在 select 语句中调整 oracle 子查询答案

【问题标题】：Tuning oracle subquery in select statement在 select 语句中调整 oracle 子查询
【发布时间】：2015-12-19 08:19:53
【问题描述】：

我有一个主表和一个参考表，如下所示。

WITH MAS as (
SELECT 10 as CUSTOMER_ID, 1 PROCESS_ID, 44 PROCESS_TYPE, 200 as AMOUNT FROM DUAL UNION ALL
SELECT 10 as CUSTOMER_ID, 1 PROCESS_ID, 44 PROCESS_TYPE, 250 as AMOUNT FROM DUAL UNION ALL
SELECT 10 as CUSTOMER_ID, 2 PROCESS_ID, 45 PROCESS_TYPE, 300 as AMOUNT FROM DUAL UNION ALL
SELECT 10 as CUSTOMER_ID, 2 PROCESS_ID, 45 PROCESS_TYPE, 350 as AMOUNT FROM DUAL 
), REFTAB as (
SELECT 44 PROCESS_TYPE, 'A' GROUP_ID FROM DUAL UNION ALL 
SELECT 44 PROCESS_TYPE, 'B' GROUP_ID FROM DUAL UNION ALL
SELECT 45 PROCESS_TYPE, 'C' GROUP_ID FROM DUAL UNION ALL 
SELECT 45 PROCESS_TYPE, 'D' GROUP_ID FROM DUAL
) SELECT ...

我的第一个正确工作的select 语句是这个：

SELECT CUSTOMER_ID,
       SUM(AMOUNT) as AMOUNT1,
       SUM(CASE WHEN PROCESS_TYPE IN (SELECT PROCESS_TYPE FROM REFTAB WHERE GROUP_ID = 'A') 
                THEN AMOUNT ELSE NULL END) as AMOUNT2,
       COUNT(CASE WHEN PROCESS_TYPE IN (SELECT PROCESS_TYPE FROM REFTAB WHERE GROUP_ID = 'D') 
                  THEN 1 ELSE NULL END) as COUNT1
   FROM MAS
  GROUP BY CUSTOMER_ID

但是，为了解决性能问题，我将其更改为 select 声明：

SELECT CUSTOMER_ID,
       SUM(AMOUNT) as AMOUNT1,
       SUM(CASE WHEN GROUP_ID = 'A' THEN AMOUNT ELSE NULL END) as AMOUNT2,
       COUNT(CASE WHEN GROUP_ID = 'D' THEN 1 ELSE NULL END) as COUNT1
   FROM MAS A
   LEFT JOIN REFTAB B ON A.PROCESS_TYPE = B.PROCESS_TYPE
  GROUP BY CUSTOMER_ID

对于AMOUNT2 和COUNT1 列，值保持不变。但是对于AMOUNT1，由于与引用表的连接，值会相乘。

我知道我可以在 GROUP_ID 上再添加 1 个左连接和一个附加连接条件。但这与使用子查询没有什么不同。

知道如何在不乘以 AMOUNT1 值的情况下使查询仅使用 1 个左连接工作吗？

【问题讨论】：

标签： sql oracle11g sqlperformance query-tuning

【解决方案1】：

我知道我可以通过添加额外的 GROUP_ID 子句再添加 1 个左连接，但它不会与子查询不同。

你会感到惊讶。在 SELECT 中有 2 个左连接而不是子查询为优化器提供了更多优化查询的方法。我仍然会尝试：

select m.customer_id,
       sum(m.amount) as amount1,
       sum(case when grpA.group_id is not null then m.amount end) as amount2,
       count(grpD.group_id) as count1
  from mas m
  left join reftab grpA
    on grpA.process_type = m.process_type
   and grpA.group_id = 'A'
  left join reftab grpD
    on grpD.process_type = m.process_type
   and grpD.group_id = 'D'
 group by m.customer_id

你也可以试试这个查询，它使用SUM()解析函数计算amount1的值在连接之前避免重复值问题：

select m.customer_id,
       m.customer_sum as amount1,
       sum(case when r.group_id = 'A' then m.amount end) as amount2,
       count(case when r.group_id = 'D' then 'X' end) as count1
  from (select customer_id,
               process_type,
               amount,
               sum(amount) over (partition by customer_id) as customer_sum
          from mas) m
  left join reftab r
    on r.process_type = m.process_type
 group by m.customer_id,
          m.customer_sum

您可以测试这两种选择，看看哪一种表现更好。

【讨论】：

【解决方案2】：

从您的原始查询开始，只需将您的 IN 查询替换为 EXISTS 语句即可显着提升。另外，请注意汇总NULLs，也许您的ELSE 语句应该是0？

SELECT CUSTOMER_ID,
       SUM(AMOUNT) as AMOUNT1,
       SUM(CASE WHEN EXISTS(SELECT 1 FROM REFTAB WHERE REFTAB.GROUP_ID = 'A' AND REFTAB.PROCESS_TYPE = MAS.PROCESS_TYPE)
                THEN AMOUNT ELSE NULL END) as AMOUNT2,
       COUNT(CASE WHEN EXISTS(SELECT 1 FROM REFTAB WHERE REFTAB.GROUP_ID = 'D' AND REFTAB.PROCESS_TYPE = MAS.PROCESS_TYPE) 
                  THEN 1 ELSE NULL END) as COUNT1
   FROM MAS
  GROUP BY CUSTOMER_ID

【讨论】：

【解决方案3】：

正常的方法是聚合group by之前的值。如果查询的其余部分正确，您还可以使用条件聚合：

SELECT CUSTOMER_ID,
       SUM(CASE WHEN seqnum = 1 THEN AMOUNT END) as AMOUNT1,
       SUM(CASE WHEN GROUP_ID = 'A' THEN AMOUNT ELSE NULL END) as AMOUNT2,
       COUNT(CASE WHEN GROUP_ID = 'D' THEN 1 ELSE NULL END) as COUNT1
FROM MAS A LEFT JOIN
     (SELECT B.*, ROW_NUMBER() OVER (PARTITION BY PROCESS_TYPE ORDER BY PROCESS_TYPE) as seqnum
      FROM REFTAB B
     ) B
     ON A.PROCESS_TYPE = B.PROCESS_TYPE
GROUP BY CUSTOMER_ID;

这会忽略连接创建的重复项。

【讨论】：