你可以使用CLLocationDistance中的构建来计算这个:

CLLocation *location1 = [[CLLocation alloc] initWithLatitude:latitude1 longitude:longitude1];
CLLocation *location2 = [[CLLocation alloc] initWithLatitude:latitude2 longitude:longitude2];
[self distanceInMetersFromLocation:location1 toLocation:location2]

- (int)distanceInMetersFromLocation:(CLLocation*)location1 toLocation:(CLLocation*)location2 {
    CLLocationDistance distanceInMeters = [location1 distanceFromLocation:location2];
    return distanceInMeters;
}

在你的例子中，如果你想要公里，只要除以1000。

下面是Haversine公式的typescript实现

static getDistanceFromLatLonInKm(lat1: number, lon1: number, lat2: number, lon2: number): number {
    var deg2Rad = deg => {
        return deg * Math.PI / 180;
    }

    var r = 6371; // Radius of the earth in km
    var dLat = deg2Rad(lat2 - lat1);   
    var dLon = deg2Rad(lon2 - lon1);
    var a =
        Math.sin(dLat / 2) * Math.sin(dLat / 2) +
        Math.cos(deg2Rad(lat1)) * Math.cos(deg2Rad(lat2)) *
        Math.sin(dLon / 2) * Math.sin(dLon / 2);
    var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    var d = r * c; // Distance in km
    return d;
}

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;

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;
        }
    }

这个链接可能对你有帮助，因为它详细介绍了使用哈弗辛公式来计算距离。

摘录:

这个脚本计算两点之间的大圆距离 也就是说，在地球表面上的最短距离-使用 “半正矢”公式。

function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2) {
  var R = 6371; // Radius of the earth in km
  var dLat = deg2rad(lat2-lat1);  // deg2rad below
  var dLon = deg2rad(lon2-lon1); 
  var a = 
    Math.sin(dLat/2) * Math.sin(dLat/2) +
    Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * 
    Math.sin(dLon/2) * Math.sin(dLon/2)
    ; 
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
  var d = R * c; // Distance in km
  return d;
}

function deg2rad(deg) {
  return deg * (Math.PI/180)
}

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;
    }

}