使用LINQ从数据库中查找附近的位置

Swa*_*ule 5 .net linq asp.net c#-4.0 sql-to-linq-conversion

我们希望在ASP.NET 2012中使用LINQ从数据库接收附近地点的列表,并希望对我们的策略有一些反馈.


我的表和假数据:

     PlaceId    Name       Latitude   Longitude 
       1          A          18.1        20.1
       2          B          18.2        20.2
       3          C          18.3        20.3
Run Code Online (Sandbox Code Playgroud)

1)在我们的项目中,客户端当前位置(纬度和经度)被视为输入

2)在服务器端,根据客户端的当前位置,我们需要使用LINQ从数据库中查找附近的位置

这是我之前使用的SQL代码,但现在我们想使用LINQ.

SELECT  name, Latitude, Longitude , 
  ( 3959 * acos( cos( radians(?) )* cos( radians( Latitude) ) * cos( radians( Longitude ) - radians(?) ) 
 + sin( radians(?) ) * sin( radians( Latitude) ) ) ) AS distance 
FROM TABLE_NAME 
HAVING distance < ? 
ORDER BY distance LIMIT 0 , 20
Run Code Online (Sandbox Code Playgroud)

[但问题是如何在LINQ中编写这样的查询.]

我的工作:

在搜索解决方案时,我遇到了这段代码

        var Value1 = 57.2957795130823D;
        var Value2 = 3958.75586574D;

        var searchWithin = 20;

    double latitude = ConversionHelper.SafeConvertToDoubleCultureInd(Latitude, 0),
          longitude = ConversionHelper.SafeConvertToDoubleCultureInd(Longitude, 0);

    var location = (from l in sdbml.Places
                    let temp = Math.Sin(Convert.ToDouble(l.Latitude) / Value1) *  Math.Sin(Convert.ToDouble(latitude) / Value1) +
                             Math.Cos(Convert.ToDouble(l.Latitude) / Value1) *
                             Math.Cos(Convert.ToDouble(latitude) / Value1) *
                             Math.Cos((Convert.ToDouble(longitude) / Value1) - (Convert.ToDouble(l.Longitude) / Value1))
                         let calMiles = (Value2 * 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 location .ToList();
Run Code Online (Sandbox Code Playgroud)

但问题是,如何引用ConversionHelper或它来自哪个命名空间.

所有建议表示赞赏.

Swa*_*ule 4

这是我最终必须解决的代码

1)创建一个类,说

距离模型.cs

public class DistanceModel
{
   public int PlaceId { get; set; }

   public string Name { get; set; }

   public double Latitute { get; set; }

   public double Longitude { get; set; }

} 
Run Code Online (Sandbox Code Playgroud)

2)然后将以下代码包含在您想要的任何文件中,例如

主页.cs

     /*Call GetAllNearestFamousPlaces() method to get list of nearby places depending 
      upon user current location.
      Note: GetAllNearestFamousPlaces() method takes 2 parameters as input
     that is GetAllNearestFamousPlaces(user_current_Latitude,user_current_Longitude) */


   public void GetAllNearestFamousPlaces(double currentLatitude,double currentLongitude)
    {
        List<DistanceModel> Caldistance = new List<DistanceModel>();
        var query = (from c in sdbml.Places
                     select c).ToList();
        foreach (var place in query)
        {
            double distance = Distance(currentLatitude, currentLongitude, place.Latitude, place.Logitude);
            if (distance < 25)          //nearbyplaces which are within 25 kms 
            {
                DistanceModel dist = new DistanceModel();
                dist.Name = place.PlaceName;
                dist.Latitute = place.Latitude;
                dist.Longitude = place.Logitude;
                dist.PlaceId = place.PlaceId;
                Caldistance.Add(getDiff);
            }
        }                      
    }

   private double Distance(double lat1, double lon1, double lat2, double lon2)
    {
        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 * 1.1515) / 0.6213711922;          //miles to kms
        return (dist);
    }

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

   private double rad2deg(double rad)
    {
        return (rad * 180.0 / Math.PI);
    }
Run Code Online (Sandbox Code Playgroud)

它对我有用,希望对您有帮助。