Question

我在画布上绘制一个基于六边形的网格。

每个六边形是一个在x / y坐标中保持6个点的对象。每个六边形对象也保持其X / Y列/行索引。

var canvas = document.getElementById("can");
    canvas.width = 200;
    canvas.height = 200;    
var ctx = canvas.getContext("2d");

var grid = []; // array that holds the Hex
var globalOffset = 30 // not important, just for smoother display atm

  function Point(x, y) {
    this.x = x;
    this.y = y;
  }

  function Hex(x, y, size) {
    this.size = 20;
    this.x = x;
    this.y = y;
    this.points = [];
    this.id = [];

    this.create = function(x, y) {
      var offSetX = (size / 2 * x) * -1
      var offSetY = 0;

      if (x % 2 == 1) {
        offSetY = Math.sqrt(3) / 2 * this.size;
      }

      var center = new Point(
        x * this.size * 2 + offSetX + globalOffset,
        y * Math.sqrt(3) / 2 * this.size * 2 + offSetY + globalOffset
      )

      this.midPoint = center;

      this.id[0] = x;
      this.id[1] = y;

      for (var i = 0; i < 6; i++) {
        var degree = 60 * i;
        var radian = Math.PI / 180 * degree;

        var point = new Point(
          center.x + size * Math.cos(radian),
          center.y + size * Math.sin(radian)
        )

        this.points.push(point);
      }
    }

    this.create(x, y);
  }
}



//Determine where was clicked
canvas.addEventListener("click", function(e) {

  var rect = canvas.getBoundingClientRect();

  var pos = {
    x: e.clientX - rect.left,
    y: e.clientY - rect.top
  }


  document.getElementById("pos").innerHTML = "click on: " + pos.x + " " + pos.y;
});


// Creating Hexagons, setting up their center point, pushing them into Grid.
function init() {
  for (var i = 0; i < 5; i++) {
    for (var j = 0; j < 4; j++) {
      var hex = new Hex(i, j, 20);
      grid.push(hex)
    }
  }

  //for each Hex in Grid, draw the Hex
  for (var hex in grid) {
    var item = grid[hex];
    ctx.beginPath();
    ctx.moveTo(item.points[0].x, item.points[0].y);

    for (var k = 1; k < item.points.length; k++) {
      ctx.lineTo(item.points[k].x, item.points[k].y);
    }

    ctx.closePath();
    ctx.stroke();

    var text = item.id;
    ctx.fillStyle = "black";
    ctx.fillText(text, item.midPoint.x - 7, item.midPoint.y - item.size / 2.2);

  }

当点击画布时，我想确定我是否点击了十六进制，如果我做了，是哪个十六进制（按列/行）。它的数学问题。

我该怎么做？

这里的完整工作示例： http://codepen.io/anon/pen/RrMzKy?editors=1111

Answer 1

如果将六边形中心视为圆的中心，则单击的六边形是其中心最接近咔嗒声的六边形。（应该可以在不测试到每个可能单元的距离的情况下优化它。）

为了解释不完整的覆盖范围，假设在可见的环绕的附加环中有更多（不可见的）六边形。

如果选择其中一个，或者距离大于圆半径，则点击不在可见的六边形上。

有些基于你自己提出的代码的重构，并避免了两个循环，因为唯一的好处是消除了单个sqrt函数：

Grid.prototype.getHexAt = function(pos) {

    var closest = null;
    var min = Infinity;

    grid.hexes.forEach(function(hex) {
        var dx = hex.center.x - pos.x;
        var dy = hex.center.y - pos.y;
        var distance = Math.sqrt(v.x * v.x + v.y * v.y);

        if (distance < hex.size && distance < min) {
            min = distance;
            closest = hex;
        }
    });

    return closest;   // may return null
}

Answer 2

有趣的是，Alnitak的建议相当不错，因为海克斯康确实是一个圆圈。这是我的功能。将mousevent位置x / y与每个六边形圆和Hexagon固有尺寸（宽度/高度）进行比较。鼠标位置x / y +六边形大小是否接近六边形中心，你可能点击了这个六边形。仍然测试所有Hexagon。如果你是＆＃34;关闭＆＃34;对于多个，获得鼠标过程x / y与所有有效六边形圆的矢量距离。具有最短矢量的六边形是您单击的那个。

Grid.prototype.getHexAt = function(pos){
    var inRange = [];
    var closest = null;

    for (var hex in grid.hexes) {
        var item = grid.hexes[hex];     
        var center = item.center;

        if (center.x + item.size > pos.x && center.x - item.size < pos.x) {
            if (center.y + item.size > pos.y && center.y - item.size < pos.y) {
                inRange.push(item);
            }
        }
    }   

    if (inRange.length > 1) {
        var pick = null;
        var dist = null;

        for (var i = 0; i < inRange.length; i++) {
            var vector = {
                x: inRange[i].center.x - pos.x,
                y: inRange[i].center.y - pos.y
            };

            if (vector.x < 0) {
                vector.x *= -1;
            }
            if (vector.y < 0) {
                vector.y *= -1;
            }

            if (pick == null || vector.x + vector.y < dist) {
                pick = inRange[i];
                dist = vector.x + vector.y;             
            }
        }

        closest = pick;
    }
    else {
        closest = inRange[0];
    }

    return closest;
};

在画布上绘制Hexagon，测试mouseclick事件与Hexagon

2 个答案: