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

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);

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;
}

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。

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

//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;
    }