Code example of using canvas to implement backgammon game

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Effect

Code

nbsp;html>


    <meta>
    <title>五子棋</title>
    <script></script>


<canvas></canvas>
<script>
    (function () {    // 画布绘制
        let canvas = document.getElementById("canvas");
        let context = canvas.getContext("2d");
        context.beginPath();
        for (let i = 0; i < 19; i++) {
            // 竖线绘制
            context.moveTo(10 + i * 20, 10);
            context.lineTo(10 + i * 20, 370);
            // 横线绘制
            context.moveTo(10, 10 + i * 20);
            context.lineTo(370, 10 + i * 20);
        }
        context.stroke();
    })();
    // 鼠标单击
    let blorwh = 0;
    // 定义用于判断落子的二维数组
    let matrix = new Array(19);
    // 进行赋值
    for(let i = 0; i < 19; i++){
        matrix[i] = new Array(19);
        for(let j = 0; j < 19; j++){
            matrix[i][j] = 0;
        }
    }
    $("#canvas").click((event) => {
        // 每次落子的时候取反
        blorwh = !blorwh;
        console.log(event.offsetX);
        let canvas = document.getElementById("canvas");
        let context = canvas.getContext("2d");
        // 保存要落子的坐标
        let arcPosX, arcPosY;
        // 保存棋子在数组中的位置
        let mtxPosX, mtxPosY;
        // 和每一条线进行比较，如果相差10个像素以内，即，靠近
        for(let x = 0; x < 19; x++){
            if(Math.abs(event.offsetX - (10 + x * 20)) < 10){
                // 获得需要骡子的x
                arcPosX = 10 + x * 20;
                mtxPosX = x;
            }
            if(Math.abs(event.offsetY - (10 + x * 20)) < 10){
                // 获得需要的y
                arcPosY = 10 + x * 20;
                mtxPosY = x;
            }
        }
        // 画出棋子
        // 落子为空，进行绘制，反之不绘制
        if(matrix[mtxPosX][mtxPosY] == 0) {
            context.beginPath();
            if (blorwh) {
                context.fillStyle = "white";
                context.arc(arcPosX, arcPosY, 10, 0, Math.PI * 2, false);
                context.stroke();
                // 白子为1
                matrix[mtxPosX][mtxPosY] = 1;
            } else {
                context.fillStyle = "black";
                context.arc(arcPosX, arcPosY, 10, 0, Math.PI * 2, false);
                // 黑子为2
                matrix[mtxPosX][mtxPosY] = 2;
            }
            context.fill();
        }
        // 获胜检测
        if(matrix[mtxPosX - 1][mtxPosY] == matrix[mtxPosX][mtxPosY] &&
            matrix[mtxPosX - 2][mtxPosY] == matrix[mtxPosX][mtxPosY]  &&
                matrix[mtxPosX -3][mtxPosY] == matrix[mtxPosX][mtxPosY]  &&
                    matrix[mtxPosY - 4][mtxPosY] == matrix[mtxPosX][mtxPosY]){
            if(matrix[mtxPosX][mtxPosY] == 1){
                alert("白方获胜");
            }else{
                alert("黑方获胜");
            }
        }
    })
</script>

Idea

Create an array to save the backgammon Just the location.
Winning or losing can be judged by traversing.
For repeated chess placement, just judge whether there are already chess pieces in the position of the saved array.
Judge if it falls on the cross line and nearby points. If the difference is less than a certain value, the chess placement will be made.

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