Oracle PL / SQL：访问当前行中的上一行计算值

Question

Trying to calculate the column's value wherein the value is getting calculated based on previous row's calculated value. (As in the image below )

which has column Z which is a calculated column as a product of X*Y, case when there is no previous row the value of Y column should be 1 else the value should be previous Z column value.

This can be achieved using Cursor's but trying to find it using SQL Query.

Answer 1

It looks like what you're trying to do is do a cumulative multiplication across all the rows.

You can do this by using a mathematical trick (aka natural logs and exponentials), where you do a cumulative sum of the natural logs of the x values, and then convert the results back to integers using the exponential function:

WITH your_table AS (SELECT 'ABC' a, 12 x FROM dual UNION ALL
                    SELECT 'BBC' a, 20 x FROM dual UNION ALL
                    SELECT 'CBC' a, 10 x FROM dual UNION ALL
                    SELECT 'XYZ' a, 5 x FROM dual)
 SELECT a,
       x,
       LAG(z, 1, 1) OVER (ORDER BY a) y,
       z
FROM   (SELECT a,
               x,
               EXP(SUM(LN(x)) OVER (ORDER BY a)) z
        FROM   your_table);

A            X          Y          Z
--- ---------- ---------- ----------
ABC         12          1         12
BBC         20         12        240
CBC         10        240       2400
XYZ          5       2400      12000

Then, if you need to see the previous value (aka your y column), you can throw a lag() around the calculated column to find the previous row's value (and if there is no previous row, assign a 1 to it).

---

ETA: If you already have a table with the information in it, adding a new row becomes a matter of finding the last row and using that to multiply with the new x value. Eg:

insert into your_table (a, x, y, z)
select 'YMX' a, 2 x, z as new_y, 2*z as new_z
from   (select z,
               row_number() over (order by a desc) rn
        from   your_table)
where  rn = 1;

This finds the latest row (we use the row_number() analytic function to label the rows, starting with the latest a value (which belongs to the last row)), and we can then use the z value from that row to find the new y value, and multiply it by the new x value to find the new z value.

Then you just need to insert this row.

Answer 2

I think the simplest method doesn't use lag() or a subquery at all. One method is simple division:

SELECT a, x, y
       ( EXP(SUM(LN(x * y)) OVER (ORDER BY a)) ) / (x * y) as z
FROM t;

Or use a window clause with coalesce() :

SELECT a, x, y
       COALESCE( EXP(SUM(LN(x * y)) OVER (ORDER BY a ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING)), 1 ) as z
FROM t;