我已经使用了https://bl.ocks.org/mbostock/8027637给出的解决方案，它返回x和y点到路径的距离，如果距离小于1px或笔划的宽度，我认为x和y点在路径上。

function closestPoint(pathNode, point) {
  var pathLength = pathNode.getTotalLength(),
      precision = 8,
      best,
      bestLength,
      bestDistance = Infinity;

  // linear scan for coarse approximation
  for (var scan, scanLength = 0, scanDistance; scanLength <= pathLength; scanLength += precision) {
    if ((scanDistance = distance2(scan = pathNode.getPointAtLength(scanLength))) < bestDistance) {
      best = scan, bestLength = scanLength, bestDistance = scanDistance;
    }
  }

  // binary search for precise estimate
  precision /= 2;
  while (precision > 0.5) {
    var before,
        after,
        beforeLength,
        afterLength,
        beforeDistance,
        afterDistance;
    if ((beforeLength = bestLength - precision) >= 0 && (beforeDistance = distance2(before = pathNode.getPointAtLength(beforeLength))) < bestDistance) {
      best = before, bestLength = beforeLength, bestDistance = beforeDistance;
    } else if ((afterLength = bestLength + precision) <= pathLength && (afterDistance = distance2(after = pathNode.getPointAtLength(afterLength))) < bestDistance) {
      best = after, bestLength = afterLength, bestDistance = afterDistance;
    } else {
      precision /= 2;
    }
  }

  best = [best.x, best.y];
  best.distance = Math.sqrt(bestDistance);
  return best;

  function distance2(p) {
    var dx = p.x - point[0],
        dy = p.y - point[1];
    return dx * dx + dy * dy;
  }
}

字符串

**SVGGeometryElement.isPointInFill（）**方法确定给定点是否在元素的填充形状内。正常命中测试规则适用;元素上的pointer-events属性的值确定点是否被认为在填充内。点参数被解释为元素的局部坐标系中的点。

var myPath = document.getElementById("mypath");
var txt = document.getElementById("txt");
var svg = document.getElementsByTagName("svg")[0];

svg.addEventListener("mousemove", function(event) {
  var mouseX = event.clientX;
  var mouseY = event.clientY;  
  
  var point = svg.createSVGPoint();
  point.x = event.clientX;
  point.y = event.clientY;  
  point=point.matrixTransform(svg.getScreenCTM().inverse());
  
  var inShape=myPath.isPointInFill( point ); 
  txt.innerText=inShape+" x:"+point.x+" y:"+point.y;
});

个字符