用t-sql编码

点为(@px， @py)，线段从(@ax， @ay)到(@bx， @by)

create function fn_sqr (@NumberToSquare decimal(18,10)) 
returns decimal(18,10)
as 
begin
    declare @Result decimal(18,10)
    set @Result = @NumberToSquare * @NumberToSquare
    return @Result
end
go

create function fn_Distance(@ax decimal (18,10) , @ay decimal (18,10), @bx decimal(18,10),  @by decimal(18,10)) 
returns decimal(18,10)
as
begin
    declare @Result decimal(18,10)
    set @Result = (select dbo.fn_sqr(@ax - @bx) + dbo.fn_sqr(@ay - @by) )
    return @Result
end
go

create function fn_DistanceToSegmentSquared(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10)) 
returns decimal(18,10)
as 
begin
    declare @l2 decimal(18,10)
    set @l2 = (select dbo.fn_Distance(@ax, @ay, @bx, @by))
    if @l2 = 0
        return dbo.fn_Distance(@px, @py, @ax, @ay)
    declare @t decimal(18,10)
    set @t = ((@px - @ax) * (@bx - @ax) + (@py - @ay) * (@by - @ay)) / @l2
    if (@t < 0) 
        return dbo.fn_Distance(@px, @py, @ax, @ay);
    if (@t > 1) 
        return dbo.fn_Distance(@px, @py, @bx, @by);
    return dbo.fn_Distance(@px, @py,  @ax + @t * (@bx - @ax),  @ay + @t * (@by - @ay))
end
go

create function fn_DistanceToSegment(@px decimal(18,10), @py decimal(18,10), @ax decimal(18,10), @ay decimal(18,10), @bx decimal(18,10), @by decimal(18,10)) 
returns decimal(18,10)
as 
begin
    return sqrt(dbo.fn_DistanceToSegmentSquared(@px, @py , @ax , @ay , @bx , @by ))
end
go

--example execution for distance from a point at (6,1) to line segment that runs from (4,2) to (2,1)
select dbo.fn_DistanceToSegment(6, 1, 4, 2, 2, 1) 
--result = 2.2360679775

--example execution for distance from a point at (-3,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(-3, -2, 0, -2, -2, 1) 
--result = 2.4961508830

--example execution for distance from a point at (0,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(0,-2, 0, -2, -2, 1) 
--result = 0.0000000000

以下是Grumdrig解决方案的一个更完整的说明。这个版本还返回最近的点本身。

#include "stdio.h"
#include "math.h"

class Vec2
{
public:
    float _x;
    float _y;

    Vec2()
    {
        _x = 0;
        _y = 0;
    }

    Vec2( const float x, const float y )
    {
        _x = x;
        _y = y;
    }

    Vec2 operator+( const Vec2 &v ) const
    {
        return Vec2( this->_x + v._x, this->_y + v._y );
    }

    Vec2 operator-( const Vec2 &v ) const
    {
        return Vec2( this->_x - v._x, this->_y - v._y );
    }

    Vec2 operator*( const float f ) const
    {
        return Vec2( this->_x * f, this->_y * f );
    }

    float DistanceToSquared( const Vec2 p ) const
    {
        const float dX = p._x - this->_x;
        const float dY = p._y - this->_y;

        return dX * dX + dY * dY;
    }

    float DistanceTo( const Vec2 p ) const
    {
        return sqrt( this->DistanceToSquared( p ) );
    }

    float DotProduct( const Vec2 p ) const
    {
        return this->_x * p._x + this->_y * p._y;
    }
};

// return minimum distance between line segment vw and point p, and the closest point on the line segment, q
float DistanceFromLineSegmentToPoint( const Vec2 v, const Vec2 w, const Vec2 p, Vec2 * const q )
{
    const float distSq = v.DistanceToSquared( w ); // i.e. |w-v|^2 ... avoid a sqrt
    if ( distSq == 0.0 )
    {
        // v == w case
        (*q) = v;

        return v.DistanceTo( p );
    }

    // 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

    const float t = ( p - v ).DotProduct( w - v ) / distSq;
    if ( t < 0.0 )
    {
        // beyond the v end of the segment
        (*q) = v;

        return v.DistanceTo( p );
    }
    else if ( t > 1.0 )
    {
        // beyond the w end of the segment
        (*q) = w;

        return w.DistanceTo( p );
    }

    // projection falls on the segment
    const Vec2 projection = v + ( ( w - v ) * t );

    (*q) = projection;

    return p.DistanceTo( projection );
}

float DistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY, float *qX, float *qY )
{
    Vec2 q;

    float distance = DistanceFromLineSegmentToPoint( Vec2( segmentX1, segmentY1 ), Vec2( segmentX2, segmentY2 ), Vec2( pX, pY ), &q );

    (*qX) = q._x;
    (*qY) = q._y;

    return distance;
}

void TestDistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY )
{
    float qX;
    float qY;
    float d = DistanceFromLineSegmentToPoint( segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, &qX, &qY );
    printf( "line segment = ( ( %f, %f ), ( %f, %f ) ), p = ( %f, %f ), distance = %f, q = ( %f, %f )\n",
            segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, d, qX, qY );
}

void TestDistanceFromLineSegmentToPoint()
{
    TestDistanceFromLineSegmentToPoint( 0, 0, 1, 1, 1, 0 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 5, 4 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 30, 15 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, -30, 15 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 10, 0, 5, 1 );
    TestDistanceFromLineSegmentToPoint( 0, 0, 0, 10, 1, 5 );
}

JavaScript中一个基于这个公式的更简洁的解决方案:

distToSegment: function (point, linePointA, linePointB){

    var x0 = point.X;
    var y0 = point.Y;

    var x1 = linePointA.X;
    var y1 = linePointA.Y;

    var x2 = linePointB.X;
    var y2 = linePointB.Y;

    var Dx = (x2 - x1);
    var Dy = (y2 - y1);

    var numerator = Math.abs(Dy*x0 - Dx*y0 - x1*y2 + x2*y1);
    var denominator = Math.sqrt(Dx*Dx + Dy*Dy);
    if (denominator == 0) {
        return this.dist2(point, linePointA);
    }

    return numerator/denominator;

}

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)
}



 

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));
}

和这个答案一样，只是用的是Visual Basic。使其可作为Microsoft Excel和VBA/宏中的用户定义函数使用。

函数返回点(x,y)到由(x1,y1)和(x2,y2)定义的线段的最近距离。

Function DistanceToSegment(x As Double, y As Double, x1 As Double, y1 As Double, x2 As Double, y2 As Double)

  Dim A As Double
  A = x - x1
  Dim B As Double
  B = y - y1
  Dim C  As Double
  C = x2 - x1
  Dim D As Double
  D = y2 - y1

  Dim dot As Double
  dot = A * C + B * D
  Dim len_sq As Double
  len_sq = C * C + D * D
  Dim param As Double
  param = -1

  If (len_sq <> 0) Then
      param = dot / len_sq
  End If

  Dim xx As Double
  Dim yy As Double

  If (param < 0) Then
    xx = x1
    yy = y1
  ElseIf (param > 1) Then
    xx = x2
    yy = y2
  Else
    xx = x1 + param * C
    yy = y1 + param * D
  End If

  Dim dx As Double
  dx = x - xx
  Dim dy As Double
  dy = y - yy

  DistanceToSegment = Math.Sqr(dx * dx + dy * dy)

End Function