这是一个简单的javascript函数，从这个链接可能是有用的。不知何故相关，但我们使用谷歌地球javascript插件而不是地图

function getApproximateDistanceUnits(point1, point2) {

    var xs = 0;
    var ys = 0;

    xs = point2.getX() - point1.getX();
    xs = xs * xs;

    ys = point2.getY() - point1.getY();
    ys = ys * ys;

    return Math.sqrt(xs + ys);
}

单位不是距离，而是相对于坐标的比率。还有其他相关的计算，你可以在这里代替getApproximateDistanceUnits函数链接

然后我使用这个函数来查看经纬度是否在半径内

function isMapPlacemarkInRadius(point1, point2, radi) {
    if (point1 && point2) {
        return getApproximateDistanceUnits(point1, point2) <= radi;
    } else {
        return 0;
    }
}

点可以定义为

 $$.getPoint = function(lati, longi) {
        var location = {
            x: 0,
            y: 0,
            getX: function() { return location.x; },
            getY: function() { return location.y; }
        };
        location.x = lati;
        location.y = longi;

        return location;
    };

然后你可以做你的事情，看看一个点是否在一个半径范围内，比如:

 //put it on the map if within the range of a specified radi assuming 100,000,000 units
        var iconpoint = Map.getPoint(pp.latitude, pp.longitude);
        var centerpoint = Map.getPoint(Settings.CenterLatitude, Settings.CenterLongitude);

        //approx ~200 units to show only half of the globe from the default center radius
        if (isMapPlacemarkInRadius(centerpoint, iconpoint, 120)) {
            addPlacemark(pp.latitude, pp.longitude, pp.name);
        }
        else {
            otherSidePlacemarks.push({
                latitude: pp.latitude,
                longitude: pp.longitude,
                name: pp.name
            });

        }

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;

这里有一个用PHP http://www.geodatasource.com/developers/php计算距离的好例子:

 function distance($lat1, $lon1, $lat2, $lon2, $unit) {

     $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;
     }
 }

PIP安装haversine

Python实现

原产地是美国毗连的中心。

from haversine import haversine, Unit
origin = (39.50, 98.35)
paris = (48.8567, 2.3508)
haversine(origin, paris, unit=Unit.MILES)

要得到以千米为单位的答案，只需设置unit= unit。千米(这是默认值)。

这是一个简单的javascript函数，从这个链接可能是有用的。不知何故相关，但我们使用谷歌地球javascript插件而不是地图

function getApproximateDistanceUnits(point1, point2) {

    var xs = 0;
    var ys = 0;

    xs = point2.getX() - point1.getX();
    xs = xs * xs;

    ys = point2.getY() - point1.getY();
    ys = ys * ys;

    return Math.sqrt(xs + ys);
}

单位不是距离，而是相对于坐标的比率。还有其他相关的计算，你可以在这里代替getApproximateDistanceUnits函数链接

然后我使用这个函数来查看经纬度是否在半径内

function isMapPlacemarkInRadius(point1, point2, radi) {
    if (point1 && point2) {
        return getApproximateDistanceUnits(point1, point2) <= radi;
    } else {
        return 0;
    }
}

点可以定义为

 $$.getPoint = function(lati, longi) {
        var location = {
            x: 0,
            y: 0,
            getX: function() { return location.x; },
            getY: function() { return location.y; }
        };
        location.x = lati;
        location.y = longi;

        return location;
    };

然后你可以做你的事情，看看一个点是否在一个半径范围内，比如:

 //put it on the map if within the range of a specified radi assuming 100,000,000 units
        var iconpoint = Map.getPoint(pp.latitude, pp.longitude);
        var centerpoint = Map.getPoint(Settings.CenterLatitude, Settings.CenterLongitude);

        //approx ~200 units to show only half of the globe from the default center radius
        if (isMapPlacemarkInRadius(centerpoint, iconpoint, 120)) {
            addPlacemark(pp.latitude, pp.longitude, pp.name);
        }
        else {
            otherSidePlacemarks.push({
                latitude: pp.latitude,
                longitude: pp.longitude,
                name: pp.name
            });

        }

在Mysql中使用以下函数传递参数，使用POINT(LONG,LAT)

CREATE FUNCTION `distance`(a POINT, b POINT)
 RETURNS double
    DETERMINISTIC
BEGIN

RETURN

GLength( LineString(( PointFromWKB(a)), (PointFromWKB(b)))) * 100000; -- To Make the distance in meters

END;