sql – 如何将SUM()每行转换成另一列

我有这张桌子

| ID_prim    | ID (FKey)   | Date         | Moved Items  |
 |:-----------|:------------|-------------:|:------------:|
 | 1003       | 12_1        |    nov 2013  |    2         |
 | 1003       | 12_2        |    okt 2013  |    3         |
 | 1003       | 12_3        |    dec 2014  |    5         |
 | 1003       | 12_4        |    feb 2015  |    10        |
 | 1003       | 12_5        |    apr 2012  |    1         |
 | 1003       | 12_11       |    jan 2011  |    5         |

我想查询同一个表,如下所示：

>通过desc命令日期
>每行总计每个“移动物品”
>如果Sum达到我想要的金额,停止查询
>我的期望金额从MAX’Summed Total'(26)开始,并减去我想要的金额(16)

像这样

| ID_prim    | ID (FKey)   | Date         | Moved Items  | Summed Total |
|:-----------|:------------|-------------:|:------------:|:------------:|
| 1003       | 12_4        |    feb 2015  |     10       |     26
| 1003       | 12_3        |    dec 2014  |     5        |     16
| 1003       | 12_3        |    nov 2013  |     2        |     11 <
| 1003       | 12_4        |    okt 2013  |     3        |     9 
| 1003       | 12_5        |    apr 2012  |     1        |     6
| 1003       | 12_11       |    jan 2011  |     5        |     5

当我达到“Summed Total”(26) – 16 = 10时,我想停止查询.所以从10>
我只会在数据库中获取这些值.

| ID_prim    | ID (FKey)   | Date         | Moved Items  | Summed Total |
|:-----------|:------------|-------------:|:------------:|:------------:|
| 1003       | 12_4        |    feb 2015  |     10       |     26
| 1003       | 12_3        |    dec 2014  |     5        |     16
| 1003       | 12_3        |    nov 2013  |     2        |     11

我所拥有的是如下

SELECT
  T1.ID_prim,T1.ID as ID (FKey),T1.Moved_Items as Moved Items,t1.Date,SUM(T2.MOVEMENTQTY) AS Summed Total
  FROM Table1 T1
  INNER JOIN Table1 T2 ON T2.ID <= T1.ID
    inner join table2 inout on T1.ID_prim = inout.ID_prim
    AND T2.ID_prim = inout.ID_prim
    AND T2.ID_prim = T1.ID_prim
  where t1.ID_prim = 1003
  and t2.ID_prim = 1003
  and inout.ISSOTRX = 'N'
  GROUP BY T1.ID_prim,T1.Moved Items,t1.Date
  HAVING SUM(T2.Moved Items) <= 16
  order by  t1.UPDATED desc

但是这笔钱真的没有办法.
任何人都可以帮助我做出Oracle DB的SQL语句,这将打印我所需的表？

解决方法

表：

SQL> SELECT * FROM t;

   ID_PRIM ID    DT             MOVED
---------- ----- --------- ----------
      1003 12_1  01-NOV-13          2
      1003 12_2  01-OCT-13          3
      1003 12_3  01-DEC-14          5
      1003 12_4  01-FEB-15         10
      1003 12_5  01-APR-12          1
      1003 12_11 01-JAN-11          5

6 rows selected.

SQL>

运行总计

SQL> SELECT t.*,SUM(moved) OVER(ORDER BY dt) sm FROM t ORDER BY dt DESC;

   ID_PRIM ID    DT             MOVED         SM
---------- ----- --------- ---------- ----------
      1003 12_4  01-FEB-15         10         26
      1003 12_3  01-DEC-14          5         16
      1003 12_1  01-NOV-13          2         11
      1003 12_2  01-OCT-13          3          9
      1003 12_5  01-APR-12          1          6
      1003 12_11 01-JAN-11          5          5

6 rows selected.

SQL>

所需输出

SQL> WITH DATA AS
  2    ( SELECT t.*,SUM(moved) OVER(ORDER BY dt) sm FROM t ORDER BY dt DESC
  3    )
  4  SELECT * FROM data WHERE sm >= 16;

   ID_PRIM ID    DT             MOVED         SM
---------- ----- --------- ---------- ----------
      1003 12_4  01-FEB-15         10         26
      1003 12_3  01-DEC-14          5         16

SQL>

请注意,nov 2013不是一个日期,它是一个字符串.由于您希望按日期排序,因此您必须始终使用TO_DATE将其明确转换为日期.无论如何,我用TO_DATE创建样本数据.

更新OP希望从运行时的求和值的MAX值中减去所需的值.

SQL> WITH DATA AS
  2    ( SELECT t.*,SUM(moved) OVER(ORDER BY dt) sm FROM t ORDER BY dt DESC
  3    )
  4  SELECT * FROM DATA t WHERE sm >
  5    (SELECT MAX(sm) FROM data
  6    ) - 16 ;

   ID_PRIM ID    DT             MOVED         SM
---------- ----- --------- ---------- ----------
      1003 12_4  01-FEB-15         10         26
      1003 12_3  01-DEC-14          5         16
      1003 12_1  01-NOV-13          2         11

SQL>

在更新的查询中,MAX(sm)返回26,然后在条件WHERE sm> MAX(sm)-16表示返回“sm”值大于26 -16即10的所有行.您可以使用替换变量在运行时输入值16.

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