Question

我正在使用此处给出的算法（很多其他人也在使用It）。但是在我的特定场景下，它显示的是多个多边形的相交。

我正在使用php，并在mysql中导入了geojson。并使用以下代码 https://assemblysys.com/php-point-in-polygon-algorithm/

<?php
/*
Description: The point-in-polygon algorithm allows you to check if a point is
inside a polygon or outside of it.
Author: Michaël Niessen (2009)
Website: http://AssemblySys.com

If you find this script useful, you can show your
appreciation by getting Michaël a cup of coffee ;)
PayPal: https://www.paypal.me/MichaelNiessen

As long as this notice (including author name and details) is included and
UNALTERED, this code is licensed under the GNU General Public License version 3:
http://www.gnu.org/licenses/gpl.html
*/

class pointLocation {
    var $pointOnVertex = true; // Check if the point sits exactly on one of the vertices?

    function pointLocation() {
    }

    function pointInPolygon($point, $polygon, $pointOnVertex = true) {
        $this->pointOnVertex = $pointOnVertex;

        // Transform string coordinates into arrays with x and y values
        $point = $this->pointStringToCoordinates($point);
        $vertices = array(); 
        foreach ($polygon as $vertex) {
            $vertices[] = $this->pointStringToCoordinates($vertex); 
        }

        // Check if the point sits exactly on a vertex
        if ($this->pointOnVertex == true and $this->pointOnVertex($point, $vertices) == true) {
            return "vertex";
        }

        // Check if the point is inside the polygon or on the boundary
        $intersections = 0; 
        $vertices_count = count($vertices);

        for ($i=1; $i < $vertices_count; $i++) {
            $vertex1 = $vertices[$i-1]; 
            $vertex2 = $vertices[$i];
            if ($vertex1['y'] == $vertex2['y'] and $vertex1['y'] == $point['y'] and $point['x'] > min($vertex1['x'], $vertex2['x']) and $point['x'] < max($vertex1['x'], $vertex2['x'])) { // Check if point is on an horizontal polygon boundary
                return "boundary";
            }
            if ($point['y'] > min($vertex1['y'], $vertex2['y']) and $point['y'] <= max($vertex1['y'], $vertex2['y']) and $point['x'] <= max($vertex1['x'], $vertex2['x']) and $vertex1['y'] != $vertex2['y']) { 
                $xinters = ($point['y'] - $vertex1['y']) * ($vertex2['x'] - $vertex1['x']) / ($vertex2['y'] - $vertex1['y']) + $vertex1['x']; 
                if ($xinters == $point['x']) { // Check if point is on the polygon boundary (other than horizontal)
                    return "boundary";
                }
                if ($vertex1['x'] == $vertex2['x'] || $point['x'] <= $xinters) {
                    $intersections++; 
                }
            } 
        } 
        // If the number of edges we passed through is odd, then it's in the polygon. 
        if ($intersections % 2 != 0) {
            return "inside";
        } else {
            return "outside";
        }
    }

    function pointOnVertex($point, $vertices) {
        foreach($vertices as $vertex) {
            if ($point == $vertex) {
                return true;
            }
        }

    }

    function pointStringToCoordinates($pointString) {
        $coordinates = explode(" ", $pointString);
        return array("x" => $coordinates[0], "y" => $coordinates[1]);
    }

}
?>

上面的代码在大多数点上都可以正常工作，但是在某些情况下，返回的是给定点存在于多个多边形中（将其绘制为80公里）。所以我的疑问是。 1-我在使用正确的算法解决我的问题吗？ 2-上面代码中下面几行的目的是什么？

$xinters = ($point['y'] - $vertex1['y']) * ($vertex2['x'] - $vertex1['x']) / ($vertex2['y'] - $vertex1['y']) + $vertex1['x'];

Answer 1

回答我自己的问题。由于我没有得到任何可靠的解决方案，因此我不得不提出一种解决方法。我从xinters中减去了point [x]，将它们放在一个数组中并选择了最短的一个。这样可以确保相交的多边形是我一直在寻找的多边形。

如何确定仅在一个多边形中是否存在一个点，而不是在相邻/其他附近的多边形中存在一个点？

1 个答案: