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我想使用强制转换过滤连接的结果。问题是原始字段的一部分不能转换为整数。如果在加入后应用过滤器，那将不是问题。这就是为什么我想知道是否有办法（可能是优化器提示或其他东西）在连接操作之后推送过滤器评估。

这是我为示例构建的查询。我希望它能够工作，但会因“ORA-01722 无效号码”而失败：

WITH "literal" AS (
    SELECT 1 AS "literal_id", 'abc' AS "literal"
    FROM "DUAL"
    UNION
    SELECT 2 AS "literal_id", '7' AS "literal"
    FROM "DUAL"
),
     "scalar" AS (
         SELECT 3 AS "scalar_id", 2 AS "literal_id"
         FROM "DUAL"
         CONNECT BY ROWNUM <= 10000
     )
SELECT *
FROM "scalar"
         JOIN "literal" USING ("literal_id")
WHERE TO_NUMBER("literal") > 6;

ORA-01722 被抛出是因为它应用于“文字”CTE，因此崩溃，因为 'abc' 显然不是数字。我们可以在执行计划中看到这一点：

查询执行计划

为了减少导致我的问题的可能性，我执行了该查询：

CREATE TABLE "literal" AS (
    SELECT 1 AS "literal_id", 'abc' AS "literal"
    FROM "DUAL"
    UNION
    SELECT 2 AS "literal_id", '7' AS "literal"
    FROM "DUAL"
);
CREATE TABLE "scalar" AS (
    SELECT 3 AS "scalar_id", 2 AS "literal_id"
    FROM "DUAL"
    CONNECT BY ROWNUM <= 10000
);
CREATE TABLE "joined" AS (
    SELECT *
    FROM "scalar"
             JOIN "literal" USING ("literal_id")
);
SELECT *
FROM "joined"
WHERE TO_NUMBER("literal") > 6;

效果很好。

那么，有没有办法重写这个查询（虽然我仍然需要这是一个单一的查询）所以它不会尝试转换 'abc' ？

作为参考，我在 Oracle Database 18c Standard Edition 2 Release 18.0.0.0.0 以及 Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 上进行了尝试

非常感谢。

4

2 回答 2

2

老派的技巧是构造查询，使得谓词不能被推入连接。如果您将连接放入内联视图并添加 a rownum，这将阻止优化器评估谓词，直到连接完成

WITH "literal" AS (
    SELECT 1 AS "literal_id", 'abc' AS "literal"
    FROM "DUAL"
    UNION
    SELECT 2 AS "literal_id", '7' AS "literal"
    FROM "DUAL"
),
     "scalar" AS (
         SELECT 3 AS "scalar_id", 2 AS "literal_id"
         FROM "DUAL"
         CONNECT BY ROWNUM <= 10000
     )
select *
  from (
SELECT "scalar_id",  "literal_id", "literal", rownum
FROM "scalar"
         JOIN "literal" USING ("literal_id")
)
WHERE TO_NUMBER("literal") > 6;

如果您使用的是 12.2 或更高版本，则可以利用函数的增强to_number功能在出现转换错误时返回 NULL。

WITH "literal" AS (
    SELECT 1 AS "literal_id", 'abc' AS "literal"
    FROM "DUAL"
    UNION
    SELECT 2 AS "literal_id", '7' AS "literal"
    FROM "DUAL"
),
     "scalar" AS (
         SELECT 3 AS "scalar_id", 2 AS "literal_id"
         FROM "DUAL"
         CONNECT BY ROWNUM <= 10000
     )
SELECT "scalar_id",  "literal_id", "literal"
FROM "scalar"
         JOIN "literal" USING ("literal_id")
WHERE to_number("literal" default null on conversion error) > 6;
于 2019-06-26T17:25:16.170 回答
1

对于where，使用条件转换。例如：

SELECT *
FROM "scalar" s JOIN
     "literal" l
      USING ("literal_id")
WHERE (CASE WHEN REGEXP_LIKE(l.literal, '[^[0-9]+$')
            THEN TO_NUMBER(l.literal)
       END) > 6;

至于你的问题，我不这么认为。Oracle 有一个非常复杂的优化器，因此它会重新安排操作以优化性能。您可以使用 CTE 和编译器提示来实现 CTE，但这似乎有点矫枉过正。

于 2019-06-26T16:46:31.613 回答