只需将以下内容添加到JavaScript代码的开头:

google.maps.LatLng.prototype.distanceFrom = function(latlng) {
  var lat = [this.lat(), latlng.lat()]
  var lng = [this.lng(), latlng.lng()]
  var R = 6378137;
  var dLat = (lat[1]-lat[0]) * Math.PI / 180;
  var dLng = (lng[1]-lng[0]) * Math.PI / 180;
  var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
  Math.cos(lat[0] * Math.PI / 180 ) * Math.cos(lat[1] * Math.PI / 180 ) *
  Math.sin(dLng/2) * Math.sin(dLng/2);
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
  var d = R * c;
  return Math.round(d);
}

然后像这样使用函数:

var loc1 = new GLatLng(52.5773139, 1.3712427);
var loc2 = new GLatLng(52.4788314, 1.7577444);
var dist = loc2.distanceFrom(loc1);
alert(dist/1000);

  /**
   * 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

//p1 and p2 are google.maps.LatLng(x,y) objects

function calcDistance(p1, p2) {
          var d = (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
          console.log(d);              
}

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。

离线解-哈弗辛算法

在Javascript中

var _eQuatorialEarthRadius = 6378.1370;
var _d2r = (Math.PI / 180.0);

function HaversineInM(lat1, long1, lat2, long2)
{
    return (1000.0 * HaversineInKM(lat1, long1, lat2, long2));
}

function HaversineInKM(lat1, long1, lat2, long2)
{
    var dlong = (long2 - long1) * _d2r;
    var dlat = (lat2 - lat1) * _d2r;
    var a = Math.pow(Math.sin(dlat / 2.0), 2.0) + Math.cos(lat1 * _d2r) * Math.cos(lat2 * _d2r) * Math.pow(Math.sin(dlong / 2.0), 2.0);
    var c = 2.0 * Math.atan2(Math.sqrt(a), Math.sqrt(1.0 - a));
    var d = _eQuatorialEarthRadius * c;

    return d;
}

var meLat = -33.922982;
var meLong = 151.083853;


var result1 = HaversineInKM(meLat, meLong, -32.236457779983745, 148.69094705162837);
var result2 = HaversineInKM(meLat, meLong, -33.609020205923713, 150.77061469270831);

C#

using System;

public class Program
{
    public static void Main()
    {
        Console.WriteLine("Hello World");

        var meLat = -33.922982;
        double meLong = 151.083853;


        var result1 = HaversineInM(meLat, meLong, -32.236457779983745, 148.69094705162837);
        var result2 = HaversineInM(meLat, meLong, -33.609020205923713, 150.77061469270831);

        Console.WriteLine(result1);
        Console.WriteLine(result2);
    }

    static double _eQuatorialEarthRadius = 6378.1370D;
    static double _d2r = (Math.PI / 180D);

    private static int HaversineInM(double lat1, double long1, double lat2, double long2)
    {
        return (int)(1000D * HaversineInKM(lat1, long1, lat2, long2));
    }

    private static  double HaversineInKM(double lat1, double long1, double lat2, double long2)
    {
        double dlong = (long2 - long1) * _d2r;
        double dlat = (lat2 - lat1) * _d2r;
        double a = Math.Pow(Math.Sin(dlat / 2D), 2D) + Math.Cos(lat1 * _d2r) * Math.Cos(lat2 * _d2r) * Math.Pow(Math.Sin(dlong / 2D), 2D);
        double c = 2D * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1D - a));
        double d = _eQuatorialEarthRadius * c;

        return d;
    }
}

参考: https://en.wikipedia.org/wiki/Great-circle_distance