实际上GMap3中似乎有一个方法。它是google。maps。geometry。spherical命名空间的静态方法。

它以两个LatLng对象作为参数，并将使用默认的地球半径6378137米，尽管在必要时可以使用自定义值覆盖默认半径。

确保你包括:

<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&v=3&libraries=geometry"></script>

在你的头部部分。

该呼吁将是:

google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);

使用PHP，你可以使用这个简单的函数来计算距离:

// 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

//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;
    }

在PHP中，使用谷歌映射距离矩阵API:

//Get the Driving(Mode) distance between two Geo-location points(Latitude, Longitude) pair.
function get_distance($lat1, $lat2, $long1, $long2)
{
    $url = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=".$lat1.",".$long1."&destinations=".$lat2.",".$long2."&mode=driving"."&units=imperial";
    //You can request distance data for different travel modes, request distance data in different units such as kilometers or miles, and estimate travel time in traffic.
    try{
        $ch = curl_init();
        curl_setopt($ch, CURLOPT_URL, $url);
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
        curl_setopt($ch, CURLOPT_PROXYPORT, 3128);
        curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0);
        curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
        $response = curl_exec($ch);
        curl_close($ch);
        $response_a = json_decode($response, true);
        //Invalid request OR Empty response
        if(isset($response_a['error_message']) || empty($response_a['rows'])) 
            throw new Exception($response_a['error_message']);
        
    } catch(Exception $e){
        //Handle error here.
        return [];
    }
    
    //The unit parameter in the request URL only affects the text displayed within distance fields. The distance fields in response also contain values that are always expressed in meters.
    $dist = $response_a['rows'][0]['elements'][0]['distance']['text'];
    $time = $response_a['rows'][0]['elements'][0]['duration']['text'];

    return ['distance' => $dist, 'time' => $time];
}

参考:距离矩阵API请求和响应

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

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