[转载]GPS定位,经纬度附近地点查询–C#实现方法 - 小小新新 - 博客园

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[转载]GPS定位,经纬度附近地点查询–C#实现方法 – 小小新新 – 博客园.

 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#

废话不多少直接上代码:

/// <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();
        }

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
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