在 PL/SQL 中不使用 For 循环来计算公式的函数

Question

对 PL/SQL 来说还是有点新，但基本上我正在尝试创建一个函数，该函数将根据一个人在过去 8 年中支付的金额来计算他们的分数。它使用每年的总和进行计算。公式为 (Year-1)/Year-2) * (Year-2/Year-3) * (Year3/Year4) 等等。更棘手的是，我需要跳过他们给出 0 的年份。

例如：

这是我目前的代码：

CREATE OR REPLACE FUNCTION formula
(idnum IN NUMBER)
RETURN NUMBER IS score NUMBER; 

-- Declare Variables
currentyear NUMBER := EXTRACT (YEAR FROM SYSDATE);
previousyear NUMBER := currentyear - 1;
yeareight NUMBER := currentyear - 8;
previoussum NUMBER := 0;
currentsum NUMBER := 0;
placeholder NUMBER := 0;
score NUMBER := 1;

BEGIN
-- Set Score to 0 if no history of payments in the last 8 years
SELECT NVL(SUM(amount), 0)
   INTO currentsum
   FROM moneytable g
   WHERE g.id_number = idnum
         AND g.fiscal_year BETWEEN yeareight AND previousyear;    

IF currentsum = 0 THEN score := 0;
  ELSE 
-- Loop to calculate Score
-- Score formula is (Year-1/Year -2) * (Year-2/Year-3) and so on for the last 8 years
-- Zeroes ignored for above calculations  
-- Score defaults to 1 if only one year has any gifts
FOR counter IN 1..8
  LOOP

  currentyear := currentyear - 1;
  placeholder := 0;

  SELECT NVL(SUM(amount), 0)
  INTO currentsum
  FROM moneytable g
  WHERE g.id_number = idnum
        AND g.fiscal_year = currentyear;

 IF currentsum = 0 THEN CONTINUE;
 ELSE placeholder := previoussum / currentsum; END IF;

 previoussum := currentsum;

 IF currentsum > 0 AND placeholder > 0 THEN score := score * placeholder; END IF;

END LOOP; 
END IF;
RETURN score;

END;

它有效并且给出了正确的分数，但是如果我尝试一次为几个人运行它，它会运行得非常慢。有没有更高效、优化的方法来创建这个函数？

Answer 1

先UNPIVOT摆脱空虚的岁月

select * from tab
  UNPIVOT (  value                             
            FOR year  IN                             
            (YEAR1, YEAR2, YEAR3, YEAR4, YEAR5, YEAR6, YEAR7, YEAR8) 
          )
where value != 0
order by 1,2;

NAME YEAR       VALUE
---- ----- ----------
Jane YEAR1         10
Jane YEAR3         20
Jane YEAR4         50
Jane YEAR7         30
Jane YEAR8         20
Rob  YEAR2         10
Rob  YEAR4         20
...

然后使用 LEAD 聚合函数计算系数（默认使用与行中相同的 VALUE 以忽略最后一个系数 - 将其设置为 1）。

with formula as (select * from tab
  UNPIVOT (  value                             
            FOR year  IN                             
            (YEAR1, YEAR2, YEAR3, YEAR4, YEAR5, YEAR6, YEAR7, YEAR8) 
          )
where value != 0)
select NAME, YEAR, VALUE,
VALUE / lead(value,1,VALUE) over (partition by NAME order by YEAR) as koeff
from formula
order by 1,2;

NAME YEAR       VALUE      KOEFF
---- ----- ---------- ----------
Jane YEAR1         10 ,5        
Jane YEAR3         20 ,4        
Jane YEAR4         50 1,66666667
Jane YEAR7         30 1,5       
Jane YEAR8         20          1
Rob  YEAR2         10 ,5        
Rob  YEAR4         20 ,2 
...

在最后一步中，使用this trick 计算系数的聚合乘积

with formula as (select * from tab
  UNPIVOT (  value                             
            FOR year  IN                             
            (YEAR1, YEAR2, YEAR3, YEAR4, YEAR5, YEAR6, YEAR7, YEAR8) 
          )
where value != 0),
formula2 as (
select NAME, YEAR, VALUE,
VALUE / lead(value,1,VALUE) over (partition by NAME order by YEAR) as koeff
from formula)
select name, 
round(EXP(SUM(LN(koeff))),6) score
from formula2
group by name
order by 1 ;


NAME      SCORE
---- ----------
Jane ,5        
Rob  ,1        
Tom  ,2 

测试数据

create table tab as
select 'Tom' name, 0 year1, 0 year2, 0 year3, 10 year4, 20 year5, 30 year6, 40 year7, 50 year8 
from dual union all
select 'Jane' name, 10 year1, 0 year2, 20 year3, 50 year4, 0 year5, 0 year6, 30 year7, 20 year8 
from dual union all
select 'Rob' name, 0 year1, 10 year2, 0 year3, 20 year4, 0 year5, 0 year6, 0 year7, 100 year8 
from dual;

Answer 2

也许最好调用过去 8 年的单个查询并将结果放在一个数组中，然后循环它，而不执行 8 个查询：

DECLARE
  TYPE arrayofnumbers IS TABLE OF NUMBER(11);
  sums arrayofnumbers;
BEGIN
  SELECT NVL(SUM(amount), 0)
  INTO sums
  FROM moneytable g
    WHERE g.id_number = idnum AND g.fiscal_year between currentyear and currentyear+7;

 FOR i IN 1 .. sums.count
  LOOP
    -- other code
    dbms_output.put_line(sums(i));

  END LOOP;
END;

Answer 3

您想要第一个非零值除以最后一个非零值。那将是：

select (case when year1 <> 0 then year1
             when year2 <> 0 then year2
             when year3 <> 0 then year3
             when year4 <> 0 then year4
             when year5 <> 0 then year5
             when year6 <> 0 then year6
             when year7 <> 0 then year7
             when year8 <> 0 then year8
        end) /
       (case when year8 <> 0 then year8
             when year7 <> 0 then year7
             when year6 <> 0 then year6
             when year5 <> 0 then year5
             when year4 <> 0 then year4
             when year3 <> 0 then year3
             when year2 <> 0 then year2
             when year1 <> 0 then year1
        end)  

Answer 4

只查询一次表并按年份分组结果。将它们按降序编号并循环八行。它可以在 SQL 中完成，也可以在函数中循环遍历分组行。 SQL 示例：

^dbfiddle

with y(rn, amt) as (
    select row_number() over (order by fiscal_year desc), sum(amount)
      from moneytable g  
      where id_number = 1
        and fiscal_year between extract (year from sysdate) - 8 
                            and extract(year from sysdate) - 1
      group by fiscal_year),
  c(rn, amt, prev, ret) as (
    select rn, amt, amt, 1 from y where rn = 1
    union all
    select y.rn, y.amt, c.amt, (c.amt/y.amt)*ret from c join y on y.rn = c.rn + 1)
select ret from c where rn = (select max(rn) from c)