Question

我正在申请我需要到附近的地方，我的网络服务将收到2个参数（小数经度，小数纬度）

我有一个表格，其中的位置保存在数据库中，包含经度和纬度字段，

我想检索最近的位置。

有人可以帮忙吗？

这是我的代码：

 var locations = from l in locations

     select l

以下是有关此内容的更多详情：我在数据库表中有2个字段（十进制（18,2）null）1个纬度，2个经度，

我有一个方法

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();
}

Answer 1

您可以先将数据库中的位置数据转换为System.Device.Location.GeoCoordinate，然后使用LINQ查找最近的数据。

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

Answer 2

要详细说明@Fung的注释，如果您正在使用Entity Framework / LINQ to Entities，如果您尝试在LINQ查询中使用GeoCoordinate.GetDistanceTo方法，那么您将获得带有消息的运行时NotSupportedException ：

LINQ to Entities无法识别方法'Double GetDistanceTo（System.Device.Location.GeoCoordinate）'方法，并且此方法无法转换为商店表达式。

使用Entity Framework版本5或6，另一种方法是使用System.Data.Spatial.DbGeography类。例如：

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();

此示例中的

_context是您以前实例化的DbContext实例。

虽然它目前是undocumented in MSDN，但DbGeography.Distance方法返回的单位似乎是米。请参阅：System.Data.Spatial DbGeography.Distance units?

Answer 3

这是解决方案

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();

Answer 4

你有一个有效的范围，除此之外“命中”并不真正相关吗？如果是这样，请使用

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

然后在这些结果的循环内找到具有最小平方距离值的命中。

Answer 5

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();

Answer 6

Netcore友好的解决方案。根据需要进行重构。

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));
}

查找经度和纬度最近的位置

6 个答案: