目前的工作是需要手机查找附近N米以内的商户,功能如下图
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
终于找到想要的文章
http://digdeeply.org/archives/06152067.html
PHP,Mysql-根据一个给定经纬度的点，进行附近地点查询–合理利用算法，效率提高2125倍
参考原文章 lz改成了C#类
废话不多少直接上代码:
1 /// <summary> 2 /// 经纬度坐标 3 /// </summary> 4 5 public class Degree 6 { 7 public Degree(double x, double y) 8 { 9 X = x; 10 Y = y; 11 } 12 private double x; 13 14 public double X 15 { 16 get { return x; } 17 set { x = value; } 18 } 19 private double y; 20 21 public double Y 22 { 23 get { return y; } 24 set { y = value; } 25 } 26 } 27 28 29 public class CoordDispose 30 { 31 private const double EARTH_RADIUS = 6378137.0;//地球半径(米) 32 33 /// <summary> 34 /// 角度数转换为弧度公式 35 /// </summary> 36 /// <param name="d"></param> 37 /// <returns></returns> 38 private static double radians(double d) 39 { 40 return d * Math.PI / 180.0; 41 } 42 43 /// <summary> 44 /// 弧度转换为角度数公式 45 /// </summary> 46 /// <param name="d"></param> 47 /// <returns></returns> 48 private static double degrees(double d) 49 { 50 return d * (180 / Math.PI); 51 } 52 53 /// <summary> 54 /// 计算两个经纬度之间的直接距离 55 /// </summary> 56 57 public static double GetDistance(Degree Degree1, Degree Degree2) 58 { 59 double radLat1 = radians(Degree1.X); 60 double radLat2 = radians(Degree2.X); 61 double a = radLat1 - radLat2; 62 double b = radians(Degree1.Y) - radians(Degree2.Y); 63 64 double s = 2 * Math.Asin(Math.Sqrt(Math.Pow(Math.Sin(a / 2), 2) + 65 Math.Cos(radLat1) * Math.Cos(radLat2) * Math.Pow(Math.Sin(b / 2), 2))); 66 s = s * EARTH_RADIUS; 67 s = Math.Round(s * 10000) / 10000; 68 return s; 69 } 70 71 /// <summary> 72 /// 计算两个经纬度之间的直接距离(google 算法) 73 /// </summary> 74 public static double GetDistanceGoogle(Degree Degree1, Degree Degree2) 75 { 76 double radLat1 = radians(Degree1.X); 77 double radLng1 = radians(Degree1.Y); 78 double radLat2 = radians(Degree2.X); 79 double radLng2 = radians(Degree2.Y); 80 81 double s = Math.Acos(Math.Cos(radLat1) * Math.Cos(radLat2) * Math.Cos(radLng1 - radLng2) + Math.Sin(radLat1) * Math.Sin(radLat2)); 82 s = s * EARTH_RADIUS; 83 s = Math.Round(s * 10000) / 10000; 84 return s; 85 } 86 87 /// <summary> 88 /// 以一个经纬度为中心计算出四个顶点 89 /// </summary> 90 /// <param name="distance">半径(米)</param> 91 /// <returns></returns> 92 public static Degree[] GetDegreeCoordinates(Degree Degree1, double distance) 93 { 94 double dlng = 2 * Math.Asin(Math.Sin(distance / (2 * EARTH_RADIUS)) / Math.Cos(Degree1.X)); 95 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 96 97 double dlat = distance / EARTH_RADIUS; 98 dlat = degrees(dlat);//一定转换成角度数 99 100 return new Degree[] { new Degree(Math.Round(Degree1.X + dlat,6), Math.Round(Degree1.Y - dlng,6)),//left-top 101 new Degree(Math.Round(Degree1.X - dlat,6), Math.Round(Degree1.Y - dlng,6)),//left-bottom 102 new Degree(Math.Round(Degree1.X + dlat,6), Math.Round(Degree1.Y + dlng,6)),//right-top 103 new Degree(Math.Round(Degree1.X - dlat,6), Math.Round(Degree1.Y + dlng,6)) //right-bottom 104 }; 105 106 } 107 }
测试方法：
1 static void Main(string[] args) 2 { 3 double a = CoordDispose.GetDistance(new Degree(116.412007, 39.947545), new Degree(116.412924, 39.947918));//116.416984,39.944959 4 double b = CoordDispose.GetDistanceGoogle(new Degree(116.412007, 39.947545), new Degree(116.412924, 39.947918)); 5 Degree[] dd = CoordDispose.GetDegreeCoordinates(new Degree(116.412007, 39.947545), 102); 6 Console.WriteLine(a+" "+b); 7 Console.WriteLine(dd[0].X + "," + dd[0].Y ); 8 Console.WriteLine(dd[3].X + "," + dd[3].Y); 9 Console.ReadLine(); 10 }
lz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