是否有一种算法可以检查标记（marker）是否位于多边形、矩形或圆形内部或外部。我尝试使用这个 链接编写了一个函数，但是没有成功。
3个回答
25
Leaflet中有一个函数可以检查这个。
Polygon.getBounds().contains(MarketLatLng);
- Thesane
4
32这只会检查该点是否在多边形的外接矩形内，而不是检查它是否被多边形本身包含。 - David K. Hess
4为了使其更智能，使用射线投射算法（可在此处解释：http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html）。
实现示例。 - szymonm
它的工作不正确，导致错误的结果。例如：应该返回false，但却返回true。L.polygon(arr).getBounds().contains([35.777846741014, 51.371061187093])，其中arr=[{lat:36.131196609314266, lng:51.34257481466077},{lat:36.0685037997188, lng:51.33158848653577},{lat:35.95241809543124, lng:51.31167576680921},{lat:35.80498248907691, lng:51.55337498555921},{lat:35.768777146507226, lng:51.65019200216077},{lat:35.767105732568375, lng:51.79026768575452},{lat:35.848407125075894, lng:51.884338120324834},{lat:36.0585127164683, lng:51.620666245324834},]; - mehrdad seyrafi
6
如果您在使用PHP，那么这个函数就可以运行。
$c = false;
$vertices_x = array(22.333,22.222,22,444); //latitude points of polygon
$vertices_y = array(75.111,75.2222,76.233); //longitude points of polygon
$points_polygon = count($vertices_x);
$longitude = 23.345; //latitude of point to be checked
$latitude = 75.123; //longitude of point to be checked
if (is_in_polygon($points_polygon, $vertices_x, $vertices_y, $longitude, $latitude)){
echo "Is in polygon!"."<br>";
}
else {
echo "Is not in polygon";
}
function is_in_polygon($points_polygon, $vertices_x, $vertices_y, $longitude_x, $latitude_y) {
$i = $j = $c = 0;
for ($i = 0, $j = $points_polygon-1; $i < $points_polygon; $j = $i++) {
if (($vertices_y[$i] > $latitude_y != ($vertices_y[$j] > $latitude_y)) && ($longitude_x < ($vertices_x[$j] - $vertices_x[$i]) * ($latitude_y - $vertices_y[$i]) / ($vertices_y[$j] - $vertices_y[$i]) + $vertices_x[$i])) {
$c = !$c;
}
}
return $c;
}
- user2393867
1
这个函数也适用于矩形吗？ - user2485649
0
<!-- <!DOCTYPE html>
<html>
<head>
<title>Simple Map</title>
<meta name="viewport" content="initial-scale=1.0, user-scalable=no">
<meta charset="utf-8">
<style>
#map-canvas {
margin: 0;
padding: 0;
width: 700px;
height: 500px;
}
</style>
<script src="https://maps.googleapis.com/maps/api/js?v=3.exp&sensor=false"></script>
<script>
var map;
var marker;
var latlong = [["21.001663","75.486069"],
["20.108977","73.914672"],
["21.1458","79.088155"],
["19.153061","77.305847"],
["20.0831","73.79095"],
["18.52043","73.856744"],
["16.774832","74.486265"],
["16.691308","74.244866"],
["19.876165","75.343314"],
["19.997453","73.789802"],
["20.532949","76.184303"],
["21.013321","75.563972"],
["18.9513","72.82831"],
["18.515752","73.182162"],
["19.075984","72.877656"],
["19.218331","72.97809"],
["19.844006","79.36266"],
["20.745319","78.602195"],
["21.267052","78.577973"],
["18.52043","73.856744"],
["19.96955","79.304654"],
["19.450585","72.799155"],
["18.52043","73.856744"],
["20.745319","78.602195"],
["18.9833","75.7667"],
["21.013321","75.563972"],
["21.1458","79.088155"],
["19.153061","77.305847"]];
function initialize() {
var mapOptions = {
zoom: 4,
center: new google.maps.LatLng(21.7679, 78.8718),
mapTypeId: google.maps.MapTypeId.ROADMAP
};
map = new google.maps.Map(document.getElementById('map-canvas'),
mapOptions);
drawCircle();
}
function drawCircle()
{
var options = {
strokeColor: '#800000',
strokeOpacity: 1.0,
strokeWeight: 1,
fillColor: '#C64D45',
fillOpacity: 0.5,
map: map,
center: new google.maps.LatLng(21.7679, 78.8718),
radius: 100000
};
circle = new google.maps.Circle(options);
var bounds= circle.getBounds();
for (var i = 0; i < latlong.length; i++){
var hello =new google.maps.LatLng(latlong[i][0], latlong[i][1]);
if(bounds.contains(hello))
{
marker= new google.maps.Marker({
position:hello,
});
marker.setMap(map);
console.log("Hi iam in bound");
}
else
{
console.log("Iam not in bound");
}
}
}
google.maps.event.addDomListener(window, 'load', initialize);
</script>
</head>
<body>
<div id="map-canvas"></div>
</body>
</html> -->
- Saurabh Yadav
1
2谢谢，但是你的代码使用的是Google APIs。我想要使用Leaflet APIs来进行检查点。 - user2446592
网页内容由stack overflow 提供, 点击上面的可以查看英文原文，
原文链接
原文链接
contains函数进行矩形边界检查的功能。http://leafletjs.com/reference.html#bounds 不过对于多边形和圆形没有用处。 - Patrick D