使用GPS經(jīng)緯度定位附近地點(某一點范圍內(nèi)查詢)

更新時間：2013年12月30日 09:49:19 作者：

目前的工作是需要手機查找附近N米以內(nèi)的商戶,致想法是已知一個中心點，一個半徑，求圓包含于圓拋物線里所有的點,經(jīng)緯度是一個點，半徑是一個距離，不能直接加減,下面提供C#的解決方法

數(shù)據(jù)庫中記錄了商家在百度標注的經(jīng)緯度（如：116.412007, 39.947545）

最初想法,以圓心點為中心點，對半徑做循環(huán)，半徑每增加一個像素（暫定1米）再對周長做循環(huán)，到數(shù)據(jù)庫中查詢對應點的商家（真是一個長時間的循環(huán)工作），上網(wǎng)百度類似的文章有了點眉目

大致想法是已知一個中心點，一個半徑，求圓包含于圓拋物線里所有的點，這樣的話就需要知道所要求的這個圓的對角線的頂點，問題來了經(jīng)緯度是一個點，半徑是一個距離，不能直接加減

復制代碼代碼如下:

/// <summary>
/// 經(jīng)緯度坐標
/// </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;//地球半徑(米)

        /// <summary>
        /// 角度數(shù)轉(zhuǎn)換為弧度公式
        /// </summary>
        /// <param name="d"></param>
        /// <returns></returns>
        private static double radians(double d)
        {
            return d * Math.PI / 180.0;
        }

        /// <summary>
        /// 弧度轉(zhuǎn)換為角度數(shù)公式
        /// </summary>
        /// <param name="d"></param>
        /// <returns></returns>
        private static double degrees(double d)
        {
            return d * (180 / Math.PI);
        }

        /// <summary>
        /// 計算兩個經(jīng)緯度之間的直接距離
        /// </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>
        /// 計算兩個經(jīng)緯度之間的直接距離(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>
        /// 以一個經(jīng)緯度為中心計算出四個頂點
        /// </summary>
        /// <param name="distance">半徑(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);//一定轉(zhuǎn)換成角度數(shù) 原PHP文章這個地方說的不清楚根本不正確后來lz又查了很多資料終于搞定了

double dlat = distance / EARTH_RADIUS;
dlat = degrees(dlat);//一定轉(zhuǎn)換成角度數(shù)

            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
            };

}
}

測試方法：

復制代碼代碼如下:

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();
        }

試了很多次誤差在1米左右

拿到圓的頂點就好辦了

數(shù)據(jù)庫要是sql 2008的可以直接進行空間索引經(jīng)緯度字段，這樣應該性能更好（沒有試過）

lz公司數(shù)據(jù)庫還老 2005的這也沒關(guān)系，關(guān)鍵是經(jīng)緯度拆分計算，這個就不用說了網(wǎng)上多的是最后上個實現(xiàn)的sql語句

復制代碼代碼如下:

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