使用GPS经纬度定位附近地点(某一点范围内查询)-白红宇

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

代码如下:

///     /// 经纬度坐标    /// 
    　　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)        ///         /// 角度数转换为弧度公式        /// 
        ///         /// 
             private static double radians(double d)        {            return d * Math.PI / 180.0;        }        ///         /// 弧度转换为角度数公式        /// 
        ///         /// 
             private static double degrees(double d)        {            return d * (180 / Math.PI);        }        ///         /// 计算两个经纬度之间的直接距离        /// 
        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;        }        ///         /// 计算两个经纬度之间的直接距离(google 算法)        /// 
        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;        }        ///         /// 以一个经纬度为中心计算出四个顶点        /// 
        /// 半径(米)        /// 
             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：

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

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