我有以下数据集:
DATA_FIM CENTRO C M ESTOQUE
2018-02-01 HD01 CD 70 539.000
2018-03-01 HD01 CD 70 511.000
2018-04-01 HD01 CD 70 468.000
2018-05-01 HD01 CD 70 447.000
2018-06-01 HD01 CD 70 382.000
2018-07-01 HD01 CD 70 348.000
2018-08-01 HD01 CD 70 285.000
2018-09-01 HD01 CD 70 245.000
2018-10-01 HD01 CD 70 221.000
2018-11-01 HD01 CD 70 207.000
2018-12-01 HD01 CD 70 122.000
2018-12-21 HD01 CD 70 101.000
2018-02-01 HD01 CD 71 164.000
2018-03-01 HD01 CD 71 147.000
2018-04-01 HD01 CD 71 124.000
2018-05-01 HD01 CD 71 107.000
2018-06-01 HD01 CD 71 78.000
1-以第二,第三,第四列为一组,我想从同一列的上一行中减去第五列的值。
为了创建新行。
[![数据] [2]] [2]
这个问题非常相似:[问题:13196190] [3],但是,在我的情况下,我有多个索引,并且不确定如何解决。
我正在使用建议的密码
select
ZBI_FAT_PRODUCT.DATA_FIM,
ZBI_DIM_EMPRESA.CENTRO,
ZBI_DIM_EMPRESA.CANAL,
ZBI_DIM_PRODUCT.MATERIAL,
ZBI_DIM_PRODUCT.TITULO,
ZBI_FAT_PRODUCT.ESTOQUE_VENDA,
LAG(ZBI_FAT_PRODUCT.ESTOQUE_VENDA, 1, ZBI_FAT_PRODUCT.ESTOQUE_VENDA) OVER (PARTITION BY ZBI_DIM_EMPRESA.CENTRO, ZBI_DIM_PRODUCT.MATERIAL, ZBI_DIM_PRODUCT.TITULO ORDER BY ZBI_FAT_PRODUCT.DATA_FIM) - ZBI_FAT_PRODUCT.ESTOQUE_VENDA AS NEW
FROM ZBI_DIM_EMPRESA INNER JOIN (ZBI_DIM_PRODUCT INNER JOIN ZBI_FAT_PRODUCT ON ZBI_DIM_PRODUCT.BK = ZBI_FAT_PRODUCT.BK_MATERIAL) ON ZBI_DIM_EMPRESA.BK = ZBI_FAT_PRODUCT.BK_EMPRESA;
答案 0 :(得分:4)
您似乎只想要lag()
:
select el.*,
(lag(estoque, 0, estoque) over (partition by centro, c, m order by date_mif) -
estoque
) as diff
from energylog el;
编辑:
对于已编辑的问题,似乎只有两列组成一个组
select el.*,
(lag(estoque, 0, value) over (partition by centro, canal, order by date_mif) -
estoque
) as diff
from energylog el;
答案 1 :(得分:1)
为简洁起见,并从评论中提取解决方案-
答案是使用LAG()
函数:
SELECT *,
LAG(ESTOQUE, 1, ESTOQUE) OVER (PARTITION BY CENTRO, C, M ORDER BY DATA_FIM) - ESTOQUE AS New
FROM EnergyLog
以下是此工作的示例:SQLFiddle