Question

我有一个NODES表和一个RESTAURANTS表及其GEOLOC和坐标。

select n.geoloc.sdo_point.y as node_y, n.geoloc.sdo_point.x as node_x,
r.state as r_state, r.geoloc.sdo_point.y as r_y, r.geoloc.sdo_point.x as r_x,
sdo_nn_distance(1)  as distance_mi
from   DC_NODE n,  RESTAURANTS r
where   sdo_nn (n.geoloc, r.geoloc, 'Unit = MILE sdo_num_res = 1',1) ='TRUE'

NODE_Y NODE_X   R_state  R_Y   R_X   distance_mi
38.      -77.       AK   56.  -132.  2752.

此处sdo_nn_distance(1)和where sdo_nn(...)应该将结果集限制为一组节点和相应的最近餐厅。 Oracle似乎忽略了最短距离子句。

我确实在节点附近有条目：

    --DC Restaurants are in the table
    select r.STATE, r.geoloc.sdo_point.y as r_y, r.geoloc.sdo_point.x as r_x
      from RESTAURANTS rwhere r.state = 'DC'

    DC  38. -77.
    DC  38. -76.
    DC  38. -77.

为什么Oracle会忽略“查找最近的”子句以及如何解决它？

Answer 1

更改了args的顺序以解决问题
sdo_nn (from_these_objects_to_choose.geoloc, for_each_of_these_objects.geoloc, 'Unit = MILE sdo_num_res = 1',1) ='TRUE'

Oracle12 sdo_nn无法返回最短距离对象

1 个答案: