在任何其他 SQL 之前先执行子查询重构

Question

我有一个非常复杂的视图，格式如下

create or replace view loan_vw as 
select * from (with loan_info as (select loan_table.*,commission_table.* 
                                   from loan_table,
                                  commission_table where 
                                  contract_id=commission_id)
                select /*complex transformations */ from loan_info
                where type <> 'PRINCIPAL'
                union all 
                select /*complex transformations */ from loan_info
                where type = 'PRINCIPAL')

现在如果我执行以下操作，则选择查询挂起

         select * from loan_vw where contract_id='HA001234TY56';

但是，如果我在子查询重构中硬编码或在同一会话中使用包级变量，查询会在一秒钟内返回

create or replace view loan_vw as 
        select * from (with loan_info as (select loan_table.*,commission_table.* 
                                           from loan_table,
                                          commission_table where 
                                          contract_id=commission_id
                                          and contract_id='HA001234TY56'
                                          )
                        select /*complex transformations */ from loan_info
                        where type <> 'PRINCIPAL'
                        union all 
                        select /*complex transformations */ from loan_info
                        where type = 'PRINCIPAL')

由于我使用业务对象，我不能使用包级变量

所以我的问题是Oracle中有一个提示告诉优化器首先检查子查询重构中loan_vw中的contract_id

根据要求，使用的分析函数如下

select value_date, item, credit_entry, item_paid
from (
  select value_date, item, credit_entry, debit_entry,
    greatest(0, least(credit_entry, nvl(sum(debit_entry) over (), 0)
      - nvl(sum(credit_entry) over (order by value_date
          rows between unbounded preceding and 1 preceding), 0))) as item_paid
  from your_table
)
where item is not null;

在遵循 Boneist 和 MarcinJ 的建议后，我删除了子查询重构 (CTE)，并编写了一个长查询，如下所示，将性能从 3 分钟提高到 0.156 秒

  create or replace view loan_vw as
  select /*complex transformations */
                               from loan_table,
                              commission_table where 
                              contract_id=commission_id
               and loan_table.type <> 'PRINCIPAL'
  union all
  select /*complex transformations */
                               from loan_table,
                              commission_table where 
                              contract_id=commission_id
               and loan_table.type = 'PRINCIPAL'

Answer 1

这些转换真的那么复杂，你必须使用UNION ALL吗？优化你看不到的东西真的很难，但你是否尝试过摆脱 CTE 并内联实现你的计算？

CREATE OR REPLACE VIEW loan_vw AS
SELECT loan.contract_id
     , CASE commission.type -- or wherever this comes from
         WHEN 'PRINCIPAL'
         THEN SUM(whatever) OVER (PARTITION BY loan.contract_id, loan.type) -- total_whatever

         ELSE SUM(something_else) OVER (PARTITION BY loan.contract_id, loan.type) -- total_something_else
      END AS whatever_something
  FROM loan_table loan 
 INNER 
  JOIN commission_table commission
    ON loan.contract_id = commission.commission_id

请注意，如果您的分析函数没有 PARTITION BY contract_id，您将根本无法在该 contract_id 列上使用索引。

Take a look at this db fiddle（您必须单击最后一个结果表上的... 以展开结果）。在这里，loan 表有一个索引 (PK) contract_id 列，还有 some_other_id，它也是唯一的，但没有索引，外部查询的谓词仍在 contract_id 上。如果您比较partition by contract 和partition by other id 的计划，您会发现partition by other id 计划中根本没有使用索引：与@ 相比，贷款表上有TABLE ACCESS 和FULL 选项987654336@ - UNIQUE SCAN 在partition by contract。这显然是因为优化器无法自行解决contract_id 和some_other_id 之间的关系，因此它需要在整个窗口上运行SUM 或AVG，而不是通过使用索引来限制窗口行数。

如果您有一个包含这些合同的维度表，您还可以尝试将其加入您的结果并从维度表中公开contract_id，而不是最有可能的巨额贷款事实表。有时，这可以通过在维度表上使用唯一索引来改进基数估计。

同样，如果没有查询甚至没有计划，优化黑盒真的很困难，所以我们不知道发生了什么。例如，CTE 或子查询可能会不必要地具体化。

Answer 2

感谢更新以包含列列表示例。

鉴于您更新的查询，我建议更改您的视图（或者可能创建第二个视图来查询单个contract_id，如果您的原始视图可用于查询多个contract_id - 当然，除非原始视图的结果仅对单个contract_ids有意义！）类似于：

CREATE OR REPLACE VIEW loan_vw AS 
WITH loan_info AS (SELECT l.*, c.* -- for future-proofing, you should list the column names explicitly; if this statement is rerun and there's a column with the same name in both tables, it'll fail.
                   FROM   loan_table l
                          INNER JOIN commission_table c ON l.contract_id = c.commission_id -- you should always alias the join condition columns for ease of maintenance.
                  )
SELECT value_date,
     item,
     credit_entry,
     debit_entry,
     GREATEST(0,
            LEAST(credit_entry,
                NVL(SUM(debit_entry) OVER (PARTITION BY contract_id), 0)
                  - NVL(SUM(credit_entry) OVER (PARTITION BY contract_id ORDER BY value_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0))) AS item_paid
FROM   loan_info
WHERE  TYPE <> 'PRINCIPAL'
UNION ALL
SELECT ...
FROM   loan_info
WHERE  TYPE = 'PRINCIPAL';

请注意，我已将您的联接转换为 ANSI 语法，因为它比旧式联接更容易理解（首先，更容易将联接条件与谓词分开！）。