JSFiddle - React, Tailwind, and code Playground
HTML
<script src="https://maps.googleapis.com/maps/api/js?v=3.exp"></script>
<script src="http://www.bdcc.co.uk/Gmaps/LatLngToOSGB.js"></script>
<div id="control"></div>
<div id="mapRoute"></div>
CSS
html, body, #mapRoute {
height: 100%;
width: 100%;
margin: 0px;
padding: 0px
}
JavaScript
var mapRoute;
var rtPoints;
var centerMAP = new google.maps.LatLng(-7.402438, 110.446957);
function gLatLngFromEN(e, n) {
var ogbLL = NEtoLL(e, n);
var pc = OGBToWGS84(ogbLL.lat, ogbLL.lon, 0);
return new google.maps.LatLng(pc.lat, pc.lon);
}
function routeMap() {
mapRoute = new google.maps.Map(document.getElementById('mapRoute'), {
center: centerMAP,
zoom: 14,
mapTypeId: google.maps.MapTypeId.SATELLITE
});
mapRoute.setCenter(gLatLngFromEN(469000, 169000), 13);
var rtPoints = new Array();
rtPoints.push(gLatLngFromEN(468000, 168000));
rtPoints.push(gLatLngFromEN(468000, 170000));
rtPoints.push(gLatLngFromEN(470000, 170000));
rtPoints.push(gLatLngFromEN(470000, 168000));
var rtPoly = new google.maps.Polyline({
path: rtPoints,
strokeColor: "#0000FF",
strokeWeight: 3,
map: mapRoute
});
var container = document.createElement("div");
container.style.fontFamily = 'Arial';
container.style.fontSize = 'XX-Small';
var ptr = document.createElement("INPUT");
ptr.style.width = "100px";
ptr.type = "Text";
ptr.readOnly = true;
ptr.id = "distPtr";
container.appendChild(ptr);
document.getElementById("control").appendChild(container);
google.maps.event.addListener(mapRoute, 'mousemove', function (point) {
document.getElementById('distPtr').value = Math.round(bdccGeoDistanceToPolyMtrs(rtPoly, point.latLng));
});
}
google.maps.event.addDomListener(window, 'load', routeMap);
// Code to find the distance in metres between a lat/lng point and a polyline of lat/lng points
// All in WGS84. Free for any use.
//
// Bill Chadwick 2007
// updated to Google Maps API v3, Lawrence Ross 2014
// Construct a bdccGeo from its latitude and longitude in degrees
function bdccGeo(lat, lon)
{
var theta = (lon * Math.PI / 180.0);
var rlat = bdccGeoGeocentricLatitude(lat * Math.PI / 180.0);
var c =...