数据库中记录了商家在百度标注的经纬度（如：116.412007, 39.947545）
最初想法,以圆心点为中心点，对半径做循环，半径每增加一个像素（暂定1米）再对周长做循环，到数据库中查询对应点的商家（真是一个长时间的循环工作），上网百度类似的文章有了点眉目
大致想法是已知一个中心点，一个半径，求圆包含于圆抛物线里所有的点，这样的话就需要知道所要求的这个圆的对角线的顶点，问题来了 经纬度是一个点，半径是一个距离，不能直接加减
/// <summary>
/// 经纬度坐标
/// </summary>
public class Degree
{
public Degree(double x, double y)
{
X = x;
Y = y;
}
private double x;
public double X
{
get { return x; }
set { x = value; }
}
private double y;
public double Y
{
get { return y; }
set { y = value; }
}
}
public class CoordDispose
{
private const double EARTH_RADIUS = 6378137.0;//地球半径(米)
/// <summary>
/// 角度数转换为弧度公式
/// </summary>
/// <param name="d"></param>
/// <returns></returns>
private static double radians(double d)
{
return d * Math.PI / 180.0;
}
/// <summary>
/// 弧度转换为角度数公式
/// </summary>
/// <param name="d"></param>
/// <returns></returns>
private static double degrees(double d)
{
return d * (180 / Math.PI);
}
/// <summary>
/// 计算两个经纬度之间的直接距离
/// </summary>
public static double GetDistance(Degree Degree1, Degree Degree2)
{
double radLat1 = radians(Degree1.X);
double radLat2 = radians(Degree2.X);
double a = radLat1 - radLat2;
double b = radians(Degree1.Y) - radians(Degree2.Y);
double s = 2 * Math.Asin(Math.Sqrt(Math.Pow(Math.Sin(a / 2), 2) +
Math.Cos(radLat1) * Math.Cos(radLat2) * Math.Pow(Math.Sin(b / 2), 2)));
s = s * EARTH_RADIUS;
s = Math.Round(s * 10000) / 10000;
return s;
}
/// <summary>
/// 计算两个经纬度之间的直接距离(google 算法)
/// </summary>
public static double GetDistanceGoogle(Degree Degree1, Degree Degree2)
{
double radLat1 = radians(Degree1.X);
double radLng1 = radians(Degree1.Y);
double radLat2 = radians(Degree2.X);
double radLng2 = radians(Degree2.Y);
double s = Math.Acos(Math.Cos(radLat1) * Math.Cos(radLat2) * Math.Cos(radLng1 - radLng2) + Math.Sin(radLat1) * Math.Sin(radLat2));
s = s * EARTH_RADIUS;
s = Math.Round(s * 10000) / 10000;
return s;
}
/// <summary>
/// 以一个经纬度为中心计算出四个顶点
/// </summary>
/// <param name="distance">半径(米)</param>
/// <returns></returns>
public static Degree[] GetDegreeCoordinates(Degree Degree1, double distance)
{
double dlng = 2 * Math.Asin(Math.Sin(distance / (2 * EARTH_RADIUS)) / Math.Cos(Degree1.X));
dlng = degrees(dlng);//一定转换成角度数 原PHP文章这个地方说的不清楚根本不正确 后来lz又查了很多资料终于搞定了
double dlat = distance / EARTH_RADIUS;
dlat = degrees(dlat);//一定转换成角度数
return new Degree[] { new Degree(Math.Round(Degree1.X + dlat,6), Math.Round(Degree1.Y - dlng,6)),//left-top
new Degree(Math.Round(Degree1.X - dlat,6), Math.Round(Degree1.Y - dlng,6)),//left-bottom
new Degree(Math.Round(Degree1.X + dlat,6), Math.Round(Degree1.Y + dlng,6)),//right-top
new Degree(Math.Round(Degree1.X - dlat,6), Math.Round(Degree1.Y + dlng,6)) //right-bottom
};
}
}
测试方法：
static void Main(string[] args)
{
double a = CoordDispose.GetDistance(new Degree(116.412007, 39.947545), new Degree(116.412924, 39.947918));//116.416984,39.944959
double b = CoordDispose.GetDistanceGoogle(new Degree(116.412007, 39.947545), new Degree(116.412924, 39.947918));
Degree[] dd = CoordDispose.GetDegreeCoordinates(new Degree(116.412007, 39.947545), 102);
Console.WriteLine(a+" "+b);
Console.WriteLine(dd[0].X + "," + dd[0].Y );
Console.WriteLine(dd[3].X + "," + dd[3].Y);
Console.ReadLine();
}
试了很多次 误差在1米左右
拿到圆的顶点就好办了
数据库要是sql 2008的可以直接进行空间索引经纬度字段，这样应该性能更好（没有试过）
lz公司数据库还老 2005的 这也没关系，关键是经纬度拆分计算，这个就不用说了 网上多的是 最后上个实现的sql语句
SELECT id,zuobiao FROM dbo.zuobiao WHERE zuobiao<>'' AND
dbo.Get_StrArrayStrOfIndex(zuobiao,',',1)>116.41021 AND
dbo.Get_StrArrayStrOfIndex(zuobiao,',',1)<116.413804 AND
dbo.Get_StrArrayStrOfIndex(zuobiao,',',2)<39.949369 AND
dbo.Get_StrArrayStrOfIndex(zuobiao,',',2)>39.945721
