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

//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;
    }

示例使用GPS的纬度/经度2个点。

var latitude1 = 39.46;
var longitude1 = -0.36;
var latitude2 = 40.40;
var longitude2 = -3.68;

var distance = google.maps.geometry.spherical.computeDistanceBetween(new google.maps.LatLng(latitude1, longitude1), new google.maps.LatLng(latitude2, longitude2));       

对于谷歌，你可以使用球面api, google.maps.geometry. sphere.computedistancebetween (latLngA, latLngB);

然而，如果球面投影或哈弗辛解的精度对你来说不够精确(例如，如果你靠近极点或计算更长的距离)，你应该使用不同的库。

关于这个主题的大部分信息我都是在维基百科上找到的。

要查看任何给定算法的精度是否足够，一个技巧是填充地球的最大和最小半径，并查看差异是否会给您的用例带来问题。更多细节可以在本文中找到

最后，谷歌api或haversine将毫无问题地满足大多数目的。

不得不这么做……动作脚本方式

//just make sure you pass a number to the function because it would accept you mother in law...
public var rad = function(x:*) {return x*Math.PI/180;}

protected  function distHaversine(p1:Object, p2:Object):Number {
    var R:int = 6371; // earth's mean radius in km
    var dLat:Number = rad(p2.lat() - p1.lat());
    var dLong:Number = rad(p2.lng() - p1.lng());

    var a:Number = 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:Number = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
    var d:Number = R * c;

    return d;
}

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