只是遇到了这个，我想我应该添加一个Lua实现。它假设点以表{x=xVal, y=yVal}给出，直线或线段由包含两个点的表给出(见下面的例子):

function distance( P1, P2 )
    return math.sqrt((P1.x-P2.x)^2 + (P1.y-P2.y)^2)
end

-- Returns false if the point lies beyond the reaches of the segment
function distPointToSegment( line, P )
    if line[1].x == line[2].x and line[1].y == line[2].y then
        print("Error: Not a line!")
        return false
    end

    local d = distance( line[1], line[2] )

    local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)

    local projection = {}
    projection.x = line[1].x + t*(line[2].x-line[1].x)
    projection.y = line[1].y + t*(line[2].y-line[1].y)

    if t >= 0 and t <= 1 then   -- within line segment?
        return distance( projection, {x=P.x, y=P.y} )
    else
        return false
    end
end

-- Returns value even if point is further down the line (outside segment)
function distPointToLine( line, P )
    if line[1].x == line[2].x and line[1].y == line[2].y then
        print("Error: Not a line!")
        return false
    end

    local d = distance( line[1], line[2] )

    local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)

    local projection = {}
    projection.x = line[1].x + t*(line[2].x-line[1].x)
    projection.y = line[1].y + t*(line[2].y-line[1].y)

    return distance( projection, {x=P.x, y=P.y} )
end

使用示例:

local P1 = {x = 0, y = 0}
local P2 = {x = 10, y = 10}
local line = { P1, P2 }
local P3 = {x = 7, y = 15}
print(distPointToLine( line, P3 ))  -- prints 5.6568542494924
print(distPointToSegment( line, P3 )) -- prints false

该算法基于求出指定直线与包含指定点的正交直线的交点，并计算其距离。在线段的情况下，我们必须检查交点是否在线段的点之间，如果不是这样，则最小距离是指定点与线段的一个端点之间的距离。这是一个c#实现。

Double Distance(Point a, Point b)
{
    double xdiff = a.X - b.X, ydiff = a.Y - b.Y;
    return Math.Sqrt((long)xdiff * xdiff + (long)ydiff * ydiff);
}

Boolean IsBetween(double x, double a, double b)
{
    return ((a <= b && x >= a && x <= b) || (a > b && x <= a && x >= b));
}

Double GetDistance(Point pt, Point pt1, Point pt2, out Point intersection)
{
    Double a, x, y, R;

    if (pt1.X != pt2.X) {
        a = (double)(pt2.Y - pt1.Y) / (pt2.X - pt1.X);
        x = (a * (pt.Y - pt1.Y) + a * a * pt1.X + pt.X) / (a * a + 1);
        y = a * x + pt1.Y - a * pt1.X; }
    else { x = pt1.X;  y = pt.Y; }

    if (IsBetween(x, pt1.X, pt2.X) && IsBetween(y, pt1.Y, pt2.Y)) {
        intersection = new Point((int)x, (int)y);
        R = Distance(intersection, pt); }
    else {
        double d1 = Distance(pt, pt1), d2 = Distance(pt, pt2);
        if (d1 < d2) { intersection = pt1; R = d1; }
        else { intersection = pt2; R = d2; }}

    return R;
}

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)
}

公认的答案行不通 (例如，0,0和(-10,2,10,2)之间的距离应为2)。

下面是工作代码:

   def dist2line2(x,y,line):
     x1,y1,x2,y2=line
     vx = x1 - x
     vy = y1 - y
     ux = x2-x1
     uy = y2-y1
     length = ux * ux + uy * uy
     det = (-vx * ux) + (-vy * uy) #//if this is < 0 or > length then its outside the line segment
     if det < 0:
       return (x1 - x)**2 + (y1 - y)**2
     if det > length:
       return (x2 - x)**2 + (y2 - y)**2
     det = ux * vy - uy * vx
     return det**2 / length
   def dist2line(x,y,line): return math.sqrt(dist2line2(x,y,line))

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

用Matlab直接实现Grumdrig

function ans=distP2S(px,py,vx,vy,wx,wy)
% [px py vx vy wx wy]
  t=( (px-vx)*(wx-vx)+(py-vy)*(wy-vy) )/idist(vx,wx,vy,wy)^2;
  [idist(px,vx,py,vy) idist(px,vx+t*(wx-vx),py,vy+t*(wy-vy)) idist(px,wx,py,wy) ];
  ans(1+(t>0)+(t>1)); % <0 0<=t<=1 t>1     
 end

function d=idist(a,b,c,d)
 d=abs(a-b+1i*(c-d));
end