如何计算谷歌地图V3中两个标记之间的距离?(类似于inV2的distanceFrom函数。)
谢谢. .
如何计算谷歌地图V3中两个标记之间的距离?(类似于inV2的distanceFrom函数。)
谢谢. .
当前回答
使用谷歌距离矩阵服务非常简单
第一步是从谷歌API控制台激活距离矩阵服务。 它返回一组位置之间的距离。 应用这个简单的函数
function initMap() {
var bounds = new google.maps.LatLngBounds;
var markersArray = [];
var origin1 = {lat:23.0203, lng: 72.5562};
//var origin2 = 'Ahmedabad, India';
var destinationA = {lat:23.0436503, lng: 72.55008939999993};
//var destinationB = {lat: 23.2156, lng: 72.6369};
var destinationIcon = 'https://chart.googleapis.com/chart?' +
'chst=d_map_pin_letter&chld=D|FF0000|000000';
var originIcon = 'https://chart.googleapis.com/chart?' +
'chst=d_map_pin_letter&chld=O|FFFF00|000000';
var map = new google.maps.Map(document.getElementById('map'), {
center: {lat: 55.53, lng: 9.4},
zoom: 10
});
var geocoder = new google.maps.Geocoder;
var service = new google.maps.DistanceMatrixService;
service.getDistanceMatrix({
origins: [origin1],
destinations: [destinationA],
travelMode: 'DRIVING',
unitSystem: google.maps.UnitSystem.METRIC,
avoidHighways: false,
avoidTolls: false
}, function(response, status) {
if (status !== 'OK') {
alert('Error was: ' + status);
} else {
var originList = response.originAddresses;
var destinationList = response.destinationAddresses;
var outputDiv = document.getElementById('output');
outputDiv.innerHTML = '';
deleteMarkers(markersArray);
var showGeocodedAddressOnMap = function(asDestination) {
var icon = asDestination ? destinationIcon : originIcon;
return function(results, status) {
if (status === 'OK') {
map.fitBounds(bounds.extend(results[0].geometry.location));
markersArray.push(new google.maps.Marker({
map: map,
position: results[0].geometry.location,
icon: icon
}));
} else {
alert('Geocode was not successful due to: ' + status);
}
};
};
for (var i = 0; i < originList.length; i++) {
var results = response.rows[i].elements;
geocoder.geocode({'address': originList[i]},
showGeocodedAddressOnMap(false));
for (var j = 0; j < results.length; j++) {
geocoder.geocode({'address': destinationList[j]},
showGeocodedAddressOnMap(true));
//outputDiv.innerHTML += originList[i] + ' to ' + destinationList[j] + ': ' + results[j].distance.text + ' in ' + results[j].duration.text + '<br>';
outputDiv.innerHTML += results[j].distance.text + '<br>';
}
}
}
});
}
其中origin1是你的位置,destinationA是目的地位置。您可以添加以上两个或多个数据。
Rad完整文档与示例
其他回答
只需将以下内容添加到JavaScript代码的开头:
google.maps.LatLng.prototype.distanceFrom = function(latlng) {
var lat = [this.lat(), latlng.lat()]
var lng = [this.lng(), latlng.lng()]
var R = 6378137;
var dLat = (lat[1]-lat[0]) * Math.PI / 180;
var dLng = (lng[1]-lng[0]) * Math.PI / 180;
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(lat[0] * Math.PI / 180 ) * Math.cos(lat[1] * Math.PI / 180 ) *
Math.sin(dLng/2) * Math.sin(dLng/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
return Math.round(d);
}
然后像这样使用函数:
var loc1 = new GLatLng(52.5773139, 1.3712427);
var loc2 = new GLatLng(52.4788314, 1.7577444);
var dist = loc2.distanceFrom(loc1);
alert(dist/1000);
/**
* Calculates the haversine distance between point A, and B.
* @param {number[]} latlngA [lat, lng] point A
* @param {number[]} latlngB [lat, lng] point B
* @param {boolean} isMiles If we are using miles, else km.
*/
function haversineDistance(latlngA, latlngB, isMiles) {
const squared = x => x * x;
const toRad = x => (x * Math.PI) / 180;
const R = 6371; // Earth’s mean radius in km
const dLat = toRad(latlngB[0] - latlngA[0]);
const dLon = toRad(latlngB[1] - latlngA[1]);
const dLatSin = squared(Math.sin(dLat / 2));
const dLonSin = squared(Math.sin(dLon / 2));
const a = dLatSin +
(Math.cos(toRad(latlngA[0])) * Math.cos(toRad(latlngB[0])) * dLonSin);
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
let distance = R * c;
if (isMiles) distance /= 1.609344;
return distance;
}
我在网上找到了一个版本,80%是正确的,但插入了错误的参数,在使用输入时不一致,这个版本完全解决了这个问题
对于谷歌,你可以使用球面api, google.maps.geometry. sphere.computedistancebetween (latLngA, latLngB);
然而,如果球面投影或哈弗辛解的精度对你来说不够精确(例如,如果你靠近极点或计算更长的距离),你应该使用不同的库。
关于这个主题的大部分信息我都是在维基百科上找到的。
要查看任何给定算法的精度是否足够,一个技巧是填充地球的最大和最小半径,并查看差异是否会给您的用例带来问题。更多细节可以在本文中找到
最后,谷歌api或haversine将毫无问题地满足大多数目的。
不得不这么做……动作脚本方式
//just make sure you pass a number to the function because it would accept you mother in law...
public var rad = function(x:*) {return x*Math.PI/180;}
protected function distHaversine(p1:Object, p2:Object):Number {
var R:int = 6371; // earth's mean radius in km
var dLat:Number = rad(p2.lat() - p1.lat());
var dLong:Number = rad(p2.lng() - p1.lng());
var a:Number = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(rad(p1.lat())) * Math.cos(rad(p2.lat())) * Math.sin(dLong/2) * Math.sin(dLong/2);
var c:Number = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d:Number = R * c;
return d;
}
离线解-哈弗辛算法
在Javascript中
var _eQuatorialEarthRadius = 6378.1370;
var _d2r = (Math.PI / 180.0);
function HaversineInM(lat1, long1, lat2, long2)
{
return (1000.0 * HaversineInKM(lat1, long1, lat2, long2));
}
function HaversineInKM(lat1, long1, lat2, long2)
{
var dlong = (long2 - long1) * _d2r;
var dlat = (lat2 - lat1) * _d2r;
var a = Math.pow(Math.sin(dlat / 2.0), 2.0) + Math.cos(lat1 * _d2r) * Math.cos(lat2 * _d2r) * Math.pow(Math.sin(dlong / 2.0), 2.0);
var c = 2.0 * Math.atan2(Math.sqrt(a), Math.sqrt(1.0 - a));
var d = _eQuatorialEarthRadius * c;
return d;
}
var meLat = -33.922982;
var meLong = 151.083853;
var result1 = HaversineInKM(meLat, meLong, -32.236457779983745, 148.69094705162837);
var result2 = HaversineInKM(meLat, meLong, -33.609020205923713, 150.77061469270831);
C#
using System;
public class Program
{
public static void Main()
{
Console.WriteLine("Hello World");
var meLat = -33.922982;
double meLong = 151.083853;
var result1 = HaversineInM(meLat, meLong, -32.236457779983745, 148.69094705162837);
var result2 = HaversineInM(meLat, meLong, -33.609020205923713, 150.77061469270831);
Console.WriteLine(result1);
Console.WriteLine(result2);
}
static double _eQuatorialEarthRadius = 6378.1370D;
static double _d2r = (Math.PI / 180D);
private static int HaversineInM(double lat1, double long1, double lat2, double long2)
{
return (int)(1000D * HaversineInKM(lat1, long1, lat2, long2));
}
private static double HaversineInKM(double lat1, double long1, double lat2, double long2)
{
double dlong = (long2 - long1) * _d2r;
double dlat = (lat2 - lat1) * _d2r;
double a = Math.Pow(Math.Sin(dlat / 2D), 2D) + Math.Cos(lat1 * _d2r) * Math.Cos(lat2 * _d2r) * Math.Pow(Math.Sin(dlong / 2D), 2D);
double c = 2D * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1D - a));
double d = _eQuatorialEarthRadius * c;
return d;
}
}
参考: https://en.wikipedia.org/wiki/Great-circle_distance