android java libgdx 如何让“纹理”跳起来并返回到起始位置

Question

So I am just trying to make my game character, which is a texture ( ball ), to jump up in the air and then return back down to the position that it started at when the screen is pressed. I was just wondering if someone could give me a code example or help me to do this with my current code which is below. I have basically just drawn the background and the ball texture and positioned the ball where I want it to start the jump. The ball texture is what I want to make jump straight up.

public class MyGdxGame extends ApplicationAdapter {
    SpriteBatch batch;
    Texture background;
    Texture ball;

    @Override
    public void create () {
        batch = new SpriteBatch();
        background = new Texture("gamebackground.png");

        ball = new Texture("ball2.png");
        ball.setFilter(Texture.TextureFilter.Nearest, Texture.TextureFilter.Nearest);
    }

    @Override
    public void render () {
        batch.begin();
        float scaleFactor = 2.0f;
        batch.draw(background, 0, 0, Gdx.graphics.getWidth(), Gdx.graphics.getHeight());
        batch.draw(ball, 80, 145, ball.getWidth() * scaleFactor, ball.getHeight() * scaleFactor);
        batch.end();
    }

    @Override
    public void dispose () {}
}

Answer 1

There are a million ways to do this.

Here's a simple (and not very flexible way). Create a Ball class that has variables for x and y position, velocity, and acceleration. Then give it an update method for applying the acceleration and velocity to the position:

public class Ball {

    public static final float GRAVITY = -100; // size depends on your world scale
    public static final float BOUNCE_DAMPENING = 0.6f;

    public final Vector2 position = new Vector2();
    public final Vector2 velocity = new Vector2();
    public final Vector2 acceleration = new Vector2(0, GRAVITY);

    public void update (){
        float dt = Gdx.graphics.getDeltaTime();
        velocity.add(acceleration.x * dt, acceleration.y * dt));
        position.add(velocity.x * dt, velocity.y * dt);

        if (position.y <= 0){ // hit ground, so bounce
            position.y = -position.y * BOUNCE_DAMPENING;
            velocity.y = -velocity.y * BOUNCE_DAMPENING;
        }
    }

}

This is a very rudimentary way of handling physics. It would be more sophisticated to use Box2D, but the above is fine if you're just learning.

Now, you need to create a ball instance and use it to track your ball position. Use the Ball object's position when drawing it. And you can react to taps to apply a velocity.

public class MyGdxGame extends ApplicationAdapter {
    SpriteBatch batch;
    Texture background;
    Texture ballTexture;
    Ball ball;

    @Override
    public void create () {
        batch = new SpriteBatch();
        background = new Texture("gamebackground.png");

        ballTexture = new Texture("ball2.png");
        ballTexture.setFilter(Texture.TextureFilter.Nearest, Texture.TextureFilter.Nearest);

        ball = new Ball();
    }

    @Override
    public void render () {
        Gdx.gl.glClear(GL20.GL_COLOR_BUFFER_BIT); // don't forget to clear screen

        if (Gdx.input.justTouched())
            ball.velocity.y += 100;
        ball.update();

        batch.begin();
        float scaleFactor = 2.0f;
        batch.draw(background, 0, 0, Gdx.graphics.getWidth(), Gdx.graphics.getHeight());
        batch.draw(ballTexture, ball.position.x, ball.position.y, ballTexture.getWidth() * scaleFactor, ballTexture.getHeight() * scaleFactor);
        batch.end();
    }

    @Override
    public void dispose () {
        batch.dispose();
        background.dispose();
        ballTexture.dispose();
    }
}

You also need to read up on pixel units vs. world units and how to solve the scale problem with Viewports. See https://xoppa.github.io/blog/pixels/ and https://github.com/libgdx/libgdx/wiki/Viewports