Question

我在oracle中的表中有数据，如下面的

A_CODE    B_M   P_id
------    ----  ------
123        A     1
123        A     2
123        B     5
678        B     3
678        C     3
678        B     4
123        BC    2

价值＆＃34; BC＆＃34;是B和C.数据没有标准化，所以我们需要把它计算为B和C.我需要按照下面的每个A_CODE显示计数

A_CODE   B_M   COUNT
-------  ---- -------
123        A    2
123        B    2
123        C    1
678        B    2
678        C    1

我如何在Oracle中执行此操作？

Answer 1

您应该使用CONNECT BY和CONNECT_BY_ROOT。我希望这会有所帮助：

SELECT A_CODE, B_M, COUNT (*)
    FROM (    SELECT A_CODE, SUBSTR (CONNECT_BY_ROOT (B_M), LEVEL, 1) B_M
                FROM your_table
          CONNECT BY LEVEL <= LENGTH (B_M))
          WHERE B_M IS NOT NULL
GROUP BY A_CODE, B_M
ORDER BY A_CODE;

Answer 2

请尝试以下：

SELECT A_CODE, B_M, COUNT(*) "COUNT" FROM
(SELECT A_CODE, B_M
FROM
  (SELECT A_CODE,
    SUBSTR(B_M,x.LVL,1) B_M
  FROM my_table t,
    (SELECT LEVEL LVL FROM dual
    CONNECT BY LEVEL <=
        (SELECT MAX(LENGTH(B_M)) FROM my_table)
    ) x
  WHERE t.B_M is not null
  )
WHERE B_M IS NOT NULL
UNION ALL
SELECT A_CODE, B_M FROM my_table WHERE B_M IS NULL
)
GROUP BY A_CODE,
  B_M
ORDER BY A_CODE, B_M;

Answer 3

除了/* Styles go here */ .success { background-color: green; } .error { background-color: red; }之外，您还可以创建一个包含所有可能值的表。那就是：

my_table

然后，您应该能够通过连接获得出现次数。类似的东西：

P_VAL
-----
  A
  B
  C

Answer 4

一种选择是将此表连接到一个数据集，该数据集为您提供所需的规范化结构。

with cte_normaliser as (
  select 'A'  B_M, 'A' 'B_M_norm from dual union all
  select 'B'  B_M, 'B' 'B_M_norm from dual union all
  select 'C'  B_M, 'C' 'B_M_norm from dual union all
  select 'BC' B_M, 'B' 'B_M_norm from dual union all
  select 'BC' B_M, 'C' 'B_M_norm from dual)
select my_table.A_CODE,
       n.B_M_norm,
       count(*)
from   my_table join
       cte_normaliser n on n.B_M = my_table.B_M
group by my_table.A_CODE,
       n.B_M_norm;

如果您拥有大量可变代码组合，那么使用这样的固定数据集可能是不可行的，并且该数据集可能需要动态构建。

我需要计算值

4 个答案: