/**
   * Calculates the haversine distance between point A, and B.
   * @param {number[]} latlngA [lat, lng] point A
   * @param {number[]} latlngB [lat, lng] point B
   * @param {boolean} isMiles If we are using miles, else km.
   */
  function haversineDistance(latlngA, latlngB, isMiles) {
    const squared = x => x * x;
    const toRad = x => (x * Math.PI) / 180;
    const R = 6371; // Earth’s mean radius in km

    const dLat = toRad(latlngB[0] - latlngA[0]);
    const dLon = toRad(latlngB[1] - latlngA[1]);

    const dLatSin = squared(Math.sin(dLat / 2));
    const dLonSin = squared(Math.sin(dLon / 2));

    const a = dLatSin +
              (Math.cos(toRad(latlngA[0])) * Math.cos(toRad(latlngB[0])) * dLonSin);
    const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    let distance = R * c;

    if (isMiles) distance /= 1.609344;

    return distance;
  }

我在网上找到了一个版本，80%是正确的，但插入了错误的参数，在使用输入时不一致，这个版本完全解决了这个问题

在PHP中，使用谷歌映射距离矩阵API:

//Get the Driving(Mode) distance between two Geo-location points(Latitude, Longitude) pair.
function get_distance($lat1, $lat2, $long1, $long2)
{
    $url = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=".$lat1.",".$long1."&destinations=".$lat2.",".$long2."&mode=driving"."&units=imperial";
    //You can request distance data for different travel modes, request distance data in different units such as kilometers or miles, and estimate travel time in traffic.
    try{
        $ch = curl_init();
        curl_setopt($ch, CURLOPT_URL, $url);
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
        curl_setopt($ch, CURLOPT_PROXYPORT, 3128);
        curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0);
        curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
        $response = curl_exec($ch);
        curl_close($ch);
        $response_a = json_decode($response, true);
        //Invalid request OR Empty response
        if(isset($response_a['error_message']) || empty($response_a['rows'])) 
            throw new Exception($response_a['error_message']);
        
    } catch(Exception $e){
        //Handle error here.
        return [];
    }
    
    //The unit parameter in the request URL only affects the text displayed within distance fields. The distance fields in response also contain values that are always expressed in meters.
    $dist = $response_a['rows'][0]['elements'][0]['distance']['text'];
    $time = $response_a['rows'][0]['elements'][0]['duration']['text'];

    return ['distance' => $dist, 'time' => $time];
}

参考:距离矩阵API请求和响应

//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;
    }

如果你想自己计算，那么你可以使用哈弗辛公式:

var rad = function(x) {
  return x * Math.PI / 180;
};

var getDistance = function(p1, p2) {
  var R = 6378137; // Earth’s mean radius in meter
  var dLat = rad(p2.lat() - p1.lat());
  var dLong = rad(p2.lng() - p1.lng());
  var a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
    Math.cos(rad(p1.lat())) * Math.cos(rad(p2.lat())) *
    Math.sin(dLong / 2) * Math.sin(dLong / 2);
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
  var d = R * c;
  return d; // returns the distance in meter
};

在我的情况下，这是最好的计算在SQL Server，因为我想采取当前位置，然后搜索所有邮政编码从当前位置一定距离内。我还有一个数据库，其中包含邮政编码及其纬度的列表。干杯

--will return the radius for a given number
create function getRad(@variable float)--function to return rad
returns float
as
begin
declare @retval float 
select @retval=(@variable * PI()/180)
--print @retval
return @retval
end
go

--calc distance
--drop function dbo.getDistance
create function getDistance(@cLat float,@cLong float, @tLat float, @tLong float)
returns float
as
begin
declare @emr float
declare @dLat float
declare @dLong float
declare @a float
declare @distance float
declare @c float

set @emr = 6371--earth mean 
set @dLat = dbo.getRad(@tLat - @cLat);
set @dLong = dbo.getRad(@tLong - @cLong);
set @a = sin(@dLat/2)*sin(@dLat/2)+cos(dbo.getRad(@cLat))*cos(dbo.getRad(@tLat))*sin(@dLong/2)*sin(@dLong/2);
set @c = 2*atn2(sqrt(@a),sqrt(1-@a))
set @distance = @emr*@c;
set @distance = @distance * 0.621371 -- i needed it in miles
--print @distance
return @distance;
end 
go


--get all zipcodes within 2 miles, the hardcoded #'s would be passed in by C#
select *
from cityzips a where dbo.getDistance(29.76,-95.38,a.lat,a.long) <3
order by zipcode

下面是this公式的c#实现

 public class DistanceAlgorithm
{
    const double PIx = 3.141592653589793;
    const double RADIO = 6378.16;

    /// <summary>
    /// This class cannot be instantiated.
    /// </summary>
    private DistanceAlgorithm() { }

    /// <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 * RADIO) * 0.62137;//distance in miles
    }

}