Google地图：在多边形内是纬度/经度？ [英] Google Map: is a lat/lng within a polygon?

问题描述

给定一对经纬度值，我如何确定这个对是否在一个多边形内？我需要在PHP中执行此操作。我发现Google Maps API有一个 containsLocation 方法： https： //developers.google.com/maps/documentation/javascript/reference 。有没有一种方法可以从PHP中利用它？

解决方案

查找点是否在多边形中的一种方法是计算很多时候从点（任何方向）绘制的线与多边形边界相交。如果它们交叉偶数次，那么这个点就在外面。

我已经从这个 指向Polygon 的文章，并使用下面的多边形进行说明。

 <？php 
 // Point-In-Polygon Algorithm 
 $ polySides = 4; //多边形有多少个角
 $ polyX = array（4,9,11,2）; //角的水平坐标
 $ polyY = array（10,7,2,2）; //角的垂直坐标
 $ x = 3.5; 
 $ y = 13.5; // 
 // $ y = 3.5之外// // 
 
函数pointInPolygon（$ polySides，$ polyX，$ polyY，$ x，$ y）{
 $ j = $ polySides-1; 
 $ oddNodes = 0;如果（$ polyY [$ i]< $ y&& $ polyY [$ j]> =（$ i $ 0; $ i <$ polySides; $ i ++） $ y 
 || $ polyY [$ j]< $ y&& $ polyY [$ i]> = $ y）{
 if（$ polyX [$ i] +（$ $ $ polyY [$ i]）/（$ polyY [$ j]  -  $ polyY [$ i]）*（$ polyX [$ j]  -  $ polyX [$ i]）<$ x）{
 $ oddNodes =！$ oddNodes; }} 
 $ j = $ i; } 
 
返回$ oddNodes; } 
 
 $ b $ if（pointInPolygon（$ polySides，$ polyX，$ polyY，$ x，$ y））{
 echoIs in polygon！; 
} 
 else echo不在多边形中; 
？> 
  

Given a pair of lat/lng values, how do I determine if the pair is within a polygon? I need to do this in PHP. I see that Google Maps API has a containsLocation method: https://developers.google.com/maps/documentation/javascript/reference. Is there a way to leverage this from PHP?

解决方案

One way to find if a point is in a polygon is to count how many times a line drawn from the point (in any direction) intersects with the polygon boundary. If they intersect an even number of times, then the point is outside.

I have implemented the C code from this Point in Polygon article in php and used the polygon below to illustrate.

<?php
//Point-In-Polygon Algorithm
$polySides  = 4; //how many corners the polygon has
$polyX    =  array(4,9,11,2);//horizontal coordinates of corners
$polyY    =  array(10,7,2,2);//vertical coordinates of corners
$x = 3.5;
$y = 13.5;//Outside
//$y = 3.5;//Inside

function pointInPolygon($polySides,$polyX,$polyY,$x,$y) {
  $j = $polySides-1 ;
  $oddNodes = 0;
  for ($i=0; $i<$polySides; $i++) {
    if ($polyY[$i]<$y && $polyY[$j]>=$y 
 ||  $polyY[$j]<$y && $polyY[$i]>=$y) {
    if ($polyX[$i]+($y-$polyY[$i])/($polyY[$j]-$polyY[$i])*($polyX[$j]-$polyX[$i])<$x)    {
    $oddNodes=!$oddNodes; }}
   $j=$i; }

  return $oddNodes; }


 if (pointInPolygon($polySides,$polyX,$polyY,$x,$y)){
  echo "Is in polygon!";
}
else echo "Is not in polygon";
?>

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