Question

我有一个如下所示的查询：

SELECT SUM(CAPITAL_RETURN) OVER (PARTITION BY PORTF_CODE, 
                                 EXTRACT(MONTH FROM END_DATE) ORDER BY END_DATE
                                 ) AS CUM_CAPITAL_RET,
 FROM
 ....

这适用于创建累积和列，但我现在需要创建一个累积乘法列。

这可能吗？

由于

Answer 1

您可以编写用户定义的聚合函数。这是一个例子：

SQL> create or replace type t_multiply as object
  2    (
  3    total number,
  4  
  5    static function ODCIAggregateInitialize(nctx IN OUT t_multiply )
  6    return number,
  7  
  8    member function ODCIAggregateIterate(self IN OUT t_multiply ,
  9                                         value number)
 10    return number,
 11  
 12    member function ODCIAggregateTerminate(self IN t_multiply,
 13                                           retVal OUT  number,
 14                                           flags IN number)
 15    return number,
 16  
 17    member function ODCIAggregateMerge(self IN OUT t_multiply,
 18                                       ctx2 IN t_multiply)
 19    return number
 20  )
 21  /

 Type created

 SQL> create or replace type body t_multiply
   2  is
   3  
   4   static function ODCIAggregateInitialize(nctx IN OUT t_multiply)
   5   return number
   6   is
   7   begin
   8     nctx := t_multiply(1);
   9     return ODCIConst.Success;
  10   end;
  11  
  12   member function ODCIAggregateIterate(self IN OUT t_multiply,
  13                                        value IN number)
  14   return number
  15   is
  16  
  17   begin
  18    total := total * value;
  19    return ODCIConst.Success;
  20   end;
  21  
  22   member function ODCIAggregateTerminate(self IN t_multiply,
  23                                          retVal OUT number,
  24                                          flags IN number)
  25   return number
  26   is
  27   begin
  28     retval := total;
  29     return ODCIConst.Success;
  30   end;
  31  
  32   member function ODCIAggregateMerge(self IN OUT t_multiply,
  33                                      ctx2 IN t_multiply)
  34   return number
  35   is
  36   begin
  37     return ODCIConst.Success;
  38   end;
  39  end;
  40  /

  Type body created

  SQL> CREATE OR REPLACE FUNCTION multiply(input in number)
    2  RETURN number
    3  PARALLEL_ENABLE AGGREGATE USING t_multiply;
    4  /

  Function created

演示：

  SQL> with t1(col1, col2) as(
    2    select 1, 1 from dual union all
    3    select 1, 2 from dual union all
    4    select 1, 3 from dual union all
    5    select 1, 4 from dual
    6  )
    7  select col1
    8       , sum(col2) over(partition by col1 order by col2) as sum1
    9       , multiply(col2) over(partition by col1 order by col2) as mult
   10    from t1
   11  ;

     COL1       SUM1       MULT
   ---------- ---------- ----------
      1          1          1
      1          3          2
      1          6          6
      1         10         24

您可以使用model子句来实现所需的结果：

 SQL> with t1(col1, col2) as(
   2    select 1, 1 from dual union all
   3    select 1, 2 from dual union all
   4    select 1, 3 from dual union all
   5    select 1, 4 from dual
   6  )
   7  select c1
   8       , col2
   9       , mult
  10   from t1
  11   model
  12   partition by (col1 as c1)
  13   dimension by (row_number() over(order by col2) rn)
  14   measures(col2, 1 as mult)
  15   rules(
  16     mult[rn] = nvl(mult[cv() - 1], 1) * col2[cv()]
  17   )
  18  ;

     C1       COL2       MULT
  ---------- ---------- ----------
     1          1          1
     1          2          2
     1          3          6
     1          4         24

Answer 2

这样做的经典方法是取值的对数之和的指数，注意正确处理负数零。

Answer 3

将@DavidAldridge的建议放入代码中（并注意处理负值）可能看起来像这样：

WITH T AS
(
    SELECT 2 AS X FROM DUAL
    UNION ALL
    SELECT -3 AS X FROM DUAL
    UNION ALL
    SELECT 4 AS X FROM DUAL
)
SELECT
    CASE
        -- If there are any zeroes, then the result is zero.
        WHEN MAX(CASE WHEN X = 0 THEN 1 END) > 0 THEN 0
        -- Otherwise, if there is an even (or zero) number of 
        -- negative numbers, the result is positive.
        WHEN MOD(COUNT(CASE WHEN X < 0 THEN 1 END), 2) = 0 THEN 1
        -- Otherwise, the result is negative.
        ELSE -1
    END
    *
    EXP
    (
        SUM
        (
            CASE
                WHEN X = 0 THEN 0
                ELSE LN(ABS(X))
            END              
        )
    )
FROM
    T;

有一点需要注意的是，使用此方法可能存在更大的舍入错误风险，而不是使用直接使用乘法的方法（如@NicholasKrasnov建议的那些）。

plsql累积乘法

3 个答案: