我想计算每个点之间的总距离,都包含经度和纬度,这些点都存储在本地数据库中,所以方案是我要计算从点a到b&b到c&c到d以及这些点的距离(纬度和经度)使用Google Place API
存储到数据库中所有点均从数据库中获取。
您无需使用places api即可计算2个位置点之间的距离>
private double distance(double lat1, double lon1, double lat2, double lon2) {
double theta = lon1 - lon2;
double dist = Math.sin(deg2rad(lat1))
* Math.sin(deg2rad(lat2))
+ Math.cos(deg2rad(lat1))
* Math.cos(deg2rad(lat2))
* Math.cos(deg2rad(theta));
dist = Math.acos(dist);
dist = rad2deg(dist);
dist = dist * 60 * 1.1515;
return (dist);
}
private double deg2rad(double deg) {
return (deg * Math.PI / 180.0);
}
private double rad2deg(double rad) {
return (rad * 180.0 / Math.PI);
}
我不确定我是否理解此问题的范围。您是否需要考虑地球的曲率?还是只有2个随机点?
class DistanceCalculator
{
public static void main (String[] args) throws java.lang.Exception
{
System.out.println(distance(32.9697, -96.80322, 29.46786, -98.53506, "M") + " Miles\n");
System.out.println(distance(32.9697, -96.80322, 29.46786, -98.53506, "K") + " Kilometers\n");
System.out.println(distance(32.9697, -96.80322, 29.46786, -98.53506, "N") + " Nautical Miles\n");
}
private static double distance(double lat1, double lon1, double lat2, double lon2, String unit) {
if ((lat1 == lat2) && (lon1 == lon2)) {
return 0;
}
else {
double theta = lon1 - lon2;
double dist = Math.sin(Math.toRadians(lat1)) * Math.sin(Math.toRadians(lat2)) + Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2)) * Math.cos(Math.toRadians(theta));
dist = Math.acos(dist);
dist = Math.toDegrees(dist);
dist = dist * 60 * 1.1515;
if (unit == "K") {
dist = dist * 1.609344;
} else if (unit == "N") {
dist = dist * 0.8684;
}
return (dist);
}
}
}
float[] distanceWidth = new float[2];
Location.distanceBetween(
startLatitude,
startLongitude,
endLatitude,
endLongitude,
distanceWidth);
//You will get float[] of results after Location.distanceBetween().
float distance = distanceHeight[0];
float distanceInKm = distance/1000;