Question

我一直致力于编写一个查询来访问我们公司非常大的数据库，以便为客户提取最大的结算金额（在本例中为A和B）。我们希望撤回过去一个月每位客户的最高A / B和过去一年的最高A / B.

我们在结算数据库中注意到的一个问题是它存储“已取消”账单的方式。它通过将第一个记帐记录的第二个负面版本添加到记帐表中来实现。像这样：

在这种情况下，41040是不正确的账单，因此添加了负面版本的记录。但是，当我尝试在此列上选择最大值时，我仍然会返回41040而不是正确的结算值50.此表似乎没有以任何方式标记这些不正确的账单过滤掉。

我目前的解决方案是将ID列的最大值作为正确的帐单。这假设一个月输入的最终账单是正确的。

这似乎带回了正确的数据，但查询在大型数据集上的运行速度非常慢，而且我没有对此表的写访问权来添加或查看索引。总共有98,007,807行和1,596,491个独特客户，是否有优化查询以提高性能？

select mth.KY_CUSTOMER_NO,max(QY_MTH_BILLED_A) as QY_MTH_BILLED_A, max(QY_MTH_B) as QY_MTH_BILLING_B, max.MAX_BILLING_A, max.MAX_BILLING_B
from (
    --Get the max A/B values for the past month
    select m.*
    from CUSTOMER_USAGE m
    where rev_year = to_number(to_char(sysdate,'yyyy'))
    and rev_mth in (to_number(to_char(add_months(sysdate, -1), 'mm')),to_number(to_char(sysdate,'mm')))
    and ID in (select max(ID) from CUSTOMER_USAGE where KY_CUSTOMER_NO = m.KY_CUSTOMER_NO group by rev_mth, rev_year)
) mth join 
(
    --Get the max A/B values for the past year
    select KY_CUSTOMER_NO, max(QY_MTH_B) as MAX_BILLING_B, max(QY_MTH_BILLED_A) as MAX_BILLING_A from CUSTOMER_USAGE m
    where DT_ADDED > current_timestamp - 365 ID in (select max(ID) from CUSTOMER_USAGE  where KY_CUSTOMER_NO = m.KY_CUSTOMER_NO group by rev_mth, rev_year)
    group by KY_CUSTOMER_NO
) max on mth.KY_CUSTOMER_NO = max.KY_CUSTOMER_NO
group by mth.KY_CUSTOMER_NO, max.MAX_BILLING_KVA, max.MAX_BILLING_KW

Answer 1

分析功能似乎是解决方案。

我遗漏了WHERE条款，因为您的示例数据不需要这些条款，但您应该能够将它们添加回最内层的内联视图中。您也可以使用EXTRACT( YEAR FROM SYSDATE )而不是转换为字符串和从字符串转换。

Oracle安装程序：

CREATE TABLE customer_usage ( id, ky_customer_no, rev_mth, rev_year, qy_mth_billed_a, qy_mth_billed_b ) AS
SELECT 1, 1, 1, 2016,  41040, 0 FROM DUAL UNION ALL
SELECT 2, 1, 1, 2016, -41040, 0 FROM DUAL UNION ALL
SELECT 3, 1, 1, 2016,     50, 0 FROM DUAL UNION ALL
SELECT 4, 1, 1, 2016,      0, 0 FROM DUAL;

<强>查询：

SELECT id,
       ky_customer_no,
       rev_mth,
       rev_year,
       qy_mth_billed_a,
       qy_mth_billed_b
FROM   (
  SELECT c.*,
         ROW_NUMBER()
           OVER ( PARTITION BY ky_customer_no, rev_year, rev_mth
                  ORDER BY total_mth_billed_a DESC ) AS rn
  FROM   (
    SELECT c.*,
           SUM( qy_mth_billed_a )
             OVER ( PARTITION BY ky_customer_no, rev_year, rev_mth, ABS( qy_mth_billed_a )
                    ORDER BY id DESC ) AS total_mth_billed_a               
    FROM   customer_usage c
  ) c
)
WHERE  rn = 1;

<强>输出：

        ID KY_CUSTOMER_NO    REV_MTH   REV_YEAR QY_MTH_BILLED_A QY_MTH_BILLED_B
---------- -------------- ---------- ---------- --------------- ---------------
         3              1          1       2016              50               0

Answer 2

我尝试了另一种方法，但使用了大部分@ MT0设置。

CREATE TABLE customer_usage ( id, ky_customer_no, rev_mth, rev_year, qy_mth_billed_a, qy_mth_billed_b ) AS
SELECT 1, 1, 1, 2016,  41040, 0 FROM DUAL UNION ALL
SELECT 2, 1, 1, 2016, -41040, 0 FROM DUAL UNION ALL
SELECT 3, 1, 1, 2016,     50, 0 FROM DUAL UNION ALL
SELECT 4, 1, 1, 2016,      0, 0 FROM DUAL;

由于我们想要摆脱那些ABS（）相等但有不同符号的值，我试过这个：

  SELECT c.KY_CUSTOMER_NO, c.REV_MTH, c.REV_YEAR, max(qy_mth_billed_a) as qy_mth_billed_a , max(QY_MTH_BILLED_B) as qy_mth_billed_b
  FROM   (
    SELECT c.*,
           max( qy_mth_billed_a )
             OVER ( PARTITION BY ky_customer_no, rev_year, rev_mth,ABS( qy_mth_billed_a )) AS max_mth_billed_a,
           min( qy_mth_billed_a )
             OVER ( PARTITION BY ky_customer_no, rev_year, rev_mth,ABS( qy_mth_billed_a )) AS min_mth_billed_a      
    FROM   customer_usage c
  ) c where max_mth_billed_a+min_mth_billed_a!=0
group by  c.KY_CUSTOMER_NO, c.REV_MTH, c.REV_YEAR;

输出相同，并且由于您遇到了一些性能问题，我尝试了两种方法：

KY_CUSTOMER_NO  REV_MTH REV_YEAR    qy_mth_billed_a qy_mth_billed_b
1   1   1   2016    50  0

修改实际上，如果你计算每个abs（值）的不同符号并且它是一个奇数，我认为它会更快地工作（只需要一个窗函数）

SELECT c.KY_CUSTOMER_NO, c.REV_MTH, c.REV_YEAR, max(qy_mth_billed_a) as qy_mth_billed_a , max(QY_MTH_BILLED_B) as qy_mth_billed_b FROM ( SELECT c.*, count(sign( qy_mth_billed_a )) OVER ( PARTITION BY ky_customer_no, rev_year, rev_mth,ABS( qy_mth_billed_a )) AS signo FROM customer_usage c ) c where mod(signo,2) =1 group by c.KY_CUSTOMER_NO, c.REV_MTH, c.REV_YEAR

由于数据集较大且负值较大，Oracle查询运行速度非常慢

2 个答案: