我需要一个Godot (GDscript)的实现，所以我写了一个基于grumdrig接受的答案:

func minimum_distance(v: Vector2, w: Vector2, p: Vector2):
    # Return minimum distance between line segment vw and point p
    var l2: float = (v - w).length_squared()  # i.e. |w-v|^2 -  avoid a sqrt
    if l2 == 0.0:
        return p.distance_to(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.
    var t: float = max(0, min(1, (p - v).dot(w - v) / l2))
    var projection: Vector2 = v + t * (w - v)  # Projection falls on the segment
    
    return p.distance_to(projection)

对于感兴趣的人，这里是Joshua的Javascript代码到Objective-C的简单转换:

- (double)distanceToPoint:(CGPoint)p fromLineSegmentBetween:(CGPoint)l1 and:(CGPoint)l2
{
    double A = p.x - l1.x;
    double B = p.y - l1.y;
    double C = l2.x - l1.x;
    double D = l2.y - l1.y;

    double dot = A * C + B * D;
    double len_sq = C * C + D * D;
    double param = dot / len_sq;

    double xx, yy;

    if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) {
        xx = l1.x;
        yy = l1.y;
    }
    else if (param > 1) {
        xx = l2.x;
        yy = l2.y;
    }
    else {
        xx = l1.x + param * C;
        yy = l1.y + param * D;
    }

    double dx = p.x - xx;
    double dy = p.y - yy;

    return sqrtf(dx * dx + dy * dy);
}

我需要这个解决方案与MKMapPoint一起工作，所以我将分享它，以防其他人需要它。只是一些小的改变，这将返回米为单位的距离:

- (double)distanceToPoint:(MKMapPoint)p fromLineSegmentBetween:(MKMapPoint)l1 and:(MKMapPoint)l2
{
    double A = p.x - l1.x;
    double B = p.y - l1.y;
    double C = l2.x - l1.x;
    double D = l2.y - l1.y;

    double dot = A * C + B * D;
    double len_sq = C * C + D * D;
    double param = dot / len_sq;

    double xx, yy;

    if (param < 0 || (l1.x == l2.x && l1.y == l2.y)) {
        xx = l1.x;
        yy = l1.y;
    }
    else if (param > 1) {
        xx = l2.x;
        yy = l2.y;
    }
    else {
        xx = l1.x + param * C;
        yy = l1.y + param * D;
    }

    return MKMetersBetweenMapPoints(p, MKMapPointMake(xx, yy));
}

Matlab代码，内置“自检”，如果他们调用函数没有参数:

function r = distPointToLineSegment( xy0, xy1, xyP )
% r = distPointToLineSegment( xy0, xy1, xyP )

if( nargin < 3 )
    selfTest();
    r=0;
else
    vx = xy0(1)-xyP(1);
    vy = xy0(2)-xyP(2);
    ux = xy1(1)-xy0(1);
    uy = xy1(2)-xy0(2);
    lenSqr= (ux*ux+uy*uy);
    detP= -vx*ux + -vy*uy;

    if( detP < 0 )
        r = norm(xy0-xyP,2);
    elseif( detP > lenSqr )
        r = norm(xy1-xyP,2);
    else
        r = abs(ux*vy-uy*vx)/sqrt(lenSqr);
    end
end


    function selfTest()
        %#ok<*NASGU>
        disp(['invalid args, distPointToLineSegment running (recursive)  self-test...']);

        ptA = [1;1]; ptB = [-1;-1];
        ptC = [1/2;1/2];  % on the line
        ptD = [-2;-1.5];  % too far from line segment
        ptE = [1/2;0];    % should be same as perpendicular distance to line
        ptF = [1.5;1.5];      % along the A-B but outside of the segment

        distCtoAB = distPointToLineSegment(ptA,ptB,ptC)
        distDtoAB = distPointToLineSegment(ptA,ptB,ptD)
        distEtoAB = distPointToLineSegment(ptA,ptB,ptE)
        distFtoAB = distPointToLineSegment(ptA,ptB,ptF)
        figure(1); clf;
        circle = @(x, y, r, c) rectangle('Position', [x-r, y-r, 2*r, 2*r], ...
            'Curvature', [1 1], 'EdgeColor', c);
        plot([ptA(1) ptB(1)],[ptA(2) ptB(2)],'r-x'); hold on;
        plot(ptC(1),ptC(2),'b+'); circle(ptC(1),ptC(2), 0.5e-1, 'b');
        plot(ptD(1),ptD(2),'g+'); circle(ptD(1),ptD(2), distDtoAB, 'g');
        plot(ptE(1),ptE(2),'k+'); circle(ptE(1),ptE(2), distEtoAB, 'k');
        plot(ptF(1),ptF(2),'m+'); circle(ptF(1),ptF(2), distFtoAB, 'm');
        hold off;
        axis([-3 3 -3 3]); axis equal;
    end

end

基于Joshua Javascript的AutoHotkeys版本:

plDist(x, y, x1, y1, x2, y2) {
    A:= x - x1
    B:= y - y1
    C:= x2 - x1
    D:= y2 - y1

    dot:= A*C + B*D
    sqLen:= C*C + D*D
    param:= dot / sqLen

    if (param < 0 || ((x1 = x2) && (y1 = y2))) {
        xx:= x1
        yy:= y1
    } else if (param > 1) {
        xx:= x2
        yy:= y2
    } else {
        xx:= x1 + param*C
        yy:= y1 + param*D
    }

    dx:= x - xx
    dy:= y - yy

    return sqrt(dx*dx + dy*dy)
}

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。

在我自己的问题线程如何计算在C, c# / .NET 2.0或Java的所有情况下一个点和线段之间的最短2D距离?当我找到一个c#的答案时，我被要求把它放在这里:所以它是从http://www.topcoder.com/tc?d1=tutorials&d2=geometry1&module=Static修改的:

//Compute the dot product AB . BC
private double DotProduct(double[] pointA, double[] pointB, double[] pointC)
{
    double[] AB = new double[2];
    double[] BC = new double[2];
    AB[0] = pointB[0] - pointA[0];
    AB[1] = pointB[1] - pointA[1];
    BC[0] = pointC[0] - pointB[0];
    BC[1] = pointC[1] - pointB[1];
    double dot = AB[0] * BC[0] + AB[1] * BC[1];

    return dot;
}

//Compute the cross product AB x AC
private double CrossProduct(double[] pointA, double[] pointB, double[] pointC)
{
    double[] AB = new double[2];
    double[] AC = new double[2];
    AB[0] = pointB[0] - pointA[0];
    AB[1] = pointB[1] - pointA[1];
    AC[0] = pointC[0] - pointA[0];
    AC[1] = pointC[1] - pointA[1];
    double cross = AB[0] * AC[1] - AB[1] * AC[0];

    return cross;
}

//Compute the distance from A to B
double Distance(double[] pointA, double[] pointB)
{
    double d1 = pointA[0] - pointB[0];
    double d2 = pointA[1] - pointB[1];

    return Math.Sqrt(d1 * d1 + d2 * d2);
}

//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double LineToPointDistance2D(double[] pointA, double[] pointB, double[] pointC, 
    bool isSegment)
{
    double dist = CrossProduct(pointA, pointB, pointC) / Distance(pointA, pointB);
    if (isSegment)
    {
        double dot1 = DotProduct(pointA, pointB, pointC);
        if (dot1 > 0) 
            return Distance(pointB, pointC);

        double dot2 = DotProduct(pointB, pointA, pointC);
        if (dot2 > 0) 
            return Distance(pointA, pointC);
    }
    return Math.Abs(dist);
} 

我不是要回答问题，而是要问问题，所以我希望我不会因为某些原因而得到数百万张反对票，而是批评。我只是想(并被鼓励)分享其他人的想法，因为这个帖子中的解决方案要么是用一些奇异的语言(Fortran, Mathematica)，要么被某人标记为错误。对我来说唯一有用的(由Grumdrig编写)是用c++编写的，没有人标记它有错误。但是它缺少被调用的方法(dot等)。