How to Find Nearest Location Using Latitude and Longitude from a Json Data

function distance(lat1, lon1, lat2, lon2, unit) {
  var radlat1 = Math.PI * lat1/180
  var radlat2 = Math.PI * lat2/180
  var theta = lon1-lon2
  var radtheta = Math.PI * theta/180
  var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
  if (dist > 1) {
    dist = 1;
  }
  dist = Math.acos(dist)
  dist = dist * 180/Math.PI
  dist = dist * 60 * 1.1515
  if (unit=="K") { dist = dist * 1.609344 }
  if (unit=="N") { dist = dist * 0.8684 }
  return dist
}

var data = [{
    "code": "0001",
    "lat": "1.28210155945393",
    "lng": "103.81722480263163",
    "location": "Stop 1"
}, {
    "code": "0003",
    "lat": "1.2777380589964",
    "lng": "103.83749709165197",
    "location": "Stop 2"
}, {
    "code": "0002",
    "lat": "1.27832046633393",
    "lng": "103.83762574759974",
    "location": "Stop 3"
}];

var html = "";
var poslat = 1.28210155945393;
var poslng = 103.81722480263163;

for (var i = 0; i < data.length; i++) {
    // if this location is within 0.1KM of the user, add it to the list
    if (distance(poslat, poslng, data[i].lat, data[i].lng, "K") <= 0.1) {
        html += '<p>' + data[i].location + ' - ' + data[i].code + '</p>';
    }
}

$('#nearbystops').append(html);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="nearbystops"></div>

Given a lat/long, find the nearest location based on a json list of lat/long

You could simply define a function to calculate the distance and use it to sort bus stops with the key argument:

from math import cos, sqrt, pi

R = 6371000 #radius of the Earth in m
def distance(lon1, lat1, lon2, lat2):
    x = (lon2 - lon1) * cos(0.5*(lat2+lat1))
    y = (lat2 - lat1)
    return (2*pi*R/360) * sqrt( x*x + y*y )

bustops = [{"BusStopCode": "00481", "RoadName": "Woodlands Rd", "Description": "BT PANJANG TEMP BUS PK", "Latitude": 1.383764, "Longitude": 103.7583},
{"BusStopCode": "01012", "RoadName": "Victoria St", "Description": "Hotel Grand Pacific", "Latitude": 1.29684825487647, "Longitude": 103.85253591654006}]

print(sorted(bustops, key= lambda d: distance(d["Longitude"], d["Latitude"], 103.5, 1.2)))
# [{'BusStopCode': '01012', 'RoadName': 'Victoria St', 'Description': 'Hotel Grand Pacific', 'Latitude': 1.29684825487647, 'Longitude': 103.85253591654006}, {'BusStopCode': '00481', 'RoadName': 'Woodlands Rd', 'Description': 'BT PANJANG TEMP BUS PK', 'Latitude': 1.383764, 'Longitude': 103.7583}]

Once this list is sorted, you can simply extract the 5 closest bus stops with [:5].
It should be fast enough, even with 5000 bus stops.

Note that if you don't care about the specific distance but only want to sort bus stops, you could use this function as key:

def distance2(lon1, lat1, lon2, lat2):
    x = (lon2 - lon1) * cos(0.5*(lat2+lat1))
    y = (lat2 - lat1)
    return x*x + y*y

How to find nearest latitude/longitude from big json array of lat/lng

In case anyone comes here looking for a solution, how I solved it is by the following code:

for (let i = 0; i < json.length; i++) {
                if ((Math.abs(json[i].coordinates[1] - start.lat)) + ((Math.abs(json.[i].coordinates[0] - start.lng))) < sumLatLng) {
                    closest = json[i];
                    sumLatLng = (Math.abs(json.[i].coordinates[1] - start.lat)) + ((Math.abs(json.[i].coordinates[0] - start.lng)))
                } else {
                    console.log("error");
                }
            }

Basically what I do is take the sum of the starting lat & lng, subtract the lat & lng of each record from the API and take the absolute of the result. If the result is smaller than the previous result, that means that its closer.

JSON longitude latitude nearest location

My solution just creates an array of objects out of the locations and distances, sorts them, then spits them out. I've kept everything as meters instead of kilometers but it will be easy to tweak. Enjoy!

While looping:

distanceObj[i] = {
    distance: hesapla(9.9608999, 49.7222842, b.Position.Longitude, b.Position.Latitude),
    location: a
};
++i;

Then sort:

distanceObj.sort(function(a,b) {
    return parseInt(a.distance) - parseInt(b.distance)
});

This will explain better: http://jsfiddle.net/mitchmalone/Wf6qy/1/

Nearest location from JSON column

If you have PostGIS installed on your Postgres server, you could try something like this

SELECT j.*,
  ST_DISTANCE(
    ST_GeogFromText('POINT(' || lng || ' ' || lat || ')'),
    ST_GeogFromText('POINT(31.08 30.025)')
    ) AS dist
FROM YourTable AS t
CROSS JOIN jsonb_to_recordset(t.data->'locations') AS j(lat int, lng int)
ORDER BY
  dist;

I strongly suggest you store geodetic information in an actual geography column.

At the very least you should normalize this JSON out to another table if you are querying it often.

How to display the nearest latitude and longitude to my location using javascript?

Javascript function to calculate distance

$nearest=1; //Limit within 1 KM.

$("#jobnews").html("<h2>Results:</h2>"); 
$.each(info.companies.company, function(index, job)
{ 
    var lat = list.location.lat; 
    var lng = list.location.lng; 
    var nam = current.location.lat; 
    var cou = current.location.lng; 
    //In below line we will get the distance between current and given coordinates.
    var d=distance(nam, cou, lat, lng, "K");
    //Check the d is within 1 KM, If so then add it to map.
    if(nearest>d)
    {
        var map = "</p><img src='maps.google.com/maps/api/staticmap?center="; + lat + "," +lng +"&zoom=17&size=400x300&key="+key+"&maptype=roadmap&visual_refresh=true&markers=‌​color:red|"+lat+","+lng+"&sensor=true'width='300' height='280'alt='Map of your location.'/>"; 
        var nearestList = "<div><h3>DISPLAY INFO: "+lis.name+link+nam+lat+lng+cou+map+"</div>"; 
    }
}  

function distance(lat1, lon1, lat2, lon2, unit) {
    var radlat1 = Math.PI * lat1/180
    var radlat2 = Math.PI * lat2/180
    var radlon1 = Math.PI * lon1/180
    var radlon2 = Math.PI * lon2/180
    var theta = lon1-lon2
    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
    if (unit=="K") { dist = dist * 1.609344 }
    if (unit=="N") { dist = dist * 0.8684 }
    return dist
}        

Passed to function:

lat1, lon1 = Latitude and Longitude of point 1 (in decimal degrees)

lat2, lon2 = Latitude and Longitude of point 2 (in decimal degrees)

unit = the unit you desire for results

where: 'M' is statute miles

'K' is kilometers (default)

'N' is nautical miles

How to find nearest location using latitude and longitude from SQL database?

SELECT id, ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin( radians( lat ) ) ) ) AS distance FROM markers HAVING distance < 25 ORDER BY distance LIMIT 0 , 20;

How can find nearest place from current location from given data.

You have your current location latitude an longitude so just find out distance between your current location and list of location address geopoints using this formula and display address with shortest distance.

private double distance(double lat1, double lon1, double lat2, double lon2) {
        // haversine great circle distance approximation, returns meters
        double theta = lon1 - lon2;
        double dist = Math.sin(deg2rad(lat1)) * Math.sin(deg2rad(lat2))
                + Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2))
                * Math.cos(deg2rad(theta));
        dist = Math.acos(dist);
        dist = rad2deg(dist);
        dist = dist * 60; // 60 nautical miles per degree of seperation
        dist = dist * 1852; // 1852 meters per nautical mile
        return (dist);
    }

    private double deg2rad(double deg) {
        return (deg * Math.PI / 180.0);
    }

    private double rad2deg(double rad) {
        return (rad * 180.0 / Math.PI);
    }

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