1）首先，以近似值过滤sqlite数据，并减少需要在Java代码中评估的数据量。为此，请使用以下过程：
为了具有确定性的阈值并更准确地过滤数据，最好在Java代码中计算以中心点的北，西，东和南为单位的4个位置，然后轻松检查小于和等于sql运算符（>，<）确定数据库中的点是否在该矩形中。radius
该方法calculateDerivedPosition(...)为你计算这些点（图片中的p1，p2，p3，p4）。
在此处输入图片说明
/**
* Calculates the end-point from a given source at a given range (meters)
* and bearing (degrees). This methods uses simple geometry equations to
* calculate the end-point.
*
* @param point
* Point of origin
* @param range
* Range in meters
* @param bearing
* Bearing in degrees
* @return End-point from the source given the desired range and bearing.
*/
public static PointF calculateDerivedPosition(PointF point,
double range, double bearing)
{
double EarthRadius = 6371000; // m
double latA = Math.toradians(point.x);
double lonA = Math.toradians(point.y);
double angularDistance = range / EarthRadius;
double trueCourse = Math.toradians(bearing);
double lat = Math.asin(
Math.sin(latA) * Math.cos(angularDistance) +
Math.cos(latA) * Math.sin(angularDistance)
* Math.cos(trueCourse));
double dlon = Math.atan2(
Math.sin(trueCourse) * Math.sin(angularDistance)
* Math.cos(latA),
Math.cos(angularDistance) - Math.sin(latA) * Math.sin(lat));
double lon = ((lonA + dlon + Math.PI) % (Math.PI * 2)) - Math.PI;
lat = Math.todegrees(lat);
lon = Math.todegrees(lon);
PointF newPoint = new PointF((float) lat, (float) lon);
return newPoint;
}
现在创建你的查询：
PointF center = new PointF(x, y);
final double mult = 1; // mult = 1.1; is more reliable
PointF p1 = calculateDerivedPosition(center, mult * radius, 0);
PointF p2 = calculateDerivedPosition(center, mult * radius, 90);
PointF p3 = calculateDerivedPosition(center, mult * radius, 180);
PointF p4 = calculateDerivedPosition(center, mult * radius, 270);
strWhere = " WHERE "
+ COL_X + " > " + String.valueOf(p3.x) + " AND "
+ COL_X + " < " + String.valueOf(p1.x) + " AND "
+ COL_Y + " < " + String.valueOf(p2.y) + " AND "
+ COL_Y + " > " + String.valueOf(p4.y);
+
COL_X是数据库中存储纬度值且COL_Y用于经度的列的名称。
因此，你可以得到一些近似于中心点的数据。
2）现在，你可以循环使用这些过滤后的数据，并使用以下方法确定它们是否真的在你的点附近（圆圈中）：
public static boolean pointIsInCircle(PointF pointForCheck, PointF center,
double radius) {
if (getDistanceBetweenTwoPoints(pointForCheck, center) <= radius)
return true;
else
return false;
}
public static double getDistanceBetweenTwoPoints(PointF p1, PointF p2) {
double R = 6371000; // m
double dLat = Math.toradians(p2.x - p1.x);
double dLon = Math.toradians(p2.y - p1.y);
double lat1 = Math.toradians(p1.x);
double lat2 = Math.toradians(p2.x);
double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.sin(dLon / 2)
* Math.sin(dLon / 2) * Math.cos(lat1) * Math.cos(lat2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
double d = R * c;
return d;
}
请享用！
我使用并定制了此参考资料并完成了它。
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