[google-maps] Google Maps Api v3 - find nearest markers

When i click on map, which will be best way to find nearest marker or markers? is there some functions in api that will help me to do that?

it's google map api v3.

This question is related to google-maps google-maps-api-3

The answer is


Use computeDistanceBetween() Google map API method to calculate near marker between your location and markers list on google map.

Steps:-

  1. Create marker on google map.
    function addMarker(location) { var marker = new google.maps.Marker({ title: 'User added marker', icon: { path: google.maps.SymbolPath.BACKWARD_CLOSED_ARROW, scale: 5 }, position: location, map: map }); }

  2. On Mouse click create event for getting lat, long of your location and pass that to find_closest_marker().

    function find_closest_marker(event) {
          var distances = [];
          var closest = -1;
          for (i = 0; i < markers.length; i++) {
            var d = google.maps.geometry.spherical.computeDistanceBetween(markers[i].position, event.latLng);
            distances[i] = d;
            if (closest == -1 || d < distances[closest]) {
              closest = i;
            }
          }
          alert('Closest marker is: ' + markers[closest].getTitle());
        }
    

visit this link follow the steps. You will able to get nearer marker to your location.


The formula above didn't work for me, but I used this without any issue. Pass your current location to the function, and loop through an array of markers to find the closest:

function find_closest_marker( lat1, lon1 ) {    
    var pi = Math.PI;
    var R = 6371; //equatorial radius
    var distances = [];
    var closest = -1;

    for( i=0;i<markers.length; i++ ) {  
        var lat2 = markers[i].position.lat();
        var lon2 = markers[i].position.lng();

        var chLat = lat2-lat1;
        var chLon = lon2-lon1;

        var dLat = chLat*(pi/180);
        var dLon = chLon*(pi/180);

        var rLat1 = lat1*(pi/180);
        var rLat2 = lat2*(pi/180);

        var a = Math.sin(dLat/2) * Math.sin(dLat/2) + 
                    Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(rLat1) * Math.cos(rLat2); 
        var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
        var d = R * c;

        distances[i] = d;
        if ( closest == -1 || d < distances[closest] ) {
            closest = i;
        }
    }

    // (debug) The closest marker is:
    console.log(markers[closest]);
}

You can use the computeDistanceBetween() method in the google.maps.geometry.spherical namespace.


Here is another function that works great for me, returns distance in kilometers:

 function distance(lat1, lng1, lat2, lng2) {
        var radlat1 = Math.PI * lat1 / 180;
        var radlat2 = Math.PI * lat2 / 180;
        var radlon1 = Math.PI * lng1 / 180;
        var radlon2 = Math.PI * lng2 / 180;
        var theta = lng1 - lng2;
        var radtheta = Math.PI * theta / 180;
        var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
        dist = Math.acos(dist);
        dist = dist * 180 / Math.PI;
        dist = dist * 60 * 1.1515;

        //Get in in kilometers
        dist = dist * 1.609344;

        return dist;
    }

Are you aware of Mysql Spatial extensions?

You could use something like MBRContains(g1,g2).


I'd like to expand on Leor's suggestion for anyone confused on how to compute the nearest location and actually provide a working solution:

I'm using markers in a markers array e.g. var markers = [];.

Then let's have our position as something like var location = new google.maps.LatLng(51.99, -0.74);

Then we simply reduce our markers against the location we have like so:

markers.reduce(function (prev, curr) {

    var cpos = google.maps.geometry.spherical.computeDistanceBetween(location.position, curr.position);
    var ppos = google.maps.geometry.spherical.computeDistanceBetween(location.position, prev.position);

    return cpos < ppos ? curr : prev;

}).position

What pops out is your closest marker LatLng object.