越來越多的Android應用都加入了“附近的人”的功能,比如微信、陌陌、淘寶等,今天分享一個demo,簡單的來實現這一功能。主要原理爲:手機端上傳gps數據到服務器,服務器從數據庫中查詢其他用戶的gps數據,分別計算2個pgs之間的距離,然後將計算好的數據返回給手機,手機進行展示。
源碼下載地址: https://github.com/feicien/studydemo
手機端項目:NearByDemo
服務器端項目:NearbyServerDemo
手機端代碼講解:
MainActivity是項目的入口Activity
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@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);
}
....
}
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查看附近的人,是需要使用用戶信息的,因此在OnCreate方法中先判斷用戶是不是第一次打開應用,如果是第一次打開應用,跳轉到LoginActivity,進行用戶信息登記:
之後便進入MainActivity:
點擊ActionBar上的附近的人,便會顯示從服務器獲取到的用戶信息(目前服務器是把所有用戶信息全部返回):
請求網絡使用的是Google在IO大會上才推出的Volley.
服務器端是使用Java web編寫的。在這裏不詳細介紹了。計算距離的邏輯是從Android的提供的接口(Location.distanceBetween)中拔來的,應該是最精確的方法了
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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);
}
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下面再提供了一種更簡單的方法(感謝@L給未來的自己)
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public static double getDistance(double lat1,double longt1 , double lat2,double longt2
) {
double PI = 3.14159265358979323; // 圓周率
double R = 6371229; // 地球的半徑
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;
}
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這裏是把地球當成圓球來處理的
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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
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經過測試,對於4千米的2點,相差爲10米左右,誤差是可以接受的,因此推薦使用該方法。