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

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);

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

使用谷歌距离矩阵服务非常简单

第一步是从谷歌API控制台激活距离矩阵服务。 它返回一组位置之间的距离。 应用这个简单的函数

function initMap() {
        var bounds = new google.maps.LatLngBounds;
        var markersArray = [];

        var origin1 = {lat:23.0203, lng: 72.5562};
        //var origin2 = 'Ahmedabad, India';
        var destinationA = {lat:23.0436503, lng: 72.55008939999993};
        //var destinationB = {lat: 23.2156, lng: 72.6369};

        var destinationIcon = 'https://chart.googleapis.com/chart?' +
            'chst=d_map_pin_letter&chld=D|FF0000|000000';
        var originIcon = 'https://chart.googleapis.com/chart?' +
            'chst=d_map_pin_letter&chld=O|FFFF00|000000';
        var map = new google.maps.Map(document.getElementById('map'), {
          center: {lat: 55.53, lng: 9.4},
          zoom: 10
        });
        var geocoder = new google.maps.Geocoder;

        var service = new google.maps.DistanceMatrixService;
        service.getDistanceMatrix({
          origins: [origin1],
          destinations: [destinationA],
          travelMode: 'DRIVING',
          unitSystem: google.maps.UnitSystem.METRIC,
          avoidHighways: false,
          avoidTolls: false
        }, function(response, status) {
          if (status !== 'OK') {
            alert('Error was: ' + status);
          } else {
            var originList = response.originAddresses;
            var destinationList = response.destinationAddresses;
            var outputDiv = document.getElementById('output');
            outputDiv.innerHTML = '';
            deleteMarkers(markersArray);

            var showGeocodedAddressOnMap = function(asDestination) {
              var icon = asDestination ? destinationIcon : originIcon;
              return function(results, status) {
                if (status === 'OK') {
                  map.fitBounds(bounds.extend(results[0].geometry.location));
                  markersArray.push(new google.maps.Marker({
                    map: map,
                    position: results[0].geometry.location,
                    icon: icon
                  }));
                } else {
                  alert('Geocode was not successful due to: ' + status);
                }
              };
            };

            for (var i = 0; i < originList.length; i++) {
              var results = response.rows[i].elements;
              geocoder.geocode({'address': originList[i]},
                  showGeocodedAddressOnMap(false));
              for (var j = 0; j < results.length; j++) {
                geocoder.geocode({'address': destinationList[j]},
                    showGeocodedAddressOnMap(true));
                //outputDiv.innerHTML += originList[i] + ' to ' + destinationList[j] + ': ' + results[j].distance.text + ' in ' +                    results[j].duration.text + '<br>';
                outputDiv.innerHTML += results[j].distance.text + '<br>';
              }
            }

          }
        });
      }

其中origin1是你的位置，destinationA是目的地位置。您可以添加以上两个或多个数据。

Rad完整文档与示例

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
};

//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;
    }