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);
}

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;

//JAVA
    public Double getDistanceBetweenTwoPoints(Double latitude1, Double longitude1, Double latitude2, Double longitude2) {
    final int RADIUS_EARTH = 6371;

    double dLat = getRad(latitude2 - latitude1);
    double dLong = getRad(longitude2 - longitude1);

    double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(getRad(latitude1)) * Math.cos(getRad(latitude2)) * Math.sin(dLong / 2) * Math.sin(dLong / 2);
    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    return (RADIUS_EARTH * c) * 1000;
    }

    private Double getRad(Double x) {
    return x * Math.PI / 180;
    }

计算距离——尤其是大距离——的主要挑战之一是解释地球的曲率。如果地球是平的，计算两点之间的距离就会像计算直线一样简单!哈弗辛公式包括一个常数(下面是R变量)，它表示地球的半径。根据你是用英里还是公里来测量，它分别等于3956英里或6367公里。 基本公式是:

Dlon = lon2 - lon1 dat = lat2 - lat1 = (sin (dlat / 2)) ^ 2 + cos (lat1) * cos (lat2) * (sin (dlon / 2)) ^ 2 C = 2 * atan2(√(a)，√(1-a)) distance = R * c(其中R为地球半径) R = 6367公里OR 3956英里

     lat1, lon1: The Latitude and Longitude of point 1 (in decimal degrees)
     lat2, lon2: The Latitude and Longitude of point 2 (in decimal degrees)
     unit: The unit of measurement in which to calculate the results where:
     'M' is statute miles (default)
     'K' is kilometers
     'N' is nautical miles

样本

function distance(lat1, lon1, lat2, lon2, unit) {
    try {
        var radlat1 = Math.PI * lat1 / 180
        var radlat2 = Math.PI * lat2 / 180
        var theta = lon1 - lon2
        var radtheta = Math.PI * theta / 180
        var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
        dist = Math.acos(dist)
        dist = dist * 180 / Math.PI
        dist = dist * 60 * 1.1515
        if (unit == "K") {
            dist = dist * 1.609344
        }
        if (unit == "N") {
            dist = dist * 0.8684
        }
        return dist
    } catch (err) {
        console.log(err);
    }
}

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);
}

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