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

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);

其他回答

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);
//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);              
}

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

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);