在linq中查找半径15米以内的位置
Find Locations within a radius of 15 meters in linq
我有一个位置 Table 并在其中保存经纬度位置。
我想在特殊位置的半径范围内查找位置
如何找到它?
var lon = 52.12457;
var lat = 58.9542154;
var locations=db.Locations.Where(m=>?)
将 JS 函数从 this answer 从我上面的评论翻译成 C# 得到以下内容
public static double GetDistance(double lat1, double lon1, double lat2, double lon2) {
var R = 6371; // Radius of the earth in km
var dLat = Deg2Rad(lat2-lat1); // deg2rad below
var dLon = Deg2Rad(lon2-lon1);
var a = Math.Sin(dLat/2) * Math.Sin(dLat/2) +
Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) *
Math.Sin(dLon/2) * Math.Sin(dLon/2);
var c = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1-a));
var d = R * c; // Distance in km
return d * 1000; //distance in m
}
public static double Deg2Rad(double deg) {
return deg * (Math.PI/180);
}
然后您可以轻松地在 Where
子句中调用
var lon = 52.12457;
var lat = 58.9542154;
var locations = db.Locations.Where(m=> GetDistance(lat, lon, m.lat, m.lon) < 15);
我有一个位置 Table 并在其中保存经纬度位置。
我想在特殊位置的半径范围内查找位置
如何找到它?
var lon = 52.12457;
var lat = 58.9542154;
var locations=db.Locations.Where(m=>?)
将 JS 函数从 this answer 从我上面的评论翻译成 C# 得到以下内容
public static double GetDistance(double lat1, double lon1, double lat2, double lon2) {
var R = 6371; // Radius of the earth in km
var dLat = Deg2Rad(lat2-lat1); // deg2rad below
var dLon = Deg2Rad(lon2-lon1);
var a = Math.Sin(dLat/2) * Math.Sin(dLat/2) +
Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) *
Math.Sin(dLon/2) * Math.Sin(dLon/2);
var c = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1-a));
var d = R * c; // Distance in km
return d * 1000; //distance in m
}
public static double Deg2Rad(double deg) {
return deg * (Math.PI/180);
}
然后您可以轻松地在 Where
子句中调用
var lon = 52.12457;
var lat = 58.9542154;
var locations = db.Locations.Where(m=> GetDistance(lat, lon, m.lat, m.lon) < 15);