XNA-Mouse.Left Button多次执行

Posted 2021-04-27

我是编程新手，但在开始之前，请先回答我，为什么更新不止一次执行，就像我是个虚拟人一样解释它。

无论如何，我试图使此代码仅运行一次，因为到目前为止，它多次执行。

 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);
    }

答案

之所以出现这种现象，是因为XNA每秒自动多次调用Update方法，大约每秒60次。与Draw相同，它可能会渲染约60 FPS。

因此，如果您按下按钮一秒钟，并且该方法被调用60次，则在这60次中ButtonState.Pressed的求值为true。

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;
}

因此，单击将仅注册一次（因为鼠标当前需要单击，但是以前必须处于释放状态）。如果有一天，您还需要保留键盘的历史记录。

我还更改了显示切换的逻辑，这样更清洁。您的注释中也有一个错误（应该为//，如果为false，则它将设置为true）。这很正常，评论经常在说谎；总是尝试尽可能少地注释，只有代码才能说明事实。

---

如果对您有帮助，这是我之前使用的帮助程序类：

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));
    }
}

它将自动注册为游戏组件，无需添加。状态将自行更新，因此只需调用helper方法。

您的最后一个if将成为：

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

这样比较干净，它提供了一种集中式方式来处理鼠标/键盘（是SRP的）。

另一答案

private KeyboardState currentKeyboardState;
private KeyboardState lastKeyboardState;

lastKeyboardState = currentKeyboardState;
currentKeyboardState = Keyboard.GetState();

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

if (IsNewKeyPress(Keys.Space))
        { //Do something }

对于键盘上的任何键，此键都有效！谢谢兄弟皮埃尔