Een cumulatief bedrag resetten?

U moet groepen van opeenvolgende dagen identificeren waarbij oos =1 of 0. Dit kan worden gedaan door de LAG-functie te gebruiken om te bepalen wanneer de oos-kolom verandert en deze vervolgens op te tellen.

with x (s_date,qty,oos,chg) as (
  select s_date,qty,oos,
         case when oos = lag(oos,1) over (order by s_date)
                then 0
                else 1
         end
  from stk
  )
select s_date,qty,oos,
       sum(chg) over (order by s_date) grp
from x;

uitgang :

|                         S_DATE | QTY | OOS | GRP |
|--------------------------------|-----|-----|-----|
| January, 01 2013 00:00:00+0000 |   0 |   1 |   1 |
| January, 02 2013 00:00:00+0000 |   0 |   1 |   1 |
| January, 03 2013 00:00:00+0000 |   0 |   1 |   1 |
| January, 04 2013 00:00:00+0000 |   5 |   0 |   2 |
| January, 05 2013 00:00:00+0000 |   0 |   1 |   3 |
| January, 06 2013 00:00:00+0000 |   0 |   1 |   3 |

Vervolgens kun je deze oos optellen, gepartitioneerd per grp-kolom om opeenvolgende oos-dagen te krijgen.

with x (s_date,qty,oos,chg) as (
  select s_date,qty,oos,
         case when oos = lag(oos,1) over (order by s_date)
                then 0
                else 1
         end
  from stk
  ),
y (s_date,qty,oos,grp) as (
  select s_date,qty,oos,
         sum(chg) over (order by s_date)
  from x
  )
select s_date,qty,oos,
       sum(oos) over (partition by grp order by s_date) cum_days_oos
from y;

uitgang:

|                         S_DATE | QTY | OOS | CUM_DAYS_OOS |
|--------------------------------|-----|-----|--------------|
| January, 01 2013 00:00:00+0000 |   0 |   1 |            1 |
| January, 02 2013 00:00:00+0000 |   0 |   1 |            2 |
| January, 03 2013 00:00:00+0000 |   0 |   1 |            3 |
| January, 04 2013 00:00:00+0000 |   5 |   0 |            0 |
| January, 05 2013 00:00:00+0000 |   0 |   1 |            1 |
| January, 06 2013 00:00:00+0000 |   0 |   1 |            2 |

Demo op sqlfiddle.