如果你正在使用python; PIP安装地质

from geopy.distance import geodesic


origin = (30.172705, 31.526725)  # (latitude, longitude) don't confuse
destination = (30.288281, 31.732326)

print(geodesic(origin, destination).meters)  # 23576.805481751613
print(geodesic(origin, destination).kilometers)  # 23.576805481751613
print(geodesic(origin, destination).miles)  # 14.64994773134371

你可以使用CLLocationDistance中的构建来计算这个:

CLLocation *location1 = [[CLLocation alloc] initWithLatitude:latitude1 longitude:longitude1];
CLLocation *location2 = [[CLLocation alloc] initWithLatitude:latitude2 longitude:longitude2];
[self distanceInMetersFromLocation:location1 toLocation:location2]

- (int)distanceInMetersFromLocation:(CLLocation*)location1 toLocation:(CLLocation*)location2 {
    CLLocationDistance distanceInMeters = [location1 distanceFromLocation:location2];
    return distanceInMeters;
}

在你的例子中，如果你想要公里，只要除以1000。

我已经创建了这个小Javascript LatLng对象，可能对某人有用。

var latLng1 = new LatLng(5, 3);
var latLng2 = new LatLng(6, 7);
var distance = latLng1.distanceTo(latLng2); 

代码:

/**
 * latLng point
 * @param {Number} lat
 * @param {Number} lng
 * @returns {LatLng}
 * @constructor
 */
function LatLng(lat,lng) {
    this.lat = parseFloat(lat);
    this.lng = parseFloat(lng);

    this.__cache = {};
}

LatLng.prototype = {
    toString: function() {
        return [this.lat, this.lng].join(",");
    },

    /**
     * calculate distance in km to another latLng, with caching
     * @param {LatLng} latLng
     * @returns {Number} distance in km
     */
    distanceTo: function(latLng) {
        var cacheKey = latLng.toString();
        if(cacheKey in this.__cache) {
            return this.__cache[cacheKey];
        }

        // the fastest way to calculate the distance, according to this jsperf test;
        // http://jsperf.com/haversine-salvador/8
        // http://stackoverflow.com/questions/27928
        var deg2rad = 0.017453292519943295; // === Math.PI / 180
        var lat1 = this.lat * deg2rad;
        var lng1 = this.lng * deg2rad;
        var lat2 = latLng.lat * deg2rad;
        var lng2 = latLng.lng * deg2rad;
        var a = (
            (1 - Math.cos(lat2 - lat1)) +
            (1 - Math.cos(lng2 - lng1)) * Math.cos(lat1) * Math.cos(lat2)
            ) / 2;
        var distance = 12742 * Math.asin(Math.sqrt(a)); // Diameter of the earth in km (2 * 6371)

        // cache the distance
        this.__cache[cacheKey] = distance;

        return distance;
    }
};

这里有一个用PHP http://www.geodatasource.com/developers/php计算距离的好例子:

 function distance($lat1, $lon1, $lat2, $lon2, $unit) {

     $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;
     }
 }

如果你想要驾驶距离/路线(张贴在这里，因为这是谷歌上两点之间距离的第一个结果，但对大多数人来说，驾驶距离更有用)，你可以使用谷歌地图距离矩阵服务:

getDrivingDistanceBetweenTwoLatLong(origin, destination) {

 return new Observable(subscriber => {
  let service = new google.maps.DistanceMatrixService();
  service.getDistanceMatrix(
    {
      origins: [new google.maps.LatLng(origin.lat, origin.long)],
      destinations: [new google.maps.LatLng(destination.lat, destination.long)],
      travelMode: 'DRIVING'
    }, (response, status) => {
      if (status !== google.maps.DistanceMatrixStatus.OK) {
        console.log('Error:', status);
        subscriber.error({error: status, status: status});
      } else {
        console.log(response);
        try {
          let valueInMeters = response.rows[0].elements[0].distance.value;
          let valueInKms = valueInMeters / 1000;
          subscriber.next(valueInKms);
          subscriber.complete();
        }
       catch(error) {
        subscriber.error({error: error, status: status});
       }
      }
    });
});
}

FSharp版本，使用里程:

let radialDistanceHaversine location1 location2 : float = 
                let degreeToRadian degrees = degrees * System.Math.PI / 180.0
                let earthRadius = 3959.0
                let deltaLat = location2.Latitude - location1.Latitude |> degreeToRadian
                let deltaLong = location2.Longitude - location1.Longitude |> degreeToRadian
                let a =
                    (deltaLat / 2.0 |> sin) ** 2.0
                    + (location1.Latitude |> degreeToRadian |> cos)
                    * (location2.Latitude |> degreeToRadian |> cos)
                    * (deltaLong / 2.0 |> sin) ** 2.0
                atan2 (a |> sqrt) (1.0 - a |> sqrt)
                * 2.0
                * earthRadius