在select语句中调整oracle子查询

Question

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，我将其更改为以下select语句:

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，该值被乘以。

我知道我可以在GROUP_ID上添加1个额外的左连接条件。但这与使用子查询没有任何不同。

任何想法如何使查询工作只有1左连接，而不乘以AMOUNT1值?

Answer 1

我知道我可以添加一个额外的GROUP_ID子句，但它不会与子查询不同。

你会很惊讶的。在SELECT中使用两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。我还是会试试:

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

您也可以尝试这个查询，它使用SUM()解析函数在连接之前计算amount1值，以避免重复值问题:

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

您可以测试这两个选项，看看哪一个性能更好。

Answer 2

从原始查询开始，简单地将IN查询替换为EXISTS语句应该会提供显着的提升。另外，对NULL s求和要小心，也许你的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

Answer 3

通常是将group by之前的值进行汇总。如果查询的其余部分是正确的，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;

忽略连接创建的重复项。