Oracle CTE性能问题

为了使比较不区分大小写，但正确显示，我们在下面编写了查询

WITH VALUE_LIST
     AS (SELECT *
           FROM (SELECT m1."ASSIGNMENT" col#0,
                        m1."ASSIGNEE_NAME" col#1,
                        ROW_NUMBER ()
                        OVER (
                           PARTITION BY m1."ASSIGNMENT", m1."ASSIGNEE_NAME"
                           ORDER BY m1."ASSIGNMENT", m1."ASSIGNEE_NAME")
                           ROW_INDEX
                   FROM PROBSUMMARYM1 m1)
          WHERE ROW_INDEX = 1)
SELECT vl.col#0, vl.col#1, GROUPED_DATA.CNT
  FROM (  SELECT UPPER (m1."ASSIGNMENT") col#0,
                 UPPER (m1."ASSIGNEE_NAME") col#1,
                 COUNT (*) CNT
            FROM PROBSUMMARYM1 m1
        GROUP BY UPPER (m1."ASSIGNMENT"), UPPER (m1."ASSIGNEE_NAME")
        ORDER BY UPPER (m1."ASSIGNMENT") ASC, UPPER (m1."ASSIGNEE_NAME") ASC)
       GROUPED_DATA,
       VALUE_LIST vl
 WHERE     (   GROUPED_DATA.col#0 = UPPER (vl.col#0)
            OR (GROUPED_DATA.col#0 IS NULL AND vl.col#0 IS NULL))
       AND (   GROUPED_DATA.col#1 = UPPER (vl.col#1)
            OR (GROUPED_DATA.col#1 IS NULL AND vl.col#1 IS NULL))

但随着组数的增加，性能显著下降。 是否有任何查询可以获得相同的查询，但没有性能问题

尝试此查询：

SELECT ASSIGNMENT, 
       ASSIGNEE_NAME,
       CNT
FROM (
  SELECT m1."ASSIGNMENT", 
         m1."ASSIGNEE_NAME",
         row_number() OVER (
               PARTITION BY m1."ASSIGNMENT", m1."ASSIGNEE_NAME"
               ORDER BY m1."ASSIGNMENT", m1."ASSIGNEE_NAME"
         )  As ROW_INDEX,
         COUNT(*) OVER (
               PARTITION BY UPPER (m1."ASSIGNMENT"), 
                            UPPER (m1."ASSIGNEE_NAME")
         ) As CNT
  FROM PROBSUMMARYM1 m1
)
WHERE ROW_INDEX = 1
;

作为旁注，问题中的这部分查询：

WITH VALUE_LIST
     AS (SELECT *
           FROM (SELECT m1."ASSIGNMENT" col#0,
                        m1."ASSIGNEE_NAME" col#1,
                        ROW_NUMBER ()
                        OVER (
                           PARTITION BY m1."ASSIGNMENT", m1."ASSIGNEE_NAME"
                           ORDER BY m1."ASSIGNMENT", m1."ASSIGNEE_NAME")
                           ROW_INDEX
                   FROM PROBSUMMARYM1 m1)
          WHERE ROW_INDEX = 1)

这是一个非常复杂的版本：

SELECT DISTINCT m1."ASSIGNMENT" col#0, 
                m1."ASSIGNEE_NAME" col#1
FROM PROBSUMMARYM1 m1

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您是否尝试过基于函数的索引

在PROBSUMMARYM1（上限（“赋值”）、上限（“受让人名称”））上创建索引somename

？