PLSQL累積乘法

Question

我有一個查詢，看起來像這樣：

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

這是罰款創建累積和列，但我現在需要創建一個累積乘法列。

這可能嗎？

感謝

Answer 1

做的傳統方式，這是通過取值的對數的總和的指數，同時注意正確處理底片零。

Answer 2

1. 您可以編寫用戶定義的聚合函數。這裏有一個例子：

   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 
   

   示範：

   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 
   

2. 你可以使用示範條款，以達到期望的結果：

   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 3

爲了把@ DavidAldridge的建議爲代碼（並小心處理負值）可能看起來像這樣：

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; 

有一點要注意，雖然是有可能用這種方法比使用方法舍入誤差更大的風險（如通過@NicholasKrasnov建議那些的）直接用乘法。