JavaSQlite获取最近的位置（经度和纬度）

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(...)

在此处输入图片说明

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 reliablePointF 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;    }

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我使用并定制了此参考资料并完成了它。