如何在svg中获取路径中最精确的x和y坐标？

Question

我在svg编辑2.7工作，我在svg的路径标签工作，我创建路径一些不规则的形状。然后我可以在路径标记上方或内部拖动任何svg对象。当我在路径标记上方或内部拖放时，它将导致该对象角度的精确最近绘制的x和y点。我需要在路径标签上拖动SVG对象我需要更改角度（转换部分：旋转）。例如，我在工作区域有绘制路径（表面），我需要拖放该路径（表面）。在这里，我附上了基于问题的图像：

在这里，我发现路径标记中的所有点。但是我需要获得精确点，而拖动的对象在该路径边界中留下。

遵循获取最近路径点的代码

function closestPoint(pathNode, point) {
  var pathLength = pathNode.getTotalLength(),
      precision = pathLength / pathNode.pathSegList.numberOfItems * .125,
      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 *= .5;
  while (precision > .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 *= .5;
    }
  }
  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;
  }
}

以上代码用于查找路径标记中的邻居行。

    var segments = pathNode.pathSegList;
      for (var i=0,len=segments.numberOfItems;i<len;++i){
      var segment = segments.getItem(i);
      switch(segment.pathSegType){
      case SVGPathSeg.PATHSEG_LINETO_ABS:
// segment is a SVGPathSegLinetoAbs object
      console.log( "Absolute Line To", segment.x, segment.y );
      break;
      case SVGPathSeg.PATHSEG_CLOSEPATH:
      break;
      }
   }

此处绝对线TO 会产生路径标记中的所有点。当用户将门对象留在第二行路径标记上时，如何在路径中获取x和y坐标。 提前致谢