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
下面是this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
}
}
其他回答
//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;
}
要在谷歌地图上计算距离,您可以使用方向API。这是最简单的方法之一。要从谷歌服务器获取数据,您可以使用Retrofit或Volley。两者都有各自的优势。看看下面的代码,我使用了改造来实现它:
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());
}
});
}
上面是用于计算距离的函数build_retrofit_and_get_response的代码。下面是对应的改装界面:
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);
}
我希望这能解释你的问题。祝一切顺利:)
来源:谷歌地图距离计算器
在我的情况下,这是最好的计算在SQL Server,因为我想采取当前位置,然后搜索所有邮政编码从当前位置一定距离内。我还有一个数据库,其中包含邮政编码及其纬度的列表。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
/**
* 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。