如何计算谷歌地图V3中两个标记之间的距离?(类似于inV2的distanceFrom函数。)
谢谢. .
如何计算谷歌地图V3中两个标记之间的距离?(类似于inV2的distanceFrom函数。)
谢谢. .
当前回答
//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);
}
其他回答
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
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
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。
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
实际上GMap3中似乎有一个方法。它是google。maps。geometry。spherical命名空间的静态方法。
它以两个LatLng对象作为参数,并将使用默认的地球半径6378137米,尽管在必要时可以使用自定义值覆盖默认半径。
确保你包括:
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&v=3&libraries=geometry"></script>
在你的头部部分。
该呼吁将是:
google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);