上面的函数在垂直线上不起作用。这是一个工作正常的函数! 与点p1 p2相交。CheckPoint为p;

public float DistanceOfPointToLine2(PointF p1, PointF p2, PointF p)
{
  //          (y1-y2)x + (x2-x1)y + (x1y2-x2y1)
  //d(P,L) = --------------------------------
  //         sqrt( (x2-x1)pow2 + (y2-y1)pow2 )

  double ch = (p1.Y - p2.Y) * p.X + (p2.X - p1.X) * p.Y + (p1.X * p2.Y - p2.X * p1.Y);
  double del = Math.Sqrt(Math.Pow(p2.X - p1.X, 2) + Math.Pow(p2.Y - p1.Y, 2));
  double d = ch / del;
  return (float)d;
}

伊莱，你选定的代码是错误的。在线段所在直线附近但远离线段一端的点将被错误地判断为接近线段。更新:上面提到的错误答案已不再被接受。

下面是一些正确的c++代码。它假设一个2d向量类vec2 {float x,y;}，本质上，带有加法、subract、缩放等运算符，以及一个距离和点积函数(即x1 x2 + y1 y2)。

float minimum_distance(vec2 v, vec2 w, vec2 p) {
  // Return minimum distance between line segment vw and point p
  const float l2 = length_squared(v, w);  // i.e. |w-v|^2 -  avoid a sqrt
  if (l2 == 0.0) return distance(p, v);   // v == w case
  // Consider the line extending the segment, parameterized as v + t (w - v).
  // We find projection of point p onto the line. 
  // It falls where t = [(p-v) . (w-v)] / |w-v|^2
  // We clamp t from [0,1] to handle points outside the segment vw.
  const float t = max(0, min(1, dot(p - v, w - v) / l2));
  const vec2 projection = v + t * (w - v);  // Projection falls on the segment
  return distance(p, projection);
}

编辑:我需要一个Javascript实现，所以在这里，没有依赖关系(或注释，但它是一个直接的端口以上)。点被表示为具有x和y属性的对象。

function sqr(x) { return x * x }
function dist2(v, w) { return sqr(v.x - w.x) + sqr(v.y - w.y) }
function distToSegmentSquared(p, v, w) {
  var l2 = dist2(v, w);
  if (l2 == 0) return dist2(p, v);
  var t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
  t = Math.max(0, Math.min(1, t));
  return dist2(p, { x: v.x + t * (w.x - v.x),
                    y: v.y + t * (w.y - v.y) });
}
function distToSegment(p, v, w) { return Math.sqrt(distToSegmentSquared(p, v, w)); }

编辑2:我需要一个Java版本，但更重要的是，我需要3d版本而不是2d版本。

float dist_to_segment_squared(float px, float py, float pz, float lx1, float ly1, float lz1, float lx2, float ly2, float lz2) {
  float line_dist = dist_sq(lx1, ly1, lz1, lx2, ly2, lz2);
  if (line_dist == 0) return dist_sq(px, py, pz, lx1, ly1, lz1);
  float t = ((px - lx1) * (lx2 - lx1) + (py - ly1) * (ly2 - ly1) + (pz - lz1) * (lz2 - lz1)) / line_dist;
  t = constrain(t, 0, 1);
  return dist_sq(px, py, pz, lx1 + t * (lx2 - lx1), ly1 + t * (ly2 - ly1), lz1 + t * (lz2 - lz1));
}

这里，在函数参数中，<px,py,pz>是问题点，线段有端点<lx1,ly1,lz1>和<lx2,ly2,lz2>。函数dist_sq(假定存在)求两点之间距离的平方。

请参见以下网站中的Matlab几何工具箱: http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html

按Ctrl +f，输入“segment”，查找线段相关函数。函数“segment_point_dist_2d.”和segment_point_dist_3d。M "是你需要的。

几何代码有C版本、c++版本、FORTRAN77版本、FORTRAN90版本和MATLAB版本。

这个答案是基于公认答案的JavaScript解决方案。 它主要只是格式更好，函数名更长，当然函数语法更短，因为它是在ES6 + CoffeeScript中。

JavaScript版本(ES6)

distanceSquared = (v, w)=> Math.pow(v.x - w.x, 2) + Math.pow(v.y - w.y, 2);
distance = (v, w)=> Math.sqrt(distanceSquared(v, w));

distanceToLineSegmentSquared = (p, v, w)=> {
    l2 = distanceSquared(v, w);
    if (l2 === 0) {
        return distanceSquared(p, v);
    }
    t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
    t = Math.max(0, Math.min(1, t));
    return distanceSquared(p, {
        x: v.x + t * (w.x - v.x),
        y: v.y + t * (w.y - v.y)
    });
}
distanceToLineSegment = (p, v, w)=> {
    return Math.sqrt(distanceToLineSegmentSquared(p, v));
}

CoffeeScript版本

distanceSquared = (v, w)-> (v.x - w.x) ** 2 + (v.y - w.y) ** 2
distance = (v, w)-> Math.sqrt(distanceSquared(v, w))

distanceToLineSegmentSquared = (p, v, w)->
    l2 = distanceSquared(v, w)
    return distanceSquared(p, v) if l2 is 0
    t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2
    t = Math.max(0, Math.min(1, t))
    distanceSquared(p, {
        x: v.x + t * (w.x - v.x)
        y: v.y + t * (w.y - v.y)
    })

distanceToLineSegment = (p, v, w)->
    Math.sqrt(distanceToLineSegmentSquared(p, v, w))

Lua: 查找线段(不是整条线)与点之间的最小距离

function solveLinearEquation(A1,B1,C1,A2,B2,C2)
--it is the implitaion of a method of solving linear equations in x and y
  local f1 = B1*C2 -B2*C1
  local f2 = A2*C1-A1*C2
  local f3 = A1*B2 -A2*B1
  return {x= f1/f3, y= f2/f3}
end


function pointLiesOnLine(x,y,x1,y1,x2,y2)
  local dx1 = x-x1
  local  dy1 = y-y1
  local dx2 = x-x2
  local  dy2 = y-y2
  local crossProduct = dy1*dx2 -dx1*dy2

if crossProduct ~= 0  then  return  false
else
  if ((x1>=x) and (x>=x2)) or ((x2>=x) and (x>=x1)) then
    if ((y1>=y) and (y>=y2)) or ((y2>=y) and (y>=y1)) then
      return true
    else return false end
  else  return false end
end
end


function dist(x1,y1,x2,y2)
  local dx = x1-x2
  local dy = y1-y2
  return math.sqrt(dx*dx + dy* dy)
 end


function findMinDistBetnPointAndLine(x1,y1,x2,y2,x3,y3)
-- finds the min  distance between (x3,y3) and line (x1,y2)--(x2,y2)
   local A2,B2,C2,A1,B1,C1
   local dx = y2-y1
   local dy = x2-x1
   if dx == 0 then A2=1 B2=0 C2=-x3 A1=0 B1=1 C1=-y1 
   elseif dy == 0 then A2=0 B2=1 C2=-y3 A1=1 B1=0 C1=-x1
   else
      local m1 = dy/dx
      local m2 = -1/m1
      A2=m2 B2=-1 C2=y3-m2*x3 A1=m1 B1=-1 C1=y1-m1*x1
   end
 local intsecPoint= solveLinearEquation(A1,B1,C1,A2,B2,C2)
if pointLiesOnLine(intsecPoint.x, intsecPoint.y,x1,y1,x2,y2) then
   return dist(intsecPoint.x, intsecPoint.y, x3,y3)
 else
   return math.min(dist(x3,y3,x1,y1),dist(x3,y3,x2,y2))
end
end

您可以尝试PHP geo-math-php的库

composer require rkondratuk/geo-math-php:^1

例子:

<?php

use PhpGeoMath\Model\GeoSegment;
use PhpGeoMath\Model\Polar3dPoint;

$polarPoint1 = new Polar3dPoint(
    40.758742779050706, -73.97855507715238, Polar3dPoint::EARTH_RADIUS_IN_METERS
);

$polarPoint2 = new Polar3dPoint(
    40.74843388072615, -73.98566565776102, Polar3dPoint::EARTH_RADIUS_IN_METERS
);

$polarPoint3 = new Polar3dPoint(
    40.74919365249446, -73.98133456388013, Polar3dPoint::EARTH_RADIUS_IN_METERS
);

$arcSegment = new GeoSegment($polarPoint1, $polarPoint2);
$nearestPolarPoint = $arcSegment->calcNearestPoint($polarPoint3);

// Shortest distance from point-3 to segment(point-1, point-2)
$geoDistance = $nearestPolarPoint->calcGeoDistanceToPoint($polarPoint3);