/**
   * 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

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
};

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
    }

}    

//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;
    }

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

使用PHP，你可以使用这个简单的函数来计算距离:

// to calculate distance between two lat & lon

function calculate_distance($lat1, $lon1, $lat2, $lon2, $unit='N') 
{ 
  $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;
      }
}

// function ends here