你如何计算谷歌地图V3中两个标记之间的距离？（类似于distanceFromV2中的功能。）
谢谢..
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465374 次
16 回答
502
如果你想自己计算，那么你可以使用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
};
于 2009-10-01T09:07:35.633 回答
320
GMap3中实际上似乎有一个方法。它是命名空间的静态方法google.maps.geometry.spherical。
它将两个LatLng对象作为参数，并将使用 6378137 米的默认地球半径，但如果需要，可以使用自定义值覆盖默认半径。
确保您包括：
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&v=3&libraries=geometry"></script>
在你的头部。
电话将是：
google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);
于 2011-02-02T12:59:31.480 回答
37
使用 GPS 纬度/经度 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));
于 2015-04-07T12:04:37.683 回答
16
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);
}
然后使用这样的功能：
var loc1 = new GLatLng(52.5773139, 1.3712427);
var loc2 = new GLatLng(52.4788314, 1.7577444);
var dist = loc2.distanceFrom(loc1);
alert(dist/1000);
于 2011-03-10T19:31:57.347 回答
14
//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);
}
于 2015-10-12T13:28:21.033 回答
11
这是this forumula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
}
}
于 2011-02-18T11:16:09.293 回答
11
使用 google，您可以使用球形 api来完成google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);。
但是，如果球面投影或半正弦解的精度对您来说不够精确（例如，如果您靠近极点或计算更长的距离），您应该使用不同的库。
我在 Wikipedia here上找到的有关该主题的大部分信息。
查看任何给定算法的精度是否足够的一个技巧是填写地球的最大和最小半径，看看差异是否会导致您的用例出现问题。更多细节可以在这篇文章中找到
最后，google api 或 hasrsine 将毫无问题地服务于大多数用途。
于 2013-01-22T10:39:20.030 回答
9
使用 PHP，您可以使用这个简单的函数计算距离：
// 计算两个纬度和经度之间的距离
函数计算距离（$lat1，$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 ($unit == "K") {
返回 ($miles * 1.609344);
} else if ($unit == "N") {
返回 ($miles * 0.8684);
} 别的 {
返回 $miles；
}
}
// 函数到此结束
于 2012-06-21T07:32:21.800 回答
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;
}
}
于 2016-01-20T19:12:23.447 回答
3
不得不这样做...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;
}
于 2012-12-13T05:06:00.213 回答
3
就我而言，最好在 SQL Server 中进行计算，因为我想获取当前位置，然后搜索距当前位置一定距离内的所有邮政编码。我还有一个数据库，其中包含邮政编码及其经纬度列表。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
于 2014-04-14T19:23:32.493 回答
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;
}
于 2014-06-05T10:51:40.407 回答
1
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。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 是目的地位置。您可以添加以上两个或更多数据。
带有示例的Rad 完整文档
于 2018-06-16T06:34:37.047 回答
1
/**
* 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;
}
我在网上找了一个版本，80%正确，但插入了错误的参数，使用输入不一致，这个版本完全修复了
于 2018-08-07T06:43:03.240 回答
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。看看我使用改造来实现它的以下代码：
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());
}
});
}
以上是计算距离的函数 build_retrofit_and_get_response 的代码。下面是对应的改造界面：
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);
}
我希望这能解释您的查询。祝一切顺利 ：）
来源：谷歌地图距离计算器
于 2016-09-22T12:29:59.593 回答
0
首先，您是指距离为整个路径的长度还是只想知道位移（直线距离）？我看没有人在这里指出距离和位移之间的区别。Spherical对于距离计算 JSON/XML 数据给出的每个路线点，至于位移，有一个使用类的内置解决方案
//calculates distance between two points in km's
function calcDistance(p1, p2) {
return (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
}
于 2021-01-18T04:06:32.890 回答