yourActor.getWidth()/2,y-yourActor.getHeight()/2);
        }});

LibGDX：如何在拖动后继续显示actor（拖出）

1 个答案: