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;

可能有一个更简单、更正确的解决方案:地球的周长在赤道上是40000公里，在格林威治(或任何经度)周期上约为37000公里。因此:

pythagoras = function (lat1, lon1, lat2, lon2) {
   function sqr(x) {return x * x;}
   function cosDeg(x) {return Math.cos(x * Math.PI / 180.0);}

   var earthCyclePerimeter = 40000000.0 * cosDeg((lat1 + lat2) / 2.0);
   var dx = (lon1 - lon2) * earthCyclePerimeter / 360.0;
   var dy = 37000000.0 * (lat1 - lat2) / 360.0;

   return Math.sqrt(sqr(dx) + sqr(dy));
};

我同意它应该被微调，我自己说过它是一个椭球，所以半径乘以余弦值是不同的。但它更准确一点。与谷歌map相比，误差明显减小。

function distance($lat1, $lon1, $lat2, $lon2) { 
    $pi80 = M_PI / 180; 
    $lat1 *= $pi80; $lon1 *= $pi80; $lat2 *= $pi80; $lon2 *= $pi80; 
    $dlat = $lat2 - $lat1; 
    $dlon = $lon2 - $lon1; 
    $a = sin($dlat / 2) * sin($dlat / 2) + cos($lat1) * cos($lat2) * sin($dlon / 2) * sin($dlon / 2);  
    $km = 6372.797 * 2 * atan2(sqrt($a), sqrt(1 - $a)); 
    return $km; 
}

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;
}

非常感谢这一切。我在Objective-C iPhone应用程序中使用了以下代码:

const double PIx = 3.141592653589793;
const double RADIO = 6371; // Mean radius of Earth in Km

double convertToRadians(double val) {

   return val * PIx / 180;
}

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

        double dlon = convertToRadians(place2.longitude - place1.longitude);
        double dlat = convertToRadians(place2.latitude - place1.latitude);

        double a = ( pow(sin(dlat / 2), 2) + cos(convertToRadians(place1.latitude))) * cos(convertToRadians(place2.latitude)) * pow(sin(dlon / 2), 2);
        double angle = 2 * asin(sqrt(a));

        return angle * RADIO;
}

纬度和经度是十进制的。我没有在asin()调用中使用min()，因为我使用的距离非常小，以至于它们不需要min()。

它给出了错误的答案，直到我传入弧度的值-现在它几乎与从苹果地图应用程序中获得的值相同:-)

额外的更新:

如果你使用的是iOS4或更高版本，那么苹果会提供一些方法来实现相同的功能:

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

    MKMapPoint  start, finish;


    start = MKMapPointForCoordinate(place1);
    finish = MKMapPointForCoordinate(place2);

    return MKMetersBetweenMapPoints(start, finish) / 1000;
}

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;