CODEWITHSUNDEEP
c#linq
Ashfaque Ali Solangi
Ashfaque Ali Solangi

Reputation: 1891

Find closest location with longitude and latitude

I am working on a application where I need to get nearby location, my web service will receive 2 parameters (decimal longitude, decimal latitude )

I have a table where the locations are saved in database with longitude and latitude fields,

I want to retrieve the nearest locations.

Can anyone help?

Here is my code:

 var locations = from l in locations

     select l

Here are further details about this : i have a 2 fields (decimal(18, 2) null) 1 latitude, 2 longitude inside a database table,

and i have a method 

public List<Locations>  GetLocation(decimal? Long, decimal? lat) 
{
var Loc = from l in Locations
  //// now here is how to get nearest location ? how to query?
  //// i have also tried Math.Abs(l.Lat - lat) its giving error about nullable decimal always hence i have seted decimal to nullable or converted to nullable
 //// also i have tried where (l.lat - Lat) * (l.lon - Long)  this is also giving error about can not convert decimal to bool
return Loc.ToList();
}

Upvotes: 34

Views: 41204

Answers (6)

maeneak
maeneak

Reputation: 613

A netcore friendly solution. Refactor as needed.

public static System.Drawing.PointF getClosestPoint(System.Drawing.PointF[] points, System.Drawing.PointF query) {
    return points.OrderBy(x => distance(query, x)).First();
}

public static double distance(System.Drawing.PointF pt1, System.Drawing.PointF pt2) {
    return Math.Sqrt((pt2.Y - pt1.Y) * (pt2.Y - pt1.Y) + (pt2.X - pt1.X) * (pt2.X - pt1.X));
}

Upvotes: 1

Rajesh-Systematix
Rajesh-Systematix

Reputation: 109

var objAllListing = (from listing in _listingWithLanguageRepository.GetAll().Where(z => z.IsActive == true)
                                     let distance = 12742 * SqlFunctions.Asin(SqlFunctions.SquareRoot(SqlFunctions.Sin(((SqlFunctions.Pi() / 180) * (listing.Listings.Latitude - sourceLatitude)) / 2) * SqlFunctions.Sin(((SqlFunctions.Pi() / 180) * (listing.Listings.Latitude - sourceLatitude)) / 2) +
                                                        SqlFunctions.Cos((SqlFunctions.Pi() / 180) * sourceLatitude) * SqlFunctions.Cos((SqlFunctions.Pi() / 180) * (listing.Listings.Latitude)) *
                                                        SqlFunctions.Sin(((SqlFunctions.Pi() / 180) * (listing.Listings.Longitude - sourceLongitude)) / 2) * SqlFunctions.Sin(((SqlFunctions.Pi() / 180) * (listing.Listings.Longitude - sourceLongitude)) / 2)))
                                     where distance <= input.Distance

                                     select new ListingFinalResult { ListingDetail = listing, Distance = distance }).ToList();//.Take(5).OrderBy(x => x.distance).ToList();

Upvotes: 1

Jon Schneider
Jon Schneider

Reputation: 26973

To elaborate on the comment by @Fung, if you are using Entity Framework / LINQ to Entities, if you try to use the GeoCoordinate.GetDistanceTo method in a LINQ query, you'll get a runtime NotSupportedException with the message:

LINQ to Entities does not recognize the method 'Double GetDistanceTo(System.Device.Location.GeoCoordinate)' method, and this method cannot be translated into a store expression.

With Entity Framework version 5 or 6, an alternative is to use the System.Data.Spatial.DbGeography class. For example:

DbGeography searchLocation = DbGeography.FromText(String.Format("POINT({0} {1})", longitude, latitude));

var nearbyLocations = 
    (from location in _context.Locations
     where  // (Additional filtering criteria here...)
     select new 
     {
         LocationID = location.ID,
         Address1 = location.Address1,
         City = location.City,
         State = location.State,
         Zip = location.Zip,
         Latitude = location.Latitude,
         Longitude = location.Longitude,
         Distance = searchLocation.Distance(
             DbGeography.FromText("POINT(" + location.Longitude + " " + location.Latitude + ")"))
     })
    .OrderBy(location => location.Distance)
    .ToList();

_context in this example is your previously-instantiated DbContext instance.

Although it's currently undocumented in MSDN, the units returned by the DbGeography.Distance method appear to be meters. See: System.Data.Spatial DbGeography.Distance units?

Upvotes: 10

Ashfaque Ali Solangi
Ashfaque Ali Solangi

Reputation: 1891

Here is Solution

var constValue = 57.2957795130823D

var constValue2 = 3958.75586574D;

var searchWithin = 20;

double latitude = ConversionHelper.SafeConvertToDoubleCultureInd(Latitude, 0),
                    longitude = ConversionHelper.SafeConvertToDoubleCultureInd(Longitude, 0);
var loc = (from l in DB.locations
let temp = Math.Sin(Convert.ToDouble(l.Latitude) / constValue) *  Math.Sin(Convert.ToDouble(latitude) / constValue) +
                                 Math.Cos(Convert.ToDouble(l.Latitude) / constValue) *
                                 Math.Cos(Convert.ToDouble(latitude) / constValue) *
                                 Math.Cos((Convert.ToDouble(longitude) / constValue) - (Convert.ToDouble(l.Longitude) / constValue))
                             let calMiles = (constValue2 * Math.Acos(temp > 1 ? 1 : (temp < -1 ? -1 : temp)))
                             where (l.Latitude > 0 && l.Longitude > 0)
                             orderby calMiles

select new location
  {
     Name = l.name
  });
  return loc .ToList();

Upvotes: 5

Fung
Fung

Reputation: 3558

You could first convert the location data in database to System.Device.Location.GeoCoordinate, then use LINQ to find the nearest one.

var coord = new GeoCoordinate(latitude, longitude);
var nearest = locations.Select(x => new GeoCoordinate(x.Latitude, x.Longitude))
                       .OrderBy(x => x.GetDistanceTo(coord))
                       .First();

Upvotes: 64

Kirk B.
Kirk B.

Reputation: 456

Do you have a valid range, outside of which the "hit" is not really relevant? If so, use

from l in locations where ((l.lat - point.lat) * (l.lat - point.lat)) + ((l.lng - point.lng) * (l.lng - point.lng)) < (range * range) select l

then find the hit with the smallest squared distance value within a loop of those results.

Upvotes: 2

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