Question

我有一个代表长岛的多边形。我想计算从长岛内的一个地址（由一个点表示）到海岸上最近的点（由该多边形的周长表示）之间的距离。

下面是我编写的查询，但它返回的距离为0，因为该点位于多边形内。在这种情况下，我需要使用其他功能还是其他方法来解决此问题？

select /*+ ordered */ 
 sdo_nn_distance (1) distance 
from ABPROD.ABT_COASTLINE_SHAPE_DATA CSD 
where sdo_nn (CSD_LOC_GEO,sdo_geometry(2001,8307,sdo_point_type(-73.1,40.8, null),null, null),'unit=mile',1) = 'TRUE' 
and CSD_LOC_ID = '166'
and rownum = 1

Answer 1

我希望能够按照sdo_nn_distance函数的方式找到一个简单的解决方案，但是我没有这么幸运，因此我不得不依靠一些变通方法。我使用纽约岛多边形进行了测试，它的执行速度相当快。我仍然需要查看切换到北美大陆多边形时返回的速度。

select * from 
(select 
sdo_geom.sdo_distance(sdo_geometry(2001,4326,null,sdo_elem_info_array(1, 1, 1),sdo_ordinate_array(/*selected coordinates*/-72.883398, 40.895885)),
                      sdo_geometry(2001,4326,null,sdo_elem_info_array(1, 1, 1),sdo_ordinate_array(X,Y)),1,'unit=Mile') distance_mile
from 
 ABPROD.ABT_COASTLINE_SHAPE_DATA             CSD,
    --Above line identifies the table that contains all of the polygons
table(SDO_UTIL.GETVERTICES(CSD.CSD_LOC_GEO)) t
    --Above line creates a list of all of the coordinates (X,Y) that make up the polygon that the selected coordinates fall within

where 
    SDO_RELATE(CSD_LOC_GEO, sdo_geometry(2001,8307,sdo_point_type(/*selected coordinates*/-72.883398, 40.895885, null),null, null), 'mask=touch+contains') = 'TRUE'
    --Above line finds the polygon that the selected coordinates fall within

and CSD_LEVEL_NBR = 1
    --Above line limits results to land shapes, rivers and lakes are excluded

order by 1 asc
    --Above line orders results by distance_mile so that row #1 is the closest distance
)
where rownum = 1
    --Above line limits results to only the closest distance

使用Oracle SQL空间函数来计算点与围绕该点的多边形的周长之间的距离？

1 个答案: