使用GPS经纬度定位附近地点(某一点范围内查询)
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
/// <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();
}
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