如何计算Google地图V3中两个标记之间的距离? (类似于V2中的distanceFrom
函数。)
谢谢..
答案 0 :(得分:426)
如果您想自己计算,那么您可以使用Haversine公式:
var rad = function(x) {
return x * Math.PI / 180;
};
var getDistance = function(p1, p2) {
var R = 6378137; // Earth’s mean radius in meter
var dLat = rad(p2.lat() - p1.lat());
var dLong = rad(p2.lng() - p1.lng());
var a = 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 = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
var d = R * c;
return d; // returns the distance in meter
};
答案 1 :(得分:299)
GMap3中似乎有一种方法。它是google.maps.geometry.spherical
命名空间的静态方法。
它将两个LatLng
个对象作为参数,并使用默认的地球半径6378137米,但必要时可以使用自定义值覆盖默认半径。
确保包括:
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&v=3&libraries=geometry"></script>
在你的头部。
电话会是:
google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);
答案 2 :(得分:34)
答案 3 :(得分:26)
使用GPS经纬度为2点的示例。
var latitude1 = 39.46;
var longitude1 = -0.36;
var latitude2 = 40.40;
var longitude2 = -3.68;
var distance = google.maps.geometry.spherical.computeDistanceBetween(new google.maps.LatLng(latitude1, longitude1), new google.maps.LatLng(latitude2, longitude2));
答案 4 :(得分:13)
只需将其添加到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);
}
然后使用这样的函数:
var loc1 = new GLatLng(52.5773139, 1.3712427);
var loc2 = new GLatLng(52.4788314, 1.7577444);
var dist = loc2.distanceFrom(loc1);
alert(dist/1000);
答案 5 :(得分:12)
//p1 and p2 are google.maps.LatLng(x,y) objects
function calcDistance(p1, p2) {
var d = (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
console.log(d);
}
答案 6 :(得分:11)
以下是此公式的c#实现
public class DistanceAlgorithm
{
const double PIx = 3.141592653589793;
const double RADIO = 6378.16;
/// <summary>
/// This class cannot be instantiated.
/// </summary>
private DistanceAlgorithm() { }
/// <summary>
/// Convert degrees to Radians
/// </summary>
/// <param name="x">Degrees</param>
/// <returns>The equivalent in radians</returns>
public static double Radians(double x)
{
return x * PIx / 180;
}
/// <summary>
/// Calculate the distance between two places.
/// </summary>
/// <param name="lon1"></param>
/// <param name="lat1"></param>
/// <param name="lon2"></param>
/// <param name="lat2"></param>
/// <returns></returns>
public static double DistanceBetweenPlaces(
double lon1,
double lat1,
double lon2,
double lat2)
{
double dlon = Radians(lon2 - lon1);
double dlat = Radians(lat2 - lat1);
double a = (Math.Sin(dlat / 2) * Math.Sin(dlat / 2)) + Math.Cos(Radians(lat1)) * Math.Cos(Radians(lat2)) * (Math.Sin(dlon / 2) * Math.Sin(dlon / 2));
double angle = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
return (angle * RADIO) * 0.62137;//distance in miles
}
}
答案 7 :(得分:11)
使用Google,您可以使用spherical api,google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);
来完成此操作。
但是,如果球形投影或半正弦解决方案的精度对您来说不够精确(例如,如果您接近极点或计算更长的距离),则应使用不同的库。
我在维基百科here上找到的关于这个主题的大部分信息。
查看任何给定算法的精度是否足够的技巧是填写地球的最大和最小半径,看看差异是否可能导致用例出现问题。可以在this article
中找到更多详细信息最终,google api或者hasrsine可以毫无问题地用于大多数目的。
答案 8 :(得分:9)
使用PHP,您可以使用这个简单的函数计算距离:
// to calculate distance between two lat & lon function calculate_distance($lat1, $lon1, $lat2, $lon2, $unit='N') { $theta = $lon1 - $lon2; $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta)); $dist = acos($dist); $dist = rad2deg($dist); $miles = $dist * 60 * 1.1515; $unit = strtoupper($unit); if ($unit == "K") { return ($miles * 1.609344); } else if ($unit == "N") { return ($miles * 0.8684); } else { return $miles; } } // function ends here
答案 9 :(得分:8)
使用Vincenty算法implemented in Javascript可以增加处理时间,从而获得良好的准确性。
答案 10 :(得分:8)
离线解决方案 - Haversine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;
}
}
答案 11 :(得分:3)
在我的情况下,最好在SQL Server中计算,因为我想获取当前位置,然后搜索距离当前位置一定距离内的所有邮政编码。我还有一个DB,其中包含一个邮政编码列表和他们的lat long。干杯
--will return the radius for a given number
create function getRad(@variable float)--function to return rad
returns float
as
begin
declare @retval float
select @retval=(@variable * PI()/180)
--print @retval
return @retval
end
go
--calc distance
--drop function dbo.getDistance
create function getDistance(@cLat float,@cLong float, @tLat float, @tLong float)
returns float
as
begin
declare @emr float
declare @dLat float
declare @dLong float
declare @a float
declare @distance float
declare @c float
set @emr = 6371--earth mean
set @dLat = dbo.getRad(@tLat - @cLat);
set @dLong = dbo.getRad(@tLong - @cLong);
set @a = sin(@dLat/2)*sin(@dLat/2)+cos(dbo.getRad(@cLat))*cos(dbo.getRad(@tLat))*sin(@dLong/2)*sin(@dLong/2);
set @c = 2*atn2(sqrt(@a),sqrt(1-@a))
set @distance = @emr*@c;
set @distance = @distance * 0.621371 -- i needed it in miles
--print @distance
return @distance;
end
go
--get all zipcodes within 2 miles, the hardcoded #'s would be passed in by C#
select *
from cityzips a where dbo.getDistance(29.76,-95.38,a.lat,a.long) <3
order by zipcode
答案 12 :(得分:3)
//JAVA
public Double getDistanceBetweenTwoPoints(Double latitude1, Double longitude1, Double latitude2, Double longitude2) {
final int RADIUS_EARTH = 6371;
double dLat = getRad(latitude2 - latitude1);
double dLong = getRad(longitude2 - longitude1);
double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(getRad(latitude1)) * Math.cos(getRad(latitude2)) * Math.sin(dLong / 2) * Math.sin(dLong / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return (RADIUS_EARTH * c) * 1000;
}
private Double getRad(Double x) {
return x * Math.PI / 180;
}
答案 13 :(得分:3)
必须这样做......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;
}
答案 14 :(得分:2)
使用 Google Distance Matrix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。 并应用这个简单的功能
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是您的位置,destinationA是destindation location.you可以添加以上两个或更多数据。
Rad Full Documentation ,带有示例
答案 15 :(得分:2)
/**
* 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80%正确的版本,但是插入了错误的参数并且与输入内容不一致,此版本已完全解决
答案 16 :(得分:1)
请参阅GLatLng对象上的distanceFrom函数;函数参数在v2和v3之间略有变化。
答案 17 :(得分:1)
然后检查路由[0] [0]。里面有一个估计的持续时间和估计行进距离。
答案 18 :(得分:0)
答案 19 :(得分:0)
要在Google地图上计算距离,您可以使用Directions API。这将是最简单的方法之一。要从Google Server获取数据,您可以使用Retrofit或Volley。两者都有自己的优势。看一下我使用改造实现它的代码:
private void build_retrofit_and_get_response(String type) {
String url = "https://maps.googleapis.com/maps/";
Retrofit retrofit = new Retrofit.Builder()
.baseUrl(url)
.addConverterFactory(GsonConverterFactory.create())
.build();
RetrofitMaps service = retrofit.create(RetrofitMaps.class);
Call<Example> call = service.getDistanceDuration("metric", origin.latitude + "," + origin.longitude,dest.latitude + "," + dest.longitude, type);
call.enqueue(new Callback<Example>() {
@Override
public void onResponse(Response<Example> response, Retrofit retrofit) {
try {
//Remove previous line from map
if (line != null) {
line.remove();
}
// This loop will go through all the results and add marker on each location.
for (int i = 0; i < response.body().getRoutes().size(); i++) {
String distance = response.body().getRoutes().get(i).getLegs().get(i).getDistance().getText();
String time = response.body().getRoutes().get(i).getLegs().get(i).getDuration().getText();
ShowDistanceDuration.setText("Distance:" + distance + ", Duration:" + time);
String encodedString = response.body().getRoutes().get(0).getOverviewPolyline().getPoints();
List<LatLng> list = decodePoly(encodedString);
line = mMap.addPolyline(new PolylineOptions()
.addAll(list)
.width(20)
.color(Color.RED)
.geodesic(true)
);
}
} catch (Exception e) {
Log.d("onResponse", "There is an error");
e.printStackTrace();
}
}
@Override
public void onFailure(Throwable t) {
Log.d("onFailure", t.toString());
}
});
}
以上是用于计算距离的函数build_retrofit_and_get_response的代码。下面是相应的Retrofit Interface:
package com.androidtutorialpoint.googlemapsdistancecalculator;
import com.androidtutorialpoint.googlemapsdistancecalculator.POJO.Example;
import retrofit.Call;
import retrofit.http.GET;
import retrofit.http.Query;
public interface RetrofitMaps {
/*
* Retrofit get annotation with our URL
* And our method that will return us details of student.
*/
@GET("api/directions/json?key=AIzaSyC22GfkHu9FdgT9SwdCWMwKX1a4aohGifM")
Call<Example> getDistanceDuration(@Query("units") String units, @Query("origin") String origin, @Query("destination") String destination, @Query("mode") String mode);
}
我希望这可以解释您的查询。一切顺利:)
答案 20 :(得分:0)
首先,你是指整个路径的长度还是你只想知道位移(直线距离)?我看没有人在这里指出距离和位移之间的区别。对于距离计算由 JSON/XML 数据给出的每个路线点,对于位移,有一个使用 Spherical
类
//calculates distance between two points in km's
function calcDistance(p1, p2) {
return (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
}