XNA - Mouse.Left Button 被多次执行

Question

我是编程的菜鸟，但在我们开始之前，请回答我为什么更新执行不止一次，并像我是个傻瓜一样解释一下。

无论如何，所以我试图让这段代码只运行一次，因为到目前为止它执行了不止一次。

 protected override void Update(GameTime gameTime)
    {
        // Allows the game to exit
        if (GamePad.GetState(PlayerIndex.One).Buttons.Back == ButtonState.Pressed)
            this.Exit();

        // TODO: Add your update logic here

         button = Mouse.GetState();




         if (button.X < buttonPosition.X || button.Y < buttonPosition.Y || button.X > buttonPosition.X + font1.MeasureString(buttonText).X ||
             button.Y > buttonPosition.Y + font1.MeasureString(buttonText).Y)
             buttonColour = new Color(0, 0, 0);//if the mouse if not hovering over the font it stays that color
         else
             buttonColour = new Color(0, 255, 255);//changes to this color if it is hovering over text

        if(button.LeftButton==ButtonState.Pressed)
         display = (display == false) ? true : false;  //if display = true it will set it to false
        //if false then it will set it to false



 }

如果需要，这就是 Draw 方法。

 protected override void Draw(GameTime gameTime)
    {
        GraphicsDevice.Clear(Color.CornflowerBlue);
        spriteBatch.Begin();
        spriteBatch.DrawString(font1, buttonText, buttonPosition, buttonColour);   //this is the button leftbutton has to click to trigger the below if statement.
        if (display)
            spriteBatch.DrawString(font1, text, position, Color.White);
        spriteBatch.End();  //it will draw this when leftbutton  clicks the above button

        // TODO: Add your drawing code here

        base.Draw(gameTime);
    }

Answer 1

您之所以会出现这种行为，是因为UpdateXNA 每秒自动多次调用该方法，大约每秒 60 次。与 相同Draw，它可能会渲染大约 60 FPS。

因此，如果您按下按钮一秒钟并且该方法被调用 60 次，ButtonState.Pressed则将评估true60 次。

要解决此问题，您需要保留按钮的历史记录。这可以通过在每次更新时存储状态来实现：

//Define this at class level
private MouseState lastMouseState = new MouseState();

protected override void Update(GameTime gameTime)
{
    // your other stuff

    MouseState currentState = Mouse.GetState(); //Get the state
    if (currentState.LeftButton == ButtonState.Pressed &&
        lastMouseState.LeftButton == ButtonState.Released) //Will be true only if the user is currently clicking, but wasn't on the previous call.
    {
        display = !display; //Toggle the state between true and false.
    }

    lastMouseState = currentState;
}

所以现在点击只会被注册一次（因为鼠标当前需要点击，但之前必须处于释放状态）。如果有一天你包含它，你也需要保留键盘的历史记录。

我还改变了你的显示切换逻辑，这样更清晰。您的评论中也有错误（应该是 //if false then it will set it to true）。这是正常的，评论经常撒谎；总是尽量少评论，只有代码才是真理。

如果这对您有帮助，这里是我之前使用的辅助类：

public class InputState : GameComponent
{
    private KeyboardState currentKeyboardState;
    private KeyboardState lastKeyboardState;
    private MouseState lastMouseState;
    private MouseState currentMouseState;

    public InputState(Game game) : base(game)
    {
        game.Components.Add(this);

        currentKeyboardState = new KeyboardState();
        lastKeyboardState = new KeyboardState();
        currentMouseState = new MouseState();
        lastMouseState = new MouseState();
    }

    public override void Update(GameTime gameTime)
    {
        lastKeyboardState = currentKeyboardState;
        currentKeyboardState = Keyboard.GetState();
        lastMouseState = currentMouseState;
        currentMouseState = Mouse.GetState();

        base.Update(gameTime);
    }

    public bool IsNewLeftClick()
    {
        return currentMouseState.LeftButton == ButtonState.Pressed &&
            lastMouseState.LeftButton == ButtonState.Released;
    }

    public bool IsNewRightClick()
    {
        return currentMouseState.RightButton == ButtonState.Pressed &&
            lastMouseState.RightButton == ButtonState.Released;
    }

    public Point GetMousePosition()
    {
        return new Point(currentMouseState.X, currentMouseState.Y);    
    }

    public bool IsNewKeyPress(params Keys[] keys)
    {
        return keys.Any(k => (currentKeyboardState.IsKeyDown(k) &&
                    lastKeyboardState.IsKeyUp(k)));
    }

    public bool IsCurrentlyPressed(params Keys[] keys)
    {
        return keys.Any(k => currentKeyboardState.IsKeyDown(k));
    }
}

它将自己注册为游戏组件，无需添加。状态会自行更新，因此只需调用辅助方法即可。

你最后if会变成：

if (inputState.IsNewLeftClick())
    display = !display;

这样更简洁一些，它提供了一种集中处理鼠标/键盘的方式（对 SRP 来说是的）。