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，不能直接加减复制代码 代码如下:/// <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：

复制代码 代码如下: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

复制代码 代码如下:SELECT id,zuobiao FROM dbo.zuobiao WHERE zuobiao<>'' AND dbo.Get_StrArrayStrOfIndex(zuobiao,',',1)>116.41021 ANDdbo.Get_StrArrayStrOfIndex(zuobiao,',',1)<116.413804 ANDdbo.Get_StrArrayStrOfIndex(zuobiao,',',2)<39.949369 ANDdbo.Get_StrArrayStrOfIndex(zuobiao,',',2)>39.945721

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