查找經度和緯度

Question

我一個應用程序，我需要得到附近的地點工作最接近的位置， 我的web服務將接收2個參數（十進制經度，數字緯度）

我有一個表，其中位置被保存在經度和緯度字段的數據庫中，

我想檢索最近的位置。

任何人都可以幫忙嗎？

這是我的代碼：

var locations = from l in locations 

    select l 

以下是有關這個進一步的細節： 我有一個2名的字段（十進制（18，2）空）1個緯度，一個數據庫表內2經度，

和我有一個方法

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

這裏是解決方案

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 2

你有一個有效的範圍內，在其外側「打」是不是真的有關？如果是，請使用

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 

然後在這些結果的循環內找到具有最小平方距離值的命中。

Answer 3

您可以先將數據庫中的位置數據轉換爲System.Device.Location.GeoCoordinate，然後使用LINQ找到最近的一個。

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

Answer 4

通過@Fung，爲了詳細說明註釋如果您正在使用實體框架/ LINQ到實體，如果你嘗試使用GeoCoordinate.GetDistanceTo方法在LINQ查詢，你會得到一個運行時NotSupportedException異常與消息：

LINQ到實體無法識別方法「雙GetDistanceTo（System.Device.Location.GeoCoordinate）」的方法，而這種方法不能被翻譯成店表達。

使用實體框架版本5或6，替代方法是使用System.Data.Spatial.DbGeography類。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(); 

_context在本例中是您先前實例化的DbContext實例。

雖然它現在是undocumented in MSDN，但DbGeography.Distance方法返回的單位看起來是米。參見：System.Data.Spatial DbGeography.Distance units?

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();