Question

我需要检查两个SVG Path元素是否相交。检查边界框与.getBBox()的交集是不准确的。我目前正在做的是用.getTotalLength()迭代两个路径，然后检查两个点.getPointAtLength()是否相等。

下面是一个片段，但正如您所看到的，这非常慢并且会阻止浏览器选项卡。必须有一种更有效的方法来检查两条路径之间的交叉点。

var path1 = document.getElementById("p1");
var path2 = document.getElementById("p2");
var time = document.getElementById("time");
var btn = document.getElementById("start");
btn.addEventListener("click", getIntersection);

function getIntersection() {
var start = Date.now();
  for (var i = 0; i < path1.getTotalLength(); i++) {
    for (var j = 0; j < path2.getTotalLength(); j++) {
      var point1 = path1.getPointAtLength(i);
      var point2 = path2.getPointAtLength(j);

      if (pointIntersect(point1, point2)) {
        var end = Date.now();
        time.innerHTML = (end - start) / 1000 + "s";
        return;
      }
    }

  }
}

function pointIntersect(p1, p2) {
  p1.x = Math.round(p1.x);
  p1.y = Math.round(p1.y);
  p2.x = Math.round(p2.x);
  p2.y = Math.round(p2.y);
  return p1.x === p2.x && p1.y === p2.y;
}

svg {
  fill: none;
  stroke: black;
}
#start {
  border: 1px solid;
  display: inline-block;
  position: absolute;
}

<div id="start">Start
</div>
<svg xmlns="http://www.w3.org/2000/svg">
  <path d="M 50 10 c 120 120 120 120 120 20 z" id="p1"></path>
  <path d="M 150 10 c 120 120 120 120 120 20 z" id="p2"></path>
</svg>
<div id="time"></div>

Answer 1

我不确定但如果你能从路径中提取矢量和曲线，那么可以用数学方法解决这个问题。但是，您可以通过缓存来自一条路径的点来优化您的功能，并减少对getTotalLength和getPointAtLength的调用次数。

function getIntersection() {
    var start = Date.now(),
    path1Length = path1.getTotalLength(),
    path2Length = path2.getTotalLength(),
    path2Points = [];

    for (var j = 0; j < path2Length; j++) {      
        path2Points.push(path2.getPointAtLength(j));
    }

    for (var i = 0; i < path1Length; i++) {  
        var point1 = path1.getPointAtLength(i);

        for (var j = 0; j < path2Points.length; j++) {
            if (pointIntersect(point1, path2Points[j])) {
                var end = Date.now();
                time.innerHTML = (end - start) / 1000 + "s";        
                return;
            }
       }
    }
}

这可以在0.07秒而不是4-5秒内计算示例路径。

jsfiddle

Answer 2

时间 0.027s

function getIntersection2() {
  function Intersect(p1, p2) {
    return p1.z!==p2.z && p1.x === p2.x && p1.y === p2.y;
  }
  var paths = [path1,path2];
    var start = Date.now(),
    pathLength = [path1.getTotalLength(),path2.getTotalLength()],
    pathPoints = [],
    inters = [];
 
  for (var i = 0; i < 2; i++) {  
    for (var j = 0; j < pathLength[i]; j++) {
      var p = paths[i].getPointAtLength(j);
      p.z=i;
      p.x=Math.round(p.x);
      p.y=Math.round(p.y);
      pathPoints.push(p);
    }
  }
  pathPoints.sort((a,b)=>a.x!=b.x?a.x-b.x:a.y!=b.y?a.y-b.y:0)
  // todos os pontos
  .forEach((a,i,m)=>i&&Intersect(m[i-1],a)?inters.push([a.x,a.y]):0)
  // somente o primeiro
  //.some((a,i,m)=>i&&Intersect(m[i-1],a)?inters.push([a.x,a.y]):0);
  result.innerHTML = inters;
        var end = Date.now();
        time.innerHTML = (end - start) / 1000 + "s";
        return;
}

Answer 3

您可以使用现有的库来执行此操作：https://github.com/signavio/svg-intersections

它以数学方式而不是迭代方式计算相交，因此在长路径下性能不应降低。

2个SVG路径的交集

3 个答案: