Question

我有以下数据集（table：stk）：

S_Date       Qty     OOS (Out of Stock - 1 true, 0 false)
01/01/2013   0       1
02/01/2013   0       1
03/01/2013   0       1
04/01/2013   5       0
05/01/2013   0       1
06/01/2013   0       1

我想要的是：

S_Date       Qty     Cumulative_Days_OOS
01/01/2013   0       1
02/01/2013   0       2
03/01/2013   0       3
04/01/2013   5       0  -- No longer out of stock
05/01/2013   0       1
06/01/2013   0       2

我到目前为止最接近的是以下SQL：

SELECT
  S_DATE, QTY, 
  SUM(OOS) OVER (PARTITION BY OOS ORDER BY S_DATE) CUMLATIVE_DAYS_OOS
FROM
  STK
GROUP BY
  S_DATE, QTY, OOS
ORDER BY
  1

这给了我以下输出：

S_Date       Qty     Cumulative_Days_OOS
01/01/2013   0       1
02/01/2013   0       2
03/01/2013   0       3
04/01/2013   5       0
05/01/2013   0       4
06/01/2013   0       5

它接近我想要的，但可以理解的是，总和是继续的。是否可以重置此累计金额并重新开始？

我尝试过在stackoverflow和google上搜索，但我不确定我应该搜索什么。

任何帮助都非常感激。

Answer 1

您需要识别oos = 1或0的连续天组。可以使用LAG函数查找oos列何时更改然后对其进行求和。

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;

输出：

|                         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 |

然后，你可以总结这个oos，用grp列分区以获得连续oos天。

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;

输出：

|                         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 |

sqlfiddle的演示。

Answer 2

首先，我们需要将行划分为组。在这种情况下，您可以使用当前行之前的0值作为组编号。然后，您可以对这些组使用SUM() OVER。要获得0 OOS = 0，您可以使用CASE语句，或只要OOS*SUM(OOS) OOS =（0,1）

这样的事情：

select T1.*,

OOS*SUM(OOS) OVER (PARTITION BY GRP ORDER BY S_DATE) CUMLATIVE_DAYS_OOS

FROM
(
select T.*,
(select count(*) from STK where S_Date<T.S_Date and OOS=0) GRP
FROM STK T
)  T1 
ORDER BY S_Date

SQLFiddle demo

重置累计金额？

2 个答案: