搜索附近人和商铺功能

时间:2023-01-03 07:15:21


越来越多的Android应用都加入了“附近的人”的功能,比如微信、陌陌、淘宝等,今天分享一个demo,简单的来实现这一功能。主要原理为:手机端上传gps数据到服务器,服务器从数据库中查询其他用户的gps数据,分别计算2个pgs之间的距离,然后将计算好的数据返回给手机,手机进行展示。

源码下载地址: ​​https://github.com/feicien/studydemo​​​ 手机端项目:NearByDemo
服务器端项目:NearbyServerDemo

手机端代码讲解:
MainActivity是项目的入口Activity

@Override
protected void onCreate(Bundle savedInstanceState) {
boolean first = getSharedPreferences( "userinfo", Context.MODE_PRIVATE ).getBoolean( "first", false);
if (!first) {
Intent intent = new Intent( this, LoginActivity.class );
startActivity(intent);
}
....
}

查看附近的人,是需要使用用户信息的,因此在OnCreate方法中先判断用户是不是第一次打开应用,如果是第一次打开应用,跳转到LoginActivity,进行用户信息登记:

之后便进入MainActivity:

点击ActionBar上的附近的人,便会显示从服务器获取到的用户信息(目前服务器是把所有用户信息全部返回):

请求网络使用的是Google在IO大会上才推出的Volley.

服务器端是使用Java web编写的。在这里不详细介绍了。计算距离的逻辑是从Android的提供的接口(Location.distanceBetween)中拔来的,应该是最精确的方法了

public static double computeDistance(double lat1, double lon1,
double lat2, double lon2) {
// Based on http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf
// using the "Inverse Formula" (section 4)

int MAXITERS = 20;
// Convert lat/long to radians
lat1 *= Math.PI / 180.0;
lat2 *= Math.PI / 180.0;
lon1 *= Math.PI / 180.0;
lon2 *= Math.PI / 180.0;

double a = 6378137.0; // WGS84 major axis
double b = 6356752.3142; // WGS84 semi-major axis
double f = (a - b) / a;
double aSqMinusBSqOverBSq = (a * a - b * b) / (b * b);

double L = lon2 - lon1;
double A = 0.0;
double U1 = Math.atan((1.0 - f) * Math.tan(lat1));
double U2 = Math.atan((1.0 - f) * Math.tan(lat2));

double cosU1 = Math.cos(U1);
double cosU2 = Math.cos(U2);
double sinU1 = Math.sin(U1);
double sinU2 = Math.sin(U2);
double cosU1cosU2 = cosU1 * cosU2;
double sinU1sinU2 = sinU1 * sinU2;

double sigma = 0.0;
double deltaSigma = 0.0;
double cosSqAlpha = 0.0;
double cos2SM = 0.0;
double cosSigma = 0.0;
double sinSigma = 0.0;
double cosLambda = 0.0;
double sinLambda = 0.0;

double lambda = L; // initial guess
for (int iter = 0; iter < MAXITERS; iter++) {
double lambdaOrig = lambda;
cosLambda = Math.cos(lambda);
sinLambda = Math.sin(lambda);
double t1 = cosU2 * sinLambda;
double t2 = cosU1 * sinU2 - sinU1 * cosU2 * cosLambda;
double sinSqSigma = t1 * t1 + t2 * t2; // (14)
sinSigma = Math.sqrt(sinSqSigma);
cosSigma = sinU1sinU2 + cosU1cosU2 * cosLambda; // (15)
sigma = Math.atan2(sinSigma, cosSigma); // (16)
double sinAlpha = (sinSigma == 0) ? 0.0 :
cosU1cosU2 * sinLambda / sinSigma; // (17)
cosSqAlpha = 1.0 - sinAlpha * sinAlpha;
cos2SM = (cosSqAlpha == 0) ? 0.0 :
cosSigma - 2.0 * sinU1sinU2 / cosSqAlpha; // (18)

double uSquared = cosSqAlpha * aSqMinusBSqOverBSq; // defn
A = 1 + (uSquared / 16384.0) * // (3)
(4096.0 + uSquared *
(-768 + uSquared * (320.0 - 175.0 * uSquared)));
double B = (uSquared / 1024.0) * // (4)
(256.0 + uSquared *
(-128.0 + uSquared * (74.0 - 47.0 * uSquared)));
double C = (f / 16.0) *
cosSqAlpha *
(4.0 + f * (4.0 - 3.0 * cosSqAlpha)); // (10)
double cos2SMSq = cos2SM * cos2SM;
deltaSigma = B * sinSigma * // (6)
(cos2SM + (B / 4.0) *
(cosSigma * (-1.0 + 2.0 * cos2SMSq) -
(B / 6.0) * cos2SM *
(-3.0 + 4.0 * sinSigma * sinSigma) *
(-3.0 + 4.0 * cos2SMSq)));

lambda = L +
(1.0 - C) * f * sinAlpha *
(sigma + C * sinSigma *
(cos2SM + C * cosSigma *
(-1.0 + 2.0 * cos2SM * cos2SM))); // (11)

double delta = (lambda - lambdaOrig) / lambda;
if (Math.abs(delta) < 1.0e-12) {
break;
}
}

return b * A * (sigma - deltaSigma);
}

下面再提供了一种更简单的方法(感谢@L给未来的自己)

public static double getDistance(double lat1,double longt1 , double lat2,double longt2
) {
double PI = 3.14159265358979323; // 圆周率
double R = 6371229; // 地球的半径

double x, y, distance;
x = (longt2 - longt1) * PI * R
* Math.cos(((lat1 + lat2) / 2) * PI / 180) / 180;
y = (lat2 - lat1) * PI * R / 180;
distance = Math.hypot(x, y);

return distance;
}

这里是把地球当成圆球来处理的

System.out.println(getDistance(34.8082342, 113.6125439, 34.8002478, 113.659779));
System.out.println(computeDistance(34.8082342, 113.6125439, 34.8002478, 113.659779));

4403.3428631300785
4412.121706417052

经过测试,对于4千米的2点,相差为10米左右,误差是可以接受的,因此推荐使用该方法。