Geolocation closest location(lat, long) from my position

Question

I want to show specific information depending on where i am.

I have five cities with different information, and i want to show that city(information) that i'm closest to.

How to i do that the simplest way, using javascript.

Ex.

If i store the cities lat, long in an array

var cities = [
  ['new york', '111111', '222222', 'blablabla']
  ['boston', '111111', '222222', 'blablabla']
  ['seattle', '111111', '222222', 'blablabla']
  ['london', '111111', '222222', 'blablabla']
]

And with my current location(lat, long) i want the city that i'm closet to.

Solution 1

Here is a basic code example using HTML5 geolocation to get the user's position. It then calls NearestCity() and calculates the distance (km) from the location to each city. I passed on using the Haversine formulae and instead used the simpler Pythagoras formulae and an equirectangular projection to adjust for the curvature in longitude lines.

// Get User's Coordinate from their Browser
window.onload = function() {
  // HTML5/W3C Geolocation
  if (navigator.geolocation) {
    navigator.geolocation.getCurrentPosition(UserLocation);
  }
  // Default to Washington, DC
  else
    NearestCity(38.8951, -77.0367);
}

// Callback function for asynchronous call to HTML5 geolocation
function UserLocation(position) {
  NearestCity(position.coords.latitude, position.coords.longitude);
}


// Convert Degress to Radians
function Deg2Rad(deg) {
  return deg * Math.PI / 180;
}

function PythagorasEquirectangular(lat1, lon1, lat2, lon2) {
  lat1 = Deg2Rad(lat1);
  lat2 = Deg2Rad(lat2);
  lon1 = Deg2Rad(lon1);
  lon2 = Deg2Rad(lon2);
  var R = 6371; // km
  var x = (lon2 - lon1) * Math.cos((lat1 + lat2) / 2);
  var y = (lat2 - lat1);
  var d = Math.sqrt(x * x + y * y) * R;
  return d;
}

var lat = 20; // user's latitude
var lon = 40; // user's longitude

var cities = [
  ["city1", 10, 50, "blah"],
  ["city2", 40, 60, "blah"],
  ["city3", 25, 10, "blah"],
  ["city4", 5, 80, "blah"]
];

function NearestCity(latitude, longitude) {
  var minDif = 99999;
  var closest;

  for (index = 0; index < cities.length; ++index) {
    var dif = PythagorasEquirectangular(latitude, longitude, cities[index][1], cities[index][2]);
    if (dif < minDif) {
      closest = index;
      minDif = dif;
    }
  }

  // echo the nearest city
  alert(cities[closest]);
}

Solution 2

With HTML5, you can pull the location of the user and then compares this example using a Haversine function (function below taken from here):

function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2) {
  var R = 6371; // Radius of the earth in km
  var dLat = deg2rad(lat2-lat1);  // deg2rad below
  var dLon = deg2rad(lon2-lon1); 
  var a = 
    Math.sin(dLat/2) * Math.sin(dLat/2) +
    Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * 
    Math.sin(dLon/2) * Math.sin(dLon/2)
    ; 
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
  var d = R * c; // Distance in km
  return d;
}

function deg2rad(deg) {
  return deg * (Math.PI/180)
}

Solution 3

You can calculate the distance by latitude with your location and the cities locations. And find the shortest and draw. To calculate you can read more in http://www.movable-type.co.uk/scripts/latlong.html

Comments

[Anders Kristoffersson]

Haven't tried anything yet, just checking if it is possible.

[Kennet]

Beware, there is a typo in the following line: var dif = PythagorasEquirectangular( lat, lon, cities[ index ][ 1 ], cities[ index ][ 2 ] ); Should be var dif = PythagorasEquirectangular( latitude, longitude, cities[ index ][ 1 ], cities[ index ][ 2 ] );

[Andrew - OpenGeoCode]

@Kennet - thanks for letting me know about the typo. I fixed it in the answer.

[AndrewLeonardi]

This. is. Awesome. Thanks @Andrew-OpenGeoCode

[user3120861]

This is great. I'd like to modify it to list possibly the closest three locations.

[Akhil S]

can anyone please explain me how this code is working, in detail?

[SDK]

var minDif = 99999; hi, please explain what is the value and what its purpose of it