下面是HSQLDB的SQL实现:

CREATE FUNCTION dist_to_segment(px double, py double, vx double, vy double, wx double, wy double)
  RETURNS double
BEGIN atomic
   declare l2 double;
   declare t double;
   declare nx double;
   declare ny double;
   set l2 =(vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
   IF l2 = 0 THEN
     RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
   ELSE
     set t = ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
     set t = GREATEST(0, LEAST(1, t));
     set nx=vx + t * (wx - vx);
     set ny=vy + t * (wy - vy);
     RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
   END IF;
END;

Postgres的实现:

CREATE FUNCTION dist_to_segment(px numeric, py numeric, vx numeric, vy numeric, wx numeric, wy numeric)
  RETURNS numeric
AS $$
   declare l2 numeric;
   declare t numeric;
   declare nx numeric;
   declare ny numeric;
BEGIN 
   l2 := (vx - wx)*(vx - wx) + (vy - wy)*(vy - wy);
   IF l2 = 0 THEN
     RETURN sqrt((vx - px)*(vx - px) + (vy - py)*(vy - py));
   ELSE
     t := ((px - vx) * (wx - vx) + (py - vy) * (wy - vy)) / l2;
     t := GREATEST(0, LEAST(1, t));
     nx := vx + t * (wx - vx);
     ny := vy + t * (wy - vy);
     RETURN sqrt((nx - px)*(nx - px) + (ny - py)*(ny - py));
   END IF;
END;
$$ LANGUAGE plpgsql;

I'm assuming you want to find the shortest distance between the point and a line segment; to do this, you need to find the line (lineA) which is perpendicular to your line segment (lineB) which goes through your point, determine the intersection between that line (lineA) and your line which goes through your line segment (lineB); if that point is between the two points of your line segment, then the distance is the distance between your point and the point you just found which is the intersection of lineA and lineB; if the point is not between the two points of your line segment, you need to get the distance between your point and the closer of two ends of the line segment; this can be done easily by taking the square distance (to avoid a square root) between the point and the two points of the line segment; whichever is closer, take the square root of that one.

in R

     #distance beetween segment ab and point c in 2D space
getDistance_ort_2 <- function(a, b, c){
  #go to complex numbers
  A<-c(a[1]+1i*a[2],b[1]+1i*b[2])
  q=c[1]+1i*c[2]
  
  #function to get coefficients of line (ab)
  getAlphaBeta <- function(A)
  { a<-Re(A[2])-Re(A[1])
    b<-Im(A[2])-Im(A[1])
    ab<-as.numeric()
    ab[1] <- -Re(A[1])*b/a+Im(A[1])
    ab[2] <-b/a
    if(Im(A[1])==Im(A[2])) ab<- c(Im(A[1]),0)
    if(Re(A[1])==Re(A[2])) ab <- NA
    return(ab)
  }
  
  #function to get coefficients of line ortogonal to line (ab) which goes through point q
  getAlphaBeta_ort<-function(A,q)
  { ab <- getAlphaBeta(A) 
  coef<-c(Re(q)/ab[2]+Im(q),-1/ab[2])
  if(Re(A[1])==Re(A[2])) coef<-c(Im(q),0)
  return(coef)
  }
  
  #function to get coordinates of interception point 
  #between line (ab) and its ortogonal which goes through point q
  getIntersection_ort <- function(A, q){
    A.ab <- getAlphaBeta(A)
    q.ab <- getAlphaBeta_ort(A,q)
    if (!is.na(A.ab[1])&A.ab[2]==0) {
      x<-Re(q)
      y<-Im(A[1])}
    if (is.na(A.ab[1])) {
      x<-Re(A[1])
      y<-Im(q)
    } 
    if (!is.na(A.ab[1])&A.ab[2]!=0) {
      x <- (q.ab[1] - A.ab[1])/(A.ab[2] - q.ab[2])
      y <- q.ab[1] + q.ab[2]*x}
    xy <- x + 1i*y  
    return(xy)
  }
  
  intersect<-getIntersection_ort(A,q)
  if ((Mod(A[1]-intersect)+Mod(A[2]-intersect))>Mod(A[1]-A[2])) {dist<-min(Mod(A[1]-q),Mod(A[2]-q))
  } else dist<-Mod(q-intersect)
  return(dist)
}



 

下面是devnullicus转换为c#的c++版本。对于我的实现，我需要知道交叉点，并找到他的解决方案。

public static bool PointSegmentDistanceSquared(PointF point, PointF lineStart, PointF lineEnd, out double distance, out PointF intersectPoint)
{
    const double kMinSegmentLenSquared = 0.00000001; // adjust to suit.  If you use float, you'll probably want something like 0.000001f
    const double kEpsilon = 1.0E-14; // adjust to suit.  If you use floats, you'll probably want something like 1E-7f
    double dX = lineEnd.X - lineStart.X;
    double dY = lineEnd.Y - lineStart.Y;
    double dp1X = point.X - lineStart.X;
    double dp1Y = point.Y - lineStart.Y;
    double segLenSquared = (dX * dX) + (dY * dY);
    double t = 0.0;

    if (segLenSquared >= -kMinSegmentLenSquared && segLenSquared <= kMinSegmentLenSquared)
    {
        // segment is a point.
        intersectPoint = lineStart;
        t = 0.0;
        distance = ((dp1X * dp1X) + (dp1Y * dp1Y));
    }
    else
    {
        // Project a line from p to the segment [p1,p2].  By considering the line
        // extending the segment, parameterized as p1 + (t * (p2 - p1)),
        // we find projection of point p onto the line. 
        // It falls where t = [(p - p1) . (p2 - p1)] / |p2 - p1|^2
        t = ((dp1X * dX) + (dp1Y * dY)) / segLenSquared;
        if (t < kEpsilon)
        {
            // intersects at or to the "left" of first segment vertex (lineStart.X, lineStart.Y).  If t is approximately 0.0, then
            // intersection is at p1.  If t is less than that, then there is no intersection (i.e. p is not within
            // the 'bounds' of the segment)
            if (t > -kEpsilon)
            {
                // intersects at 1st segment vertex
                t = 0.0;
            }
            // set our 'intersection' point to p1.
            intersectPoint = lineStart;
            // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
            // we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
        }
        else if (t > (1.0 - kEpsilon))
        {
            // intersects at or to the "right" of second segment vertex (lineEnd.X, lineEnd.Y).  If t is approximately 1.0, then
            // intersection is at p2.  If t is greater than that, then there is no intersection (i.e. p is not within
            // the 'bounds' of the segment)
            if (t < (1.0 + kEpsilon))
            {
                // intersects at 2nd segment vertex
                t = 1.0;
            }
            // set our 'intersection' point to p2.
            intersectPoint = lineEnd;
            // Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
            // we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
        }
        else
        {
            // The projection of the point to the point on the segment that is perpendicular succeeded and the point
            // is 'within' the bounds of the segment.  Set the intersection point as that projected point.
            intersectPoint = new PointF((float)(lineStart.X + (t * dX)), (float)(lineStart.Y + (t * dY)));
        }
        // return the squared distance from p to the intersection point.  Note that we return the squared distance
        // as an optimization because many times you just need to compare relative distances and the squared values
        // works fine for that.  If you want the ACTUAL distance, just take the square root of this value.
        double dpqX = point.X - intersectPoint.X;
        double dpqY = point.Y - intersectPoint.Y;

        distance = ((dpqX * dpqX) + (dpqY * dpqY));
    }

    return true;
}

嘿，我昨天才写的。它在Actionscript 3.0中，基本上是Javascript，尽管你可能没有相同的Point类。

//st = start of line segment
//b = the line segment (as in: st + b = end of line segment)
//pt = point to test
//Returns distance from point to line segment.  
//Note: nearest point on the segment to the test point is right there if we ever need it
public static function linePointDist( st:Point, b:Point, pt:Point ):Number
{
    var nearestPt:Point; //closest point on seqment to pt

    var keyDot:Number = dot( b, pt.subtract( st ) ); //key dot product
    var bLenSq:Number = dot( b, b ); //Segment length squared

    if( keyDot <= 0 )  //pt is "behind" st, use st
    {
        nearestPt = st  
    }
    else if( keyDot >= bLenSq ) //pt is "past" end of segment, use end (notice we are saving twin sqrts here cuz)
    {
        nearestPt = st.add(b);
    }
    else //pt is inside segment, reuse keyDot and bLenSq to get percent of seqment to move in to find closest point
    {
        var keyDotToPctOfB:Number = keyDot/bLenSq; //REM dot product comes squared
        var partOfB:Point = new Point( b.x * keyDotToPctOfB, b.y * keyDotToPctOfB );
        nearestPt = st.add(partOfB);
    }

    var dist:Number = (pt.subtract(nearestPt)).length;

    return dist;
}

此外，这里有一个关于这个问题的相当完整和可读的讨论:notejot.com

本想在GLSL中这样做，但如果可能的话，最好避免所有这些条件。使用clamp()可以避免两种端点情况:

// find closest point to P on line segment AB:
vec3 closest_point_on_line_segment(in vec3 P, in vec3 A, in vec3 B) {
    vec3 AP = P - A, AB = B - A;
    float l = dot(AB, AB);
    if (l <= 0.0000001) return A;    // A and B are practically the same
    return AP - AB*clamp(dot(AP, AB)/l, 0.0, 1.0);  // do the projection
}

如果您可以确定A和B彼此不会非常接近，则可以简化为删除If()。事实上，即使A和B是相同的，我的GPU仍然给出了这个无条件版本的正确结果(但这是使用pre-OpenGL 4.1，其中GLSL除零是未定义的):

// find closest point to P on line segment AB:
vec3 closest_point_on_line_segment(in vec3 P, in vec3 A, in vec3 B) {
    vec3 AP = P - A, AB = B - A;
    return AP - AB*clamp(dot(AP, AB)/dot(AB, AB), 0.0, 1.0);
}

计算距离是很简单的——GLSL提供了一个distance()函数，你可以在这个最近的点和P。

灵感来自Iñigo Quilez的胶囊距离函数代码