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
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
};
其他回答
首先,你指的是整个路径的长度,还是你只想知道位移(直线距离)?我看没人指出距离和位移的区别。对于距离计算JSON/XML数据给出的每个路由点,至于位移,有一个内置的解决方案使用球面类
//calculates distance between two points in km's
function calcDistance(p1, p2) {
return (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
}
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
};
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);
//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);
}