javascript - 确定点是否在(或足够接近)线段上

我想确定鼠标单击点是否位于SVG折线上。我找到this Python code来确定一个点是否位于其他两个点之间，并在JavaScript中重新实现了它。

function isOnLine(xp, yp, x1, y1, x2, y2){

    var p = new Point(x, y);
    var epsilon = 0.01;

    var crossProduct = (yp - y1) * (x2 - x1) - (xp - x1) * (y2 - y1);
    if(Math.abs(crossProduct) > epsilon)
        return false;
    var dotProduct = (xp - x1) * (x2 - x1) + (yp - y1)*(y2 - y1);
    if(dotProduct < 0)
        return false;
    var squaredLengthBA = (x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1);
    if(dotProduct > squaredLengthBA)
        return false;

    return true;
}

但这不能按我想要的方式工作，因为我永远都不会把鼠标指针完全放在线上。因此，我需要像“假想粗线”这样的东西来获得一定的误差幅度：



有任何想法吗？

最佳答案

cross product除以线的长度即可得出点与线的距离。因此，只需将其与某个阈值进行比较：

function isOnLine (xp, yp, x1, y1, x2, y2, maxDistance) {
    var dxL = x2 - x1, dyL = y2 - y1;  // line: vector from (x1,y1) to (x2,y2)
    var dxP = xp - x1, dyP = yp - y1;  // point: vector from (x1,y1) to (xp,yp)

    var squareLen = dxL * dxL + dyL * dyL;  // squared length of line
    var dotProd   = dxP * dxL + dyP * dyL;  // squared distance of point from (x1,y1) along line
    var crossProd = dyP * dxL - dxP * dyL;  // area of parallelogram defined by line and point

    // perpendicular distance of point from line
    var distance = Math.abs(crossProd) / Math.sqrt(squareLen);

    return (distance <= maxDistance && dotProd >= 0 && dotProd <= squareLen);
}

附言上面的代码实际上通过将线段的任一边加maxDistance来将线段扩展为一个框，并接受该框中的任何单击。如果要将其应用于折线（即，多个线段首尾相连），则可能会发现这些框之间存在间隙，其中两个线段以一定角度相交：



解决此问题的一种简单自然的方法是也接受端点在maxDistance半径之内的任何单击，基本上是用（半）圆形端盖填充虚构的框：



这是实现此目的的一种方法：

function isOnLineWithEndCaps (xp, yp, x1, y1, x2, y2, maxDistance) {
    var dxL = x2 - x1, dyL = y2 - y1;  // line: vector from (x1,y1) to (x2,y2)
    var dxP = xp - x1, dyP = yp - y1;  // point: vector from (x1,y1) to (xp,yp)
    var dxQ = xp - x2, dyQ = yp - y2;  // extra: vector from (x2,y2) to (xp,yp)

    var squareLen = dxL * dxL + dyL * dyL;  // squared length of line
    var dotProd   = dxP * dxL + dyP * dyL;  // squared distance of point from (x1,y1) along line
    var crossProd = dyP * dxL - dxP * dyL;  // area of parallelogram defined by line and point

    // perpendicular distance of point from line
    var distance = Math.abs(crossProd) / Math.sqrt(squareLen);

    // distance of (xp,yp) from (x1,y1) and (x2,y2)
    var distFromEnd1 = Math.sqrt(dxP * dxP + dyP * dyP);
    var distFromEnd2 = Math.sqrt(dxQ * dxQ + dyQ * dyQ);

    // if the point lies beyond the ends of the line, check if
    // it's within maxDistance of the closest end point
    if (dotProd < 0) return distFromEnd1 <= maxDistance;
    if (dotProd > squareLen) return distFromEnd2 <= maxDistance;

    // else check if it's within maxDistance of the line
    return distance <= maxDistance;
}

关于javascript - 确定点是否在(或足够接近)线段上，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/34474336/