我想每隔x秒改變菜單的背景圖片。我正在使用libGDX scene2D.ui來製作菜單。 TestScreen類擴展了AbstractScreen,它是一個實現libGDX Screen類的抽象類。 問題:在通過堆棧中的Table對象將圖像加載到舞臺後,將圖像引用更改爲不同的圖像不起作用。 Stage.draw()不關心它,就好像它複製了我的原始圖像。我想保持背景爲Image類並通過stage.draw()渲染。libgdx背景圖片變化
爲了進一步複雜化,如果我在render()方法中將圖像更改爲另一個圖像,則image.setVisible(false)也會停止工作。
public class TestScreen extends AbstractScreen {
private Stage stage;
private Image background;
private boolean ChangeBackground = true;
private final float refreshTime = 2.0f; // refresh to new image every 2 seconds.
private float counter = refreshTime;
public TestScreen(Game game) {
super(game);
}
@Override
public void render(float deltaTime) {
Gdx.gl.glClearColor(0.0f, 0.0f, 0.0f, 1.0f);
Gdx.gl.glClear(GL10.GL_COLOR_BUFFER_BIT);
if(ChangeBackground){
counter -= deltaTime;
if(counter < 0){
counter = refreshTime;
// Assets class has the 12 images loaded as "Image" objects already.
// I simple want to change the reference to other (already loaded in memory images) ...
// and make stage render the new image.
background = Assets.instance.wallpapers[(int) (Math.random()*12)]; // The image should change.
//background.setVisible(false);
}
}
stage.act(deltaTime);
stage.draw();
}
@Override
public void resize(int width, int height) {
stage.setViewport(Constants.VIEWPORT_GUI_WIDTH, Constants.VIEWPORT_GUI_HEIGHT, false);
}
@Override
public void show() {
stage = new Stage();
Gdx.input.setInputProcessor(stage);
makeStage();
}
@Override
public void hide() {
stage.dispose();
}
@Override
public void pause() {
}
// Builds the Background later and adds it to a stage through a stack.
// This is how it's done in my game. I made this test bench to demonstrate.
private void makeStage() {
Table BackGroundLayer = new Table();
background = Assets.instance.wallpapers[(int) (Math.random()*12)];
BackGroundLayer.add(background);
Stack layers = new Stack();
layers.setSize(800, 480);
layers.add(BackGroundLayer);
stage.clear();
stage.addActor(layers);
}
}
嘗試調用Table.invalidate()。 – Viacheslav
試過這個,它沒有工作。我在BackGroundLayer中創建了一個類對象,並在render方法中更改了Image之後調用了.invalidate()方法。下面的解決方案很好地工作:) – Artash