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

现在我的解决方案...... (Javascript)

这是非常快的，因为我试图避免任何数学。战俘的功能。

如你所见，在函数的最后，我得到了直线的距离。

代码来自lib http://www.draw2d.org/graphiti/jsdoc/#!/例子

/**
 * Static util function to determine is a point(px,py) on the line(x1,y1,x2,y2)
 * A simple hit test.
 * 
 * @return {boolean}
 * @static
 * @private
 * @param {Number} coronaWidth the accepted corona for the hit test
 * @param {Number} X1 x coordinate of the start point of the line
 * @param {Number} Y1 y coordinate of the start point of the line
 * @param {Number} X2 x coordinate of the end point of the line
 * @param {Number} Y2 y coordinate of the end point of the line
 * @param {Number} px x coordinate of the point to test
 * @param {Number} py y coordinate of the point to test
 **/
graphiti.shape.basic.Line.hit= function( coronaWidth, X1, Y1,  X2,  Y2, px, py)
{
  // Adjust vectors relative to X1,Y1
  // X2,Y2 becomes relative vector from X1,Y1 to end of segment
  X2 -= X1;
  Y2 -= Y1;
  // px,py becomes relative vector from X1,Y1 to test point
  px -= X1;
  py -= Y1;
  var dotprod = px * X2 + py * Y2;
  var projlenSq;
  if (dotprod <= 0.0) {
      // px,py is on the side of X1,Y1 away from X2,Y2
      // distance to segment is length of px,py vector
      // "length of its (clipped) projection" is now 0.0
      projlenSq = 0.0;
  } else {
      // switch to backwards vectors relative to X2,Y2
      // X2,Y2 are already the negative of X1,Y1=>X2,Y2
      // to get px,py to be the negative of px,py=>X2,Y2
      // the dot product of two negated vectors is the same
      // as the dot product of the two normal vectors
      px = X2 - px;
      py = Y2 - py;
      dotprod = px * X2 + py * Y2;
      if (dotprod <= 0.0) {
          // px,py is on the side of X2,Y2 away from X1,Y1
          // distance to segment is length of (backwards) px,py vector
          // "length of its (clipped) projection" is now 0.0
          projlenSq = 0.0;
      } else {
          // px,py is between X1,Y1 and X2,Y2
          // dotprod is the length of the px,py vector
          // projected on the X2,Y2=>X1,Y1 vector times the
          // length of the X2,Y2=>X1,Y1 vector
          projlenSq = dotprod * dotprod / (X2 * X2 + Y2 * Y2);
      }
  }
    // Distance to line is now the length of the relative point
    // vector minus the length of its projection onto the line
    // (which is zero if the projection falls outside the range
    //  of the line segment).
    var lenSq = px * px + py * py - projlenSq;
    if (lenSq < 0) {
        lenSq = 0;
    }
    return Math.sqrt(lenSq)<coronaWidth;
};

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

GLSL版:

// line (a -> b ) point p[enter image description here][1]
float distanceToLine(vec2 a, vec2 b, vec2 p) {
    float aside = dot((p - a),(b - a));
    if(aside< 0.0) return length(p-a);
    float bside = dot((p - b),(a - b));
    if(bside< 0.0) return length(p-b);
    vec2 pointOnLine = (bside*a + aside*b)/pow(length(a-b),2.0);
    return length(p - pointOnLine);
}

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)
}



 

省道和颤振的解决方法:

import 'dart:math' as math;
 class Utils {
   static double shortestDistance(Point p1, Point p2, Point p3){
      double px = p2.x - p1.x;
      double py = p2.y - p1.y;
      double temp = (px*px) + (py*py);
      double u = ((p3.x - p1.x)*px + (p3.y - p1.y)* py) /temp;
      if(u>1){
        u=1;
      }
      else if(u<0){
        u=0;
      }
      double x = p1.x + u*px;
      double y = p1.y + u*py;
      double dx = x - p3.x;
      double dy = y - p3.y;
      double dist = math.sqrt(dx*dx+dy*dy);
      return dist;
   }
}

class Point {
  double x;
  double y;
  Point(this.x, this.y);
}