Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
581 views
in Technique[技术] by (71.8m points)

php - Google Map: is a lat/lng within a polygon?

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?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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.

polygon

<?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";
?>

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...