Então, vamos liberar o
MODEL cláusula (um dispositivo cujo mistério só é superado por sua potência) sobre este problema:with data as (
select date'2012-01-01' as mm, 800 as a from dual union all
select date'2012-02-01' as mm, 1900 as a from dual union all
select date'2012-03-01' as mm, 1750 as a from dual union all
select date'2012-04-01' as mm, -20000 as a from dual union all
select date'2012-05-01' as mm, 900 as a from dual union all
select date'2012-06-01' as mm, 3900 as a from dual union all
select date'2012-07-01' as mm, -2600 as a from dual union all
select date'2012-08-01' as mm, -2600 as a from dual union all
select date'2012-09-01' as mm, 2100 as a from dual union all
select date'2012-10-01' as mm, -2400 as a from dual union all
select date'2012-11-01' as mm, 1100 as a from dual union all
select date'2012-12-01' as mm, 1300 as a from dual
)
select mm, a, b
from (
-- Add a dummy value for b, making it available to the MODEL clause
select mm, a, 0 b
from data
)
-- Generate a ROW_NUMBER() dimension, in order to access rows by RN
model dimension by (row_number() over (order by mm) rn)
-- Spreadsheet values / measures involved in calculations are mm, a, b
measures (mm, a, b)
-- A single rule will do. Any value of B should be calculated according to
-- GREATEST([previous value of B] + [current value of A], 0)
rules (
b[any] = greatest(nvl(b[cv(rn) - 1], 0) + a[cv(rn)], 0)
)
O rendimento acima:
MM A B
01.01.2012 800 800
01.02.2012 1900 2700
01.03.2012 1750 4450
01.04.2012 -20000 0
01.05.2012 900 900
01.06.2012 3900 4800
01.07.2012 -2600 2200
01.08.2012 -2600 0
01.09.2012 2100 2100
01.10.2012 -2400 0
01.11.2012 1100 1100
01.12.2012 1300 2400
