Question

好吧，我制作了自己的小菜单，但问题很小。当我使用'w'和's'键只有播放按钮，并选择高分按钮。但是当我使用箭头键时，它们会跳过高分数按钮，只是在播放和退出之间交替。

*我正在使用libgdx

public class levelSelection implements Screen {
           private SpriteBatch spriteBatch;
           private Texture playB;
           private Texture exitB;
           private Texture hScoreB;
           private Texture backGround;
           MyGdxGame game;
           private int selectedButton;
           BitmapFont  font;

    public levelSelection(MyGdxGame game) {
        this.game = game;
    }

    public void render(float delta) {
        Gdx.gl.glClearColor( 0f, 0f, 0f, 1f );
        Gdx.gl.glClear(GL10.GL_COLOR_BUFFER_BIT); 

        spriteBatch.begin();
        spriteBatch.draw(backGround, 0, 0);
        spriteBatch.draw(playB, 250, 450);
        spriteBatch.draw(exitB, 250, 350);
        spriteBatch.draw(hScoreB, 250, 400);
        spriteBatch.end();

        //Start Navigation Between menu buttons

            if( selectedButton == 2 && Gdx.input.isKeyPressed(Input.Keys.DOWN) || Gdx.input.isKeyPressed(Input.Keys.S)){
                selectedButton = 3;
                System.out.println("Quit button is selected");

            }
            if(selectedButton == 3 && Gdx.input.isKeyPressed(Input.Keys.UP)|| Gdx.input.isKeyPressed(Input.Keys.W)){
               selectedButton = 2;
               System.out.println("High Scores button is selected");

    }
            if(selectedButton == 2 && Gdx.input.isKeyPressed(Input.Keys.UP)|| Gdx.input.isKeyPressed(Input.Keys.W)){
                selectedButton = 1;
                System.out.println("Play button is selected");
            }
             if(selectedButton == 1 && Gdx.input.isKeyPressed(Input.Keys.DOWN)|| Gdx.input.isKeyPressed(Input.Keys.S)){
                 selectedButton = 2;
                 System.out.println("High Scores button is selected");
             }

             //end navigation between menu buttons

                 if(selectedButton == 1 && Gdx.input.isKeyPressed(Input.Keys.ENTER)){
                     game.setScreen(game.GameScreen);
                 }
                 if(selectedButton == 3 && Gdx.input.isKeyPressed(Input.Keys.ENTER)){
                     game.dispose();
                 }

                 //Draw text according to selected button
                if(selectedButton == 2){
                    spriteBatch.begin();
                    font.draw(spriteBatch, "High Score Button is Selected!", 15, 15);
                    spriteBatch.end();
                }
                if(selectedButton == 3){
                    spriteBatch.begin();
                    font.draw(spriteBatch, "Quit Button is Selected!", 15, 15);
                    spriteBatch.end();
                }
                if(selectedButton == 1){
                    spriteBatch.begin();
                    font.draw(spriteBatch, "Play Button is Selected!", 15, 15);
                    spriteBatch.end();
                }
            }
}

Answer 1

true || false = true
false || true = true

让我们将它与selectedButton == 3结合起来（假设是假）我们没有按下第一个keydown而是第二个keydown。然后我们确实遇到这种情况false && false || true - ＆gt; false || true = true。这不应该发生。它没有被选中，但关键ifstatemen说true所以它不正确。（System.out.println(false && false || true);）

这是正确的解决方案。你总是需要检查两个conidions并用or语句连接它们：

        if((selectedButton == 2 && Gdx.input.isKeyPressed(Input.Keys.DOWN)) || (selectedButton == 2 && Gdx.input.isKeyPressed(Input.Keys.S))){
            selectedButton = 3;
            System.out.println("Quit button is selected");

        }

这将创建以下布尔状态false || false && false ||true - ＆gt; false && true - ＆gt; false。

只要提一个更甜蜜的解决方案，你也可以使用XOR而不是2＆amp;＆amp;用||连接。

    if(selectedButton == 2 && Gdx.input.isKeyPressed(Input.Keys.DOWN) ^ Gdx.input.isKeyPressed(Input.Keys.S)){
        selectedButton = 3;
        System.out.println("Quit button is selected");

    }

或支撑钥匙或钥匙。

如果这不完全正确，我很抱歉。我不关心java的布尔顺序。

菜单无法正常工作

1 个答案: