Java鼠标点击问题

Question

我一直在尝试使用几种静态方法制作这款tick tack toe游戏。我相对来说还是个新手，不能完全理解鼠标点击的概念。我读过关于鼠标点击和鼠标事件，但它不完全有意义，我得到很多错误，当我尝试。最初我有一部分，它在自己的方法中获取鼠标信息，但后来我不知道如何同时返回x和y值。我在下面添加了填充数组的方法。现在我把它弄乱了，并设法让它们在自己的方法中，但仍然有问题运行程序。(它们不必在它们自己的方法中，我只是认为它会简化事情)当我运行这个程序时，它所做的就是打印无限多的行，告诉我点击了哪一行和哪一列，并在第一行和第一列上放一个0，不管我是否点击。此外，它似乎也不会在玩家之间切换回合。如果有人能帮助我，我将不胜感激。谢谢!

import java.util.*;
public class Game {
  public static int x;
  public static int y;
  public static double a;
  public static double b;
  public static int empty = 0;
  public static int Cross = 1;
  public static int Oh = -1;
  public static double[][] board = new double[3][3];
  public static int currentPlayer;
  public static int Point;

  public static void main (String args[]) {

    drawBoard();

    Fill();
  }
    public static void drawBoard(){
    StdDraw.setXscale(0,9);
    StdDraw.setYscale(0,9);
    StdDraw.setPenRadius(.01);
    StdDraw.setPenColor(StdDraw.BLACK);
    StdDraw.line(0,3,9,3);
    StdDraw.line(0,6,9,6);
    StdDraw.line(3,0,3,9);
    StdDraw.line(6,0,6,9);
  } //end draw board


  //get mouse click and turn into array spot
  public static void Mouse(){
    while(true){
      if (StdDraw.mousePressed()){
        a = StdDraw.mouseX();
        b = StdDraw.mouseY();
        System.out.println( a + " " + b);
      }

      //set column
      if ( 0<=a && a< 3){
        x =  0;}
      if ( 3<=a && a<6){
        x = 1;}
      if ( 6<=a && a< 9){
        x = 2;}
      //set row
      if ( 0<=b && b< 3){
        y = 0;}
      if ( 3<=b && b< 6){
        y = (int)1;}
      if ( 6<=b && b< 9){
        y = 2;}
      System.out.println("You clicked in Row" + x + "and column" +y);
    }
  }
      public static void Fill(){
      //fill array
        Mouse();
      boolean validInput = false;
      do{    
      for (int i = 0 ; i <=9 ; i++){
    if (i % 2 == 0){
          currentPlayer = Cross;
        }
        else {
          currentPlayer = Oh;
        }}
          if (0 <= x && x<=2 && 0 <=y && y <= 2 && board[x][y] == 0){
            //fill array spot
        board[x][y] = currentPlayer;
        //check game status and print board
        GameStatus();
        PrintBoard();
        validInput = true; //input is good, exit the loop
        }
        else { 
          System.out.println("This move is not valid. Try again.");
          }
      }while (!validInput);
    }







    public static void PrintBoard(){
  for (int j = 0; j<=2; j++){
    for (int k = 0; k<=2; k++){
      if (board[j][k] == 0){
        //do nothing leave empty
      }
      if (board[j][k] == 1){
        double l = ((j+1) * 3) - 1.5;
        double m = ((k+1) * 3) - 1.5;
        //print x
        StdDraw.text(l,m,"X");}
      if (board[j][k] == -1){
        double l = ((j+1) * 3) - 1.5;
        double m = ((k+1) * 3) - 1.5;
        //print O
        StdDraw.text(l,m,"O");}
    }
  }
    }

  public static void GameStatus(){
        //check for win
        if (// First column
            board[0][0] == currentPlayer
              && board[0][1] == currentPlayer
              && board[0][2] == currentPlayer
              //second column
              || board[1][0] == currentPlayer
              && board[1][1] == currentPlayer
              && board[1][2] == currentPlayer
              //third column
              || board[2][0] == currentPlayer
              && board[2][1] == currentPlayer
              && board[2][2] == currentPlayer
              //first row
              ||board[0][0] == currentPlayer
              && board[1][0] == currentPlayer
              && board[2][0] == currentPlayer
              //second row
              || board[0][1] == currentPlayer
              && board[1][1] == currentPlayer
              && board[2][1] == currentPlayer
              //third row
              || board[0][2] == currentPlayer
              && board[1][2] == currentPlayer
              && board[2][2] == currentPlayer
              //diagonal 1
              || board[0][2] == currentPlayer
              && board[1][1] == currentPlayer
              && board[2][0] == currentPlayer
              // diagonal 2
              || board[2][2] == currentPlayer
              && board[1][1] == currentPlayer
              && board[0][0] == currentPlayer){
          //X win
          while (currentPlayer==1){
            StdDraw.text(0.5, 0.5, "X Won!");}
          //O win
          while (currentPlayer==-1){
            StdDraw.text(0.5, 0.5, "O Won!");}
          return;
        }
        //draw
        if  (board[0][0] != 0 
               && board[0][1] != 0
               && board[0][2] != 0
               && board[1][0] != 0
               && board[1][1] != 0
               && board[1][2] != 0
               && board[2][0] != 0
               && board[2][1] != 0
               && board[2][2] != 0){
          StdDraw.text(0.5, 0.5, "Cat's Game!");
          return;}
        //still playing
        else {
          System.out.println("Keep Playing.");
          //keep playing
        }
  }//Ends playerMove
}// end game

Answer 1

首先，有一个"while(true)"语句。这是一个无限循环，你需要一些东西来告诉它跳出来。所以你才会看到垃圾短信。

其次，在哪里StdDraw对象你是用来获得你的"mousePressed"事件?我推荐"MOUSE_CLICKED"事件，如果你只是得到一个点击，但即使你停留在mousePressed，你需要确保它不会检测到它，如果鼠标没有被按下。

Answer 2

它继续打印行，因为您在Mouse()方法中创建了一个无限循环，并且它每秒检查if语句中的条件数百万次。

我建议创建一个单独的方法，例如gamelloop()，并从gamelloop()中调用所有其他方法。

在stdlib包文档中我们读到:

[键盘和鼠标输入]你应该在动画循环中使用这些方法，在尝试轮询鼠标的当前状态之前等待一段时间。

那么在你的游戏中你可以尝试这样做:

gameLoop() {
    isPlaying = true;
    drawBoard();
    ...;
    while (isPlaying) {
        isPressed = false;
        checkMousePressed();
        if (isPressed) {
            checkMove();
            ...;
        }  
        Thread.sleep(100); // 100 ms, so you repeat the loop 10 times/sec.
    }
    printWinner();
}

另外，方法名的一个好习惯是使用动词，而不是名词。因此，您说明了方法的作用，换句话说，您说明了它的目的，因此更容易遵循程序的逻辑。如果将其称为mouse()，则不清楚该方法应该做什么。