Question

我有一个列（COLUMN3）在某些情况下出现故障。我需要保留COLUMN1订单，然后显示新列“NUM_ITEM”的正确顺序。我不知道该怎么做。我试过了：

SELECT CODIMP,COLUMN1,COLUMN2,COLUMN3, 
SUM(CASE WHEN COLUMN2 = ('C170') THEN 1 
ELSE 0 END) OVER (PARTITION BY COLUMN2 ORDER BY TO_NUMBER(COLUMN1)) NUM_ITEM

FROM SPED_DATA 
WHERE CODIMP = 81 AND COLUMN2 IN ('C100','C170')

GROUP BY CODIMP,COLUMN1,COLUMN2,COLUMN3
ORDER BY TO_NUMBER(COLUMN1);

结果如下：

CODIMP  COLUMN1 COLUMN2 COLUMN3 NUM_ITEM    EXPECTED
81      2741    C100    0       0           0
81      2742    C170    4       1           1
81      2743    C170    2       2           2
81      2744    C170    1       3           3
81      2745    C170    3       4           4
81      2746    C100    0       0           0
81      2747    C170    2       5           1
81      2748    C170    1       6           2
81      2749    C100    0       0           0
81      2750    C170    1       7           1
81      2751    C100    0       0           0
81      2752    C170    1       8           1

“EXPECTED”列是我需要查看“NUM_ITEM”列的方式。

谢谢！

Answer 1

使用行数差异方法，您可以对组进行分类，然后对连续的行使用row_number。

select CODIMP,COLUMN1,COLUMN2,COLUMN3
,case when column2=170 then row_number() over(partition by codimp,column2,grp order by to_number(column1)) 
else 0 end as num_item
from (
SELECT CODIMP,COLUMN1,COLUMN2,COLUMN3, 
ROW_NUMBER() OVER (PARTITION BY CODIMP ORDER BY TO_NUMBER(COLUMN1))
-ROW_NUMBER() OVER (PARTITION BY CODIMP,COLUMN2 ORDER BY TO_NUMBER(COLUMN1)) NUM_ITEM as grp
FROM SPED_DATA 
WHERE CODIMP = 81 AND COLUMN2 IN ('C100','C170')
) t

Answer 2

看起来您需要将值分成组，其中第2列为100.您可以使用累计总和来执行此操作。然后，在此之后，使用row_number()：

select sd.*,
       row_number() over (partition by grp order by column1) as expected
from (select sd.*,
             sum(case when column2 = 'C100' then 1 else 0 end) over (order by column1) as grp
      from sped_data sd
      where column2 in ('C100', 'C170') and codimp = 81
     ) sd;

我不确定您是否确实需要订购结果。似乎order by column1足以满足结果集。

ORDER COLUMN使用SUM（）OVER（）

2 个答案: