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;

我在这里发布了我的工作示例。

在MySQL中列出表中指定点(我们使用一个随机点- lat:45.20327, long:23.7806)之间距离小于50 KM的所有点(表中字段为coord_lat和coord_long):

列出所有距离<50，单位:公里(地球半径6371公里):

SELECT denumire, (6371 * acos( cos( radians(45.20327) ) * cos( radians( coord_lat ) ) * cos( radians( 23.7806 ) - radians(coord_long) ) + sin( radians(45.20327) ) * sin( radians(coord_lat) ) )) AS distanta 
FROM obiective 
WHERE coord_lat<>'' 
    AND coord_long<>'' 
HAVING distanta<50 
ORDER BY distanta desc

上面的例子是在MySQL 5.0.95和5.5.16 (Linux)中测试的。

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;
}

下面是移植到Java的已接受的答案实现，以备任何人需要。

package com.project529.garage.util;


/**
 * Mean radius.
 */
private static double EARTH_RADIUS = 6371;

/**
 * Returns the distance between two sets of latitudes and longitudes in meters.
 * <p/>
 * Based from the following JavaScript SO answer:
 * http://stackoverflow.com/questions/27928/calculate-distance-between-two-latitude-longitude-points-haversine-formula,
 * which is based on https://en.wikipedia.org/wiki/Haversine_formula (error rate: ~0.55%).
 */
public double getDistanceBetween(double lat1, double lon1, double lat2, double lon2) {
    double dLat = toRadians(lat2 - lat1);
    double dLon = toRadians(lon2 - lon1);

    double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
            Math.cos(toRadians(lat1)) * Math.cos(toRadians(lat2)) *
                    Math.sin(dLon / 2) * Math.sin(dLon / 2);
    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    double d = EARTH_RADIUS * c;

    return d;
}

public double toRadians(double degrees) {
    return degrees * (Math.PI / 180);
}

如果你想要驾驶距离/路线(张贴在这里，因为这是谷歌上两点之间距离的第一个结果，但对大多数人来说，驾驶距离更有用)，你可以使用谷歌地图距离矩阵服务:

getDrivingDistanceBetweenTwoLatLong(origin, destination) {

 return new Observable(subscriber => {
  let service = new google.maps.DistanceMatrixService();
  service.getDistanceMatrix(
    {
      origins: [new google.maps.LatLng(origin.lat, origin.long)],
      destinations: [new google.maps.LatLng(destination.lat, destination.long)],
      travelMode: 'DRIVING'
    }, (response, status) => {
      if (status !== google.maps.DistanceMatrixStatus.OK) {
        console.log('Error:', status);
        subscriber.error({error: status, status: status});
      } else {
        console.log(response);
        try {
          let valueInMeters = response.rows[0].elements[0].distance.value;
          let valueInKms = valueInMeters / 1000;
          subscriber.next(valueInKms);
          subscriber.complete();
        }
       catch(error) {
        subscriber.error({error: error, status: status});
       }
      }
    });
});
}

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。