在没有空间类型实体框架的情况下获取半径内的所有经度和纬度

Question

我们有一个以经纬度为列的邮政编码表，由于生产需要，无法更改为空间数据类型。

正在使用STDistance()，但出现错误：

LINQ to Entities 无法识别该方法

因此我必须调用toList 然后只使用STDistance()，这会将每个邮政编码都拉入系统并导致性能下降。

private static List<string> GetPostalCodesWithinRange( DbContext db, double latitude, double longitude)
{
    var yourLocation = SqlGeography.Point(latitude, longitude, 4326);

    var query = from postal in db.POSTALS
                                 .Where(x => x.LATITUDE != null || 
                                             x.LONGITUDE != null)
                                 .ToList()
    let distance = SqlGeography.Point((double)postal.LATITUDE.Value, (double)postal.LONGITUDE.Value, 4326).STDistance(yourLocation)
                    .Value
                where postal.LATITUDE != null && postal.LONGITUDE != null &&  distance < 3000
                orderby distance
                select postal.POSTAL_CODE;

    return query.Distinct().ToList();
}

目前PostalstoList()方法只过滤掉空的纬度和经度

db.POSTALS.Where(x => x.LATITUDE != null || x.LONGITUDE != null).ToList()

由于我们只检查 3km 以内的经纬度，现在我们想优化查询，只拉取位于 3km 正方形边界内的经度和纬度，然后使用上述方法仅按半径过滤。

请问有什么方法可以得到距离中心1.5公里的角落longlat点？

Answer 1

SQL Server 地理函数“STBuffer”将缓冲区距离添加到输入地理。您可以向 STBuffer 输入一个点，您将得到一个近似于以该点为中心的圆的多边形。然后，您可以找到该多边形的最小和最大 LAT/LONG。

（这在很远的距离上并不准确，因为地球是一个球体，但对于 1.5 公里的半径，它已经足够近了；它也会在接近 180 度经度时分解，因为接近这个你去 -180度。）

忽略这些警告，这里有一个函数可以做到这一点：

public static void GetBorderBounds(SqlGeography point, double distanceMeters, out double minLat, out double minLong, out double maxLat, out double maxLong)
{
    var buffer = point.STBuffer(distanceMeters);

    minLat = (double)point.Lat;
    maxLat = (double)point.Lat;
    minLong = (double)point.Long;
    maxLong = (double)point.Long;
    for (int i = 1; i <= buffer.STNumPoints(); i++)
    {
        var p = buffer.STPointN(i);
        if (p.Lat < minLat) { minLat = (double)p.Lat; }
        if (p.Long < minLong) { minLong = (double)p.Long; }
        if (p.Lat > minLat) { maxLat = (double)p.Lat; }
        if (p.Long > maxLong) { maxLong = (double)p.Long; }
    }
}

以下是将此函数添加到原始查询的方法：

private static List<string> GetPostalCodesWithinRange(DbContext db, double latitude, double longitude)
{
    var yourLocation = SqlGeography.Point(latitude, longitude, 4326);

    double minLat1, minLong1, maxLat1, maxLong1;
    GetBorderBounds(yourLocation, 1500, out minLat1, out minLong1, out maxLat1, out maxLong1);

    var query = from postal in db.POSTALS
                                 .Where(x => x.LATITUDE != null ||
                                             x.LONGITUDE != null)
                                 .ToList()
                let distance = SqlGeography.Point((double)postal.LATITUDE.Value, (double)postal.LONGITUDE.Value, 4326).STDistance(yourLocation)
                                .Value
                where postal.LATITUDE != null && postal.LONGITUDE != null && distance < 3000
                && postal.LATITUDE > minLat1 && postal.LATITUDE < maxLat1
                && postal.LONGITUDE > minLong1 && postal.LONGITUDE < maxLong1
                orderby distance
                select postal.POSTAL_CODE;

    return query.Distinct().ToList();
}

Answer 2

在我使用 STBuffer 发布我的第一个答案后，我意识到有一种更简单的方法可以让角点与该点保持一定距离。

我保留原来的 STBuffer 答案，因为它没有错，它可以作为使用 STBuffer 和迭代 Geography 对象中的点的示例。

这里是 GetBorderBounds 的简单实现：

public static void GetBorderBounds2(SqlGeography point, double distanceMeters, out double minLat, out double minLong, out double maxLat, out double maxLong)
{
    var metresPerDegreeLat = point.STDistance(SqlGeography.Point((double)point.Lat + 1.0, (double)point.Long, 4326));
    var metresPerDegreeLong = point.STDistance(SqlGeography.Point((double)point.Lat, (double)point.Long + 1.0, 4326));
    minLat = (double)(point.Lat - distanceMeters / metresPerDegreeLat);
    maxLat = (double)(point.Lat + distanceMeters / metresPerDegreeLat);

    minLong = (double)(point.Long - distanceMeters / metresPerDegreeLong);
    maxLong = (double)(point.Long + distanceMeters / metresPerDegreeLong);
}

此实施在北极和南极附近出现故障。此外，与第一个解决方案一样，如果边界越过 180 度子午线，它会给出错误的答案。

如果你所有的点都远离两极或第 180 条子午线，这没关系。

如果您的点可能靠近两极或第 180 条子午线，这里有一个防弹实现：

public static void GetBorderBounds3(SqlGeography point, double distanceMeters, out double minLat, out double minLong, out double maxLat, out double maxLong)
{
    // Near the North pole: 
    // Select whole circle of longitude from North pole to latitude south of point by distance
    if (point.Lat >= 89)
    {
        minLat = (double)(point.Lat - distanceMeters / point.STDistance(SqlGeography.Point((double)point.Lat - 1.0, (double)point.Long, 4326)));
        maxLat = 90;
        minLong = -180;
        maxLong = 180;
        return;
    }

    // Near the South pole: 
    // Select whole circle of longitude from South pole to latitude north of point by distance metres
    if (point.Lat <= -89)
    {
        minLat = -90;
        maxLat = (double)(point.Lat + distanceMeters / point.STDistance(SqlGeography.Point((double)point.Lat + 1.0, (double)point.Long, 4326)));
        minLong = -180;
        maxLong = 180;
        return;
    }

    var metresPerDegreeLat = point.STDistance(SqlGeography.Point((double)point.Lat + 1.0, (double)point.Long, 4326));
    var metresPerDegreeLong = point.STDistance(SqlGeography.Point((double)point.Lat, (double)point.Long + 1.0, 4326));
    minLat = (double)(point.Lat - distanceMeters / metresPerDegreeLat);
    maxLat = (double)(point.Lat + distanceMeters / metresPerDegreeLat);
    minLong = (double)(point.Long - distanceMeters / metresPerDegreeLong);
    maxLong = (double)(point.Long + distanceMeters / metresPerDegreeLong);

    // If we cross the 180th meridian, select the whole circle of longitude:
    if (minLong < -180 || maxLong > 180.0)
    {
        minLong = -180;
        maxLong = 180;
    }
}