2 条 SVG 路径的交集

Question

I need to check if two SVG Path elements intersect. Checking for intersection of the bounding boxes with .getBBox() is too inaccurate. What I'm currently doing is iterating both paths with .getTotalLength() and then checking if two points .getPointAtLength() are equal.

Below is a snippet, but as you can see this is very slow and blocks the browser tab. There must be a more efficient method to check for intersections between two paths.

 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

You can use an existing library to do this: https://github.com/signavio/svg-intersections

It calculates the intersections mathematically rather than iteratively so the performance shouldn't degrade with very long paths.

Answer 2

I'm not sure but it may be possible to solve this mathematically if you could extract the vectors and curves from the paths. However, your function can be optimized by caching the points from one path, and reducing the number of calls to getTotalLength and getPointAtLength .

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;
            }
       }
    }
}

This can calculate the example paths in around 0.07 seconds instead of 4-5 seconds.

jsfiddle

Answer 3

time 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;
}