10

我刚刚开始使用 java 和 libgdx 并拥有此代码，非常简单地将精灵打印到屏幕上。这非常有效，我从中学到了很多。

package com.MarioGame;

import com.badlogic.gdx.ApplicationListener;
import com.badlogic.gdx.Gdx;
import com.badlogic.gdx.files.FileHandle;
import com.badlogic.gdx.graphics.GL10;
import com.badlogic.gdx.graphics.Texture;
import com.badlogic.gdx.graphics.g2d.Sprite;
import com.badlogic.gdx.graphics.g2d.SpriteBatch;
import com.badlogic.gdx.graphics.g2d.TextureRegion;
import com.badlogic.gdx.InputProcessor;

public class Game implements ApplicationListener {
private SpriteBatch batch;
private Texture marioTexture;
private Sprite mario;
private int marioX;
private int marioY;

@Override
public void create() {
    batch = new SpriteBatch();
    FileHandle marioFileHandle = Gdx.files.internal("mario.png"); 
    marioTexture = new Texture(marioFileHandle);
    mario = new Sprite(marioTexture, 0, 158, 32, 64);
    marioX = 0;
    marioY = 0;
}

@Override
public void render() {
    Gdx.gl.glClear(GL10.GL_COLOR_BUFFER_BIT);
    batch.begin();
    batch.draw(mario, marioX, marioY);
    batch.end();
}


@Override
public void resume() {
}

@Override
public void resize(int width, int height) {
}

@Override
public void pause() {
}

@Override
public void dispose() {
}

}

marioX当用户按下D键盘时，我将如何修改该值？

4

4 回答 4

32

对于您手头的任务，您甚至可能不需要实现 InputProcessor。您可以像这样在您的 render() 方法中使用 Input.isKeyPressed() 方法。

float marioSpeed = 10.0f; // 10 pixels per second.
float marioX;
float marioY;

public void render() {
   if(Gdx.input.isKeyPressed(Keys.DPAD_LEFT)) 
      marioX -= Gdx.graphics.getDeltaTime() * marioSpeed;
   if(Gdx.input.isKeyPressed(Keys.DPAD_RIGHT)) 
      marioX += Gdx.graphics.getDeltaTime() * marioSpeed;
   if(Gdx.input.isKeyPressed(Keys.DPAD_UP)) 
      marioY += Gdx.graphics.getDeltaTime() * marioSpeed;
   if(Gdx.input.isKeyPressed(Keys.DPAD_DOWN)) 
      marioY -= Gdx.graphics.getDeltaTime() * marioSpeed;

   Gdx.gl.glClear(GL10.GL_COLOR_BUFFER_BIT);
   batch.begin();
   batch.draw(mario, (int)marioX, (int)marioY);
   batch.end();
}

另请注意，我制作了马里奥浮动的位置坐标并将运动更改为基于时间的运动。marioSpeed 是 mario 每秒应沿任意方向行进的像素数。Gdx.graphics.getDeltaTime() 以秒为单位返回自上次调用 render() 以来经过的时间。在大多数情况下，实际上不需要强制转换为 int。

顺便说一句，我们在http://www.badlogicgames.com/forum有论坛，您也可以在其中询问 libgdx 特定问题！

hth，马里奥

于 2011-04-29T14:36:32.027 回答
2

在您的游戏循环方法（可能是渲染或制作）中需要添加一个输入处理程序（实现InputProcessor）。该类InputProcessor具有以下方法：

public boolean keyDown (int keycode);

/**
 * Called when a key was released
 * 
 * @param keycode one of the constants in {@link Input.Keys}
 * @return whether the input was processed
 */
public boolean keyUp (int keycode);

/**
 * Called when a key was typed
 * 
 * @param character The character
 * @return whether the input was processed
 */
public boolean keyTyped (char character);

并且该类Input.Keys有很多静态变量作为键码。例如：

        public static final int D = 32;
        public static final int A = 29;
        public static final int S = 47;
        public static final int W = 51;

因此，实现 key* 方法并检查是否按下了某个键并从 mario 递增或递减 X/Y。

编辑：检查此链接： http: //code.google.com/p/libgdx/source/browse/trunk/gdx/src/com/badlogic/gdx/InputProcessor.java http://code.google.com/p /libgdx/source/browse/trunk/gdx/src/com/badlogic/gdx/Input.java

或者，在主干中，演示如何处理输入： http ://code.google.com/p/libgdx/source/browse/#svn%2Ftrunk%2Fdemos

希望帮助

于 2011-04-29T14:21:39.370 回答
2

您可以使用该界面KeyListener来检测键盘动作。

public class Game implements ApplicationListener, KeyListener {

@Override
public void create() {
    //Important
    this.addKeyListener(this);

    // TODO Auto-generated method stub
    batch = new SpriteBatch();

    FileHandle marioFileHandle = Gdx.files.internal("mario.png"); 
    marioTexture = new Texture(marioFileHandle);
    mario = new Sprite(marioTexture, 0, 158, 32, 64);

    marioX = 0;
    marioY = 0;


}

    @Override
    public void keyPressed(KeyEvent e) {
        if (e.getKeyCode() == 68) { //it's the 'D' key
            //Move your mario
        }
    }

}
于 2011-04-29T14:23:35.473 回答
1

我的首选方法是使用 InputController 存储所有标准键以供检查。

import com.badlogic.gdx.Input.Keys;
import com.badlogic.gdx.InputProcessor;
import com.badlogic.gdx.math.Vector2;

public class KeyboardController  implements InputProcessor {
    public boolean left,right,up,down;
    public boolean isMouse1Down, isMouse2Down,isMouse3Down;
    public boolean isDragged;
    public Vector2 mouseLocation = new Vector2(0,0);

    @Override
    public boolean keyDown(int keycode) {
        boolean keyProcessed = false;
        switch (keycode) // switch code base on the variable keycode
        {
            case Keys.LEFT:     // if keycode is the same as Keys.LEFT a.k.a 21
                left = true;    // do this
                keyProcessed = true;    // we have reacted to a keypress 
                break;
            case Keys.RIGHT:    // if keycode is the same as Keys.LEFT a.k.a 22
                right = true;   // do this
                keyProcessed = true;    // we have reacted to a keypress 
                break;
            case Keys.UP:       // if keycode is the same as Keys.LEFT a.k.a 19
                up = true;      // do this
                keyProcessed = true;    // we have reacted to a keypress 
                break;
            case Keys.DOWN:     // if keycode is the same as Keys.LEFT a.k.a 20
                down = true;    // do this
                keyProcessed = true;    // we have reacted to a keypress
        }
        return keyProcessed;    //  return our peyProcessed flag
    }
    @Override
    public boolean keyUp(int keycode) {
        boolean keyProcessed = false;
        switch (keycode) // switch code base on the variable keycode
        {
            case Keys.LEFT:     // if keycode is the same as Keys.LEFT a.k.a 21
                left = false;   // do this
                keyProcessed = true;    // we have reacted to a keypress 
                break;
            case Keys.RIGHT:    // if keycode is the same as Keys.LEFT a.k.a 22
                right = false;  // do this
                keyProcessed = true;    // we have reacted to a keypress 
                break;
            case Keys.UP:       // if keycode is the same as Keys.LEFT a.k.a 19
                up = false;     // do this
                keyProcessed = true;    // we have reacted to a keypress 
                break;
            case Keys.DOWN:     // if keycode is the same as Keys.LEFT a.k.a 20
                down = false;   // do this
                keyProcessed = true;    // we have reacted to a keypress
        }
        return keyProcessed;    //  return our peyProcessed flag
    }
    @Override
    public boolean keyTyped(char character) {
        return false;
    }
    @Override
    public boolean touchDown(int screenX, int screenY, int pointer, int button) {
        if(button == 0){
            isMouse1Down = true;
        }else if(button == 1){
            isMouse2Down = true;
        }else if(button == 2){
            isMouse3Down = true;
        }
        mouseLocation.x = screenX;
        mouseLocation.y = screenY;
        return false;
    }

    @Override
    public boolean touchUp(int screenX, int screenY, int pointer, int button) {
        isDragged = false;
        //System.out.println(button);
        if(button == 0){
            isMouse1Down = false;
        }else if(button == 1){
            isMouse2Down = false;
        }else if(button == 2){
            isMouse3Down = false;
        }
        mouseLocation.x = screenX;
        mouseLocation.y = screenY;
        return false;
    }

    @Override
    public boolean touchDragged(int screenX, int screenY, int pointer) {
        isDragged = true;
        mouseLocation.x = screenX;
        mouseLocation.y = screenY;
        return false;
    }

    @Override
    public boolean mouseMoved(int screenX, int screenY) {
        mouseLocation.x = screenX;
        mouseLocation.y = screenY;
        return false;
    }
    @Override
    public boolean scrolled(int amount) {
        return false;
    }
}

然后我需要做的就是在游戏的 create 方法中使用 KeyboardController

controller = new KeyboardController();

然后告诉 GDX 用它来监听事件

Gdx.input.setInputProcessor(controller);

最后，如果我想检查是否按下了某个键，我可以去

if(controller.left){
    player.x -= 1;
}
于 2017-07-19T15:06:13.580 回答