在SQL Server 2008中沿路径移动点

Question

我有一个地理字段存储在我的数据库中,持有一个线串路径.

我想n沿着这个线串移动一个点米,然后返回目的地.

例如,我希望目标点从其开始处沿着线串500米.

这是一个例子 - 什么是YourFunctionHere？或者,还有另一种方式吗？

DECLARE @g geography;
SET @g = geography::STGeomFromText('LINESTRING(-122.360 47.656, -122.343 47.656, -122.310 47.690)', 4326);
SELECT @g.YourFunctionHere(100).ToString();

Answer 1

这有点棘手,但肯定是可能的.

让我们从一个点到另一个点计算轴承.给定起点,方位和距离,以下函数将返回目标点:

CREATE FUNCTION [dbo].[func_MoveTowardsPoint](@start_point geography,
                                              @end_point   geography,  
                                              @distance    int)  /* Meters */   
RETURNS geography
AS
BEGIN
    DECLARE @ang_dist float = @distance / 6371000.0;  /* Earth's radius */
    DECLARE @bearing  decimal(18,15);
    DECLARE @lat_1    decimal(18,15) = Radians(@start_point.Lat);
    DECLARE @lon_1    decimal(18,15) = Radians(@start_point.Long);
    DECLARE @lat_2    decimal(18,15) = Radians(@end_point.Lat);
    DECLARE @lon_diff decimal(18,15) = Radians(@end_point.Long - @start_point.Long);
    DECLARE @new_lat  decimal(18,15);
    DECLARE @new_lon  decimal(18,15);
    DECLARE @result   geography;

    /* First calculate the bearing */

    SET @bearing = ATN2(sin(@lon_diff) * cos(@lat_2),
                        (cos(@lat_1) * sin(@lat_2)) - 
                        (sin(@lat_1) * cos(@lat_2) * 
                        cos(@lon_diff)));

    /* Then use the bearing and the start point to find the destination */

    SET @new_lat = asin(sin(@lat_1) * cos(@ang_dist) + 
                        cos(@lat_1) * sin(@ang_dist) * cos(@bearing));

    SET @new_lon = @lon_1 + atn2( sin(@bearing) * sin(@ang_dist) * cos(@lat_1), 
                                  cos(@ang_dist) - sin(@lat_1) * sin(@lat_2));

    /* Convert from Radians to Decimal */

    SET @new_lat = Degrees(@new_lat);
    SET @new_lon = Degrees(@new_lon);

    /* Return the geography result */

    SET @result = 
        geography::STPointFromText('POINT(' + CONVERT(varchar(64), @new_lon) + ' ' + 
                                              CONVERT(varchar(64), @new_lat) + ')', 
                                   4326);

    RETURN @result;
END

我知道你需要一个以线串作为输入的函数,而不仅仅是起点和终点.该点必须沿着连接线段的路径移动,并且必须继续围绕路径的"角"移动.这可能看起来很复杂,但我认为它可以解决如下:

1. 遍历线串的每个点STPointN(),从x = 1到x = STNumPoints().
2. 找到STDistance()迭代中当前点到下一个点之间的距离:@linestring.STPointN(x).STDistance(@linestring.STPointN(x+1))

3. 如果上述距离>输入距离'n':

   ...然后目标点位于此点和下一点之间.只需将 func_MoveTowardsPoint经过点x作为起点,将点x + 1作为终点,并将距离设为n.返回结果并中断迭代.

   其他:

   ...目标点位于迭代中下一个点的路径中.从距离'n'中减去点x和点x + 1之间的距离.使用修改的距离继续迭代.

您可能已经注意到我们可以轻松地递归地实现上述内容,而不是迭代地实现上述内容.

我们开始做吧:

CREATE FUNCTION [dbo].[func_MoveAlongPath](@path geography, 
                                           @distance int, 
                                           @index int = 1)   
RETURNS geography
AS
BEGIN
    DECLARE @result       geography = null;
    DECLARE @num_points   int = @path.STNumPoints();
    DECLARE @dist_to_next float;

    IF @index < @num_points
    BEGIN
        /* There is still at least one point further from the point @index
           in the linestring. Find the distance to the next point. */

        SET @dist_to_next = @path.STPointN(@index).STDistance(@path.STPointN(@index + 1));

        IF @distance <= @dist_to_next 
        BEGIN
            /* @dist_to_next is within this point and the next. Return
              the destination point with func_MoveTowardsPoint(). */

            SET @result = [dbo].[func_MoveTowardsPoint](@path.STPointN(@index),
                                                        @path.STPointN(@index + 1),
                                                        @distance);
        END
        ELSE
        BEGIN
            /* The destination is further from the next point. Subtract
               @dist_to_next from @distance and continue recursively. */

            SET @result = [dbo].[func_MoveAlongPath](@path, 
                                                     @distance - @dist_to_next,
                                                     @index + 1);
        END
    END
    ELSE
    BEGIN
        /* There is no further point. Our distance exceeds the length 
           of the linestring. Return the last point of the linestring.
           You may prefer to return NULL instead. */

        SET @result = @path.STPointN(@index);
    END

    RETURN @result;
END

有了它,是时候做一些测试了.让我们使用问题中提供的原始线串,我们将在350米,3500米和7000米处请求目标点:

DECLARE @g geography;
SET @g = geography::STGeomFromText('LINESTRING(-122.360 47.656, 
                                               -122.343 47.656, 
                                               -122.310 47.690)', 4326);

SELECT [dbo].[func_MoveAlongPath](@g, 350, DEFAULT).ToString();
SELECT [dbo].[func_MoveAlongPath](@g, 3500, DEFAULT).ToString();
SELECT [dbo].[func_MoveAlongPath](@g, 7000, DEFAULT).ToString();

我们的测试返回以下结果:

POINT (-122.3553270591861 47.6560002502638)
POINT (-122.32676470116748 47.672728464582583)
POINT (-122.31 47.69)

请注意,我们请求的最后一个距离(7000米)超过了线串的长度,因此我们返回了最后一个点.在这种情况下,如果您愿意,可以轻松修改函数以返回NULL.