Oracle空间函数SDO_CS。转换(值)结果非常慢

Question

我有一个自定义视图，它从几何列中查询空间数据并提取纬度/经度值。但是，检索过程非常慢，最多需要5到10分钟才能检索到视图数据。

以下是我的观点：

CREATE OR REPLACE FORCE VIEW PoleData
(
   G3E_FID,
   X_COORD,
   Y_COORD,
   LATITUDE,
   LONGITUDE
)
AS
   SELECT P.g3e_fid,
          T2.X * 1000 AS x_coord,
          T2.Y * 1000 AS y_coord,
          T.Y AS latitude,
          T.X AS longitude
     FROM PolePoint P,
          TABLE (
             SDO_UTIL.GETVERTICES (SDO_CS.TRANSFORM (P.G3E_GEOMETRY, 8265))) T,
          TABLE (SDO_UTIL.GETVERTICES (P.G3E_GEOMETRY)) T2
    WHERE P.ltt_id = 0
   UNION
   SELECT P.g3e_fid,
          T2.X * 1000 AS x_coord,
          T2.Y * 1000 AS y_coord,
          T.Y AS latitude,
          T.X AS longitude
     FROM PoleDetailPoint P,
          TABLE (
             SDO_UTIL.GETVERTICES (SDO_CS.TRANSFORM (P.G3E_GEOMETRY, 8265))) T,
          TABLE (SDO_UTIL.GETVERTICES (P.G3E_GEOMETRY)) T2
    WHERE P.ltt_id = 0;

G3E_GEOMETRY列的类型SDO_GEOMETRY。PolePoint 表有 1，310，629 行，而 PoleDetailPoint 有 100 行。此表中的数据每天更新，而视图用于报告目的。

我尝试使用status=清理参数重建空间索引。但这没有任何区别。

我们的版本是甲骨文11.2.0.3。

任何关于检索此类视图/数据的提示都将不胜感激。或者其他我可以用来更快实现的空间功能？

Answer 1

您是否尝试过将cast to table元素移到with子句中？

通常它们非常慢，特别是如果优化器决定为连接表中的每一行运行它们（检查结果集是否已更改）

试试这个:

with t as (select * from table (sdo_util.getvertices (sdo_cs.transform (p.g3e_geometry, 8265)))),
     t2 as (select * from table (sdo_util.getvertices (p.g3e_geometry)))
select p.g3e_fid,
       t2.x * 1000 as x_coord,
       t2.y * 1000 as y_coord,
       t.y as latitude,
       t.x as longitude
  from polepoint p, t, t2
 where p.ltt_id = 0
union 
select p.g3e_fid,
       t2.x * 1000 as x_coord,
       t2.y * 1000 as y_coord,
       t.y as latitude,
       t.x as longitude
  from poledetailpoint p, t, t2
 where p.ltt_id = 0;

另外，您确定您需要< code >“union”(强制排序过滤器)而不是< code >“union all”？(例如，您是否期望重复的结果？如果没有使用< code>union all)

问候Olafur

Answer 2

尝试使用 UNION ALL 而不是 UNION：

SELECT P.g3e_fid,
       T2.X * 1000 AS x_coord,
       T2.Y * 1000 AS y_coord,
       T.Y AS latitude,
       T.X AS longitude
  FROM PolePoint P,
       TABLE (
          SDO_UTIL.GETVERTICES (SDO_CS.TRANSFORM (P.G3E_GEOMETRY, 8265))) T,
       TABLE (SDO_UTIL.GETVERTICES (P.G3E_GEOMETRY)) T2
 WHERE P.ltt_id = 0
UNION ALL
SELECT P.g3e_fid,
       T2.X * 1000 AS x_coord,
       T2.Y * 1000 AS y_coord,
       T.Y AS latitude,
       T.X AS longitude
  FROM PoleDetailPoint P,
       TABLE (
          SDO_UTIL.GETVERTICES (SDO_CS.TRANSFORM (P.G3E_GEOMETRY, 8265))) T,
       TABLE (SDO_UTIL.GETVERTICES (P.G3E_GEOMETRY)) T2
 WHERE P.ltt_id = 0;

性能下降的另一个潜在来源是使用两个调用来SDO_UTIL。GET_VERTICES一个直接在P.G3E_GEOMETRY上，另一个在P.G3E_GEOMETRY的变换上，您基本上将拥有两个顶点列表的交叉乘积，因此，例如，如果特定P.G3E_GEOMETRY包含5个顶点，那么您最终将得到5 * 5条记录，对于该5个顶点P.G3E_GEOMETRY，T和T2的25种可能组合中的每一条都有一条。我不知道顶点顺序是否由SDO_CS维持。TRANSFORM 函数，但如果是，则可以通过将 and t1.id = t2.id 谓词添加到查询的每一半来提高性能：

SELECT P.g3e_fid,
       T2.X * 1000 AS x_coord,
       T2.Y * 1000 AS y_coord,
       T.Y AS latitude,
       T.X AS longitude
  FROM PolePoint P,
       TABLE (
          SDO_UTIL.GETVERTICES (SDO_CS.TRANSFORM (P.G3E_GEOMETRY, 8265))) T,
       TABLE (SDO_UTIL.GETVERTICES (P.G3E_GEOMETRY)) T2
 WHERE P.ltt_id = 0
   AND T.ID = T2.ID
UNION ALL
SELECT P.g3e_fid,
       T2.X * 1000 AS x_coord,
       T2.Y * 1000 AS y_coord,
       T.Y AS latitude,
       T.X AS longitude
  FROM PoleDetailPoint P,
       TABLE (
          SDO_UTIL.GETVERTICES (SDO_CS.TRANSFORM (P.G3E_GEOMETRY, 8265))) T,
       TABLE (SDO_UTIL.GETVERTICES (P.G3E_GEOMETRY)) T2
 WHERE P.ltt_id = 0
   AND T.ID = T2.ID;