下面是SQL实现，以km为单位计算距离，

SELECT UserId, ( 3959 * acos( cos( radians( your latitude here ) ) * cos( radians(latitude) ) * 
cos( radians(longitude) - radians( your longitude here ) ) + sin( radians( your latitude here ) ) * 
sin( radians(latitude) ) ) ) AS distance FROM user HAVING
distance < 5  ORDER BY distance LIMIT 0 , 5;

要获得通过编程语言实现的更多细节，您可以浏览这里给出的php脚本

这个脚本[在PHP中]计算两点之间的距离。

public static function getDistanceOfTwoPoints($source, $dest, $unit='K') {
        $lat1 = $source[0];
        $lon1 = $source[1];
        $lat2 = $dest[0];
        $lon2 = $dest[1];

        $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 ($unit == "M")
        {
            return ($miles * 1.609344 * 1000);
        }
        else if ($unit == "N") {
            return ($miles * 0.8684);
        } 
        else {
            return $miles;
        }
    }

下面是一个c#实现:

static class DistanceAlgorithm
{
    const double PIx = 3.141592653589793;
    const double RADIUS = 6378.16;

    /// <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 * RADIUS;
    }

}

下面是一个Scala实现:

  def calculateHaversineDistance(lat1: Double, lon1: Double, lat2: Double, lon2: Double): Double = {
    val long2 = lon2 * math.Pi / 180
    val lat2 = lat2 * math.Pi / 180
    val long1 = lon1 * math.Pi / 180
    val lat1 = lat1 * math.Pi / 180

    val dlon = long2 - long1
    val dlat = lat2 - lat1
    val a = math.pow(math.sin(dlat / 2), 2) + math.cos(lat1) * math.cos(lat2) * math.pow(math.sin(dlon / 2), 2)
    val c = 2 * math.atan2(Math.sqrt(a), math.sqrt(1 - a))
    val haversineDistance = 3961 * c // 3961 = radius of earth in miles
    haversineDistance
  }

下面是postgres SQL中的一个示例(以公里为单位，为英里版本，将1.609344替换为0.8684版本)

CREATE OR REPLACE FUNCTION public.geodistance(alat float, alng float, blat  

float, blng  float)
  RETURNS float AS
$BODY$
DECLARE
    v_distance float;
BEGIN

    v_distance = asin( sqrt(
            sin(radians(blat-alat)/2)^2 
                + (
                    (sin(radians(blng-alng)/2)^2) *
                    cos(radians(alat)) *
                    cos(radians(blat))
                )
          )
        ) * cast('7926.3352' as float) * cast('1.609344' as float) ;


    RETURN v_distance;
END 
$BODY$
language plpgsql VOLATILE SECURITY DEFINER;
alter function geodistance(alat float, alng float, blat float, blng float)
owner to postgres;

仅限飞镖：

import 'dart:math' show cos, sqrt, asin;

double calculateDistance(LatLng l1, LatLng l2) {
  const p = 0.017453292519943295;
  final a = 0.5 -
      cos((l2.latitude - l1.latitude) * p) / 2 +
      cos(l1.latitude * p) *
          cos(l2.latitude * p) *
          (1 - cos((l2.longitude - l1.longitude) * p)) /
          2;
  return 12742 * asin(sqrt(a));
}