Oracle 组和数据透视 - Oracle 中的动态数据透视

Posted 2023-05-09

【中文标题】Oracle 组和数据透视 - Oracle 中的动态数据透视

【英文标题】：Oracle Group and Pivot - Dynamic Pivot in Oracle

【发布时间】：2021-02-26 07:36:22

【问题描述】：

我有一张这样的桌子：

+-----+------+------------+
| SN  | CASE | CASE_VALUE |
+-----+------+------------+
|  A  |  AA  |      1     |
|  A  |  AB  |      5     |
|  A  |  AC  |      3     |
|  A  |  AD  |      4     |
|  B  |  BA  |      5     |
|  B  |  BB  |      7     |
|  B  |  BC  |      5     |
|  B  |  BD  |      1     |
+-----+------+------------+

不知道有没有办法获得

+-----+--------+--------------+--------+--------------+--------+--------------+--------+--------------+
| SN  | CASE_1 | CASE_1_VALUE | CASE_2 | CASE_2_VALUE | CASE_3 | CASE_3_VALUE | CASE_4 | CASE_4_VALUE |
+-----+--------+--------------+--------+--------------+--------+--------------+--------+--------------+
|  A  |   AA   |      1       |   AB   |      5       |   AC   |      3       |   AD   |      4       |
|  B  |   BA   |      5       |   BB   |      7       |   BC   |      5       |   BD   |      1       |
+-----+--------+--------------+--------+--------------+--------+--------------+--------+--------------+

这四种情况不需要顺序

28/02/2021 编辑

如果案例名称之间没有顺序？ 喜欢吹

+-----+------+------------+
| SN  | CASE | CASE_VALUE |
+-----+------+------------+
|  A  |  AB  |      1     |
|  A  |  CD  |      5     |
|  A  |  IJ  |      3     |
|  A  |  GH  |      4     |
|  B  |  OP  |      5     |
|  B  |  EF  |      7     |
|  B  |  MN  |      5     |
|  B  |  KJ  |      1     |
+-----+------+------------+

【参考方案1】：

一种选择是使用条件聚合来根据需要进行旋转，例如

SELECT sn,
       MAX(CASE WHEN SUBSTR(case,2,1) = 'A' THEN case END) AS case_1,
       MAX(CASE WHEN SUBSTR(case,2,1) = 'A' THEN case_value END) AS case_1_value,
       MAX(CASE WHEN SUBSTR(case,2,1) = 'B' THEN case END) AS case_2,
       MAX(CASE WHEN SUBSTR(case,2,1) = 'B' THEN case_value END) AS case_2_value,
       MAX(CASE WHEN SUBSTR(case,2,1) = 'C' THEN case END) AS case_3,
       MAX(CASE WHEN SUBSTR(case,2,1) = 'C' THEN case_value END) AS case_3_value,
       MAX(CASE WHEN SUBSTR(case,2,1) = 'D' THEN case END) AS case_4,
       MAX(CASE WHEN SUBSTR(case,2,1) = 'D' THEN case_value END) AS case_4_value                     
  FROM t
 GROUP BY sn 

Demo

在上述情况下，枢轴是静态的。您可以创建一个返回SYS_REFCURSOR 类型的函数，如下所示，以获得动态枢轴

CREATE OR REPLACE FUNCTION Get_Pivoted_Cols RETURN SYS_REFCURSOR IS
  v_recordset SYS_REFCURSOR;
  v_sql       VARCHAR2(32767);
  v_cols      VARCHAR2(32767);   
BEGIN
  SELECT LISTAGG( 'MAX(CASE WHEN SUBSTR(case,2,1) = '''||cs||''' THEN case END )  AS  "case_'||rn||'",
                   MAX(CASE WHEN SUBSTR(case,2,1) = '''||cs||''' THEN case_value END )  AS  "case_'||rn||'_value"', ',')
           WITHIN GROUP ( ORDER BY rn )                     
    INTO v_cols
    FROM ( SELECT DISTINCT SUBSTR(case,2,1) AS cs, ROW_NUMBER() OVER (PARTITION BY sn ORDER BY case) AS rn
             FROM t );
   
  v_sql :='SELECT sn,'|| v_cols ||' FROM t GROUP BY sn';

  OPEN v_recordset FOR v_sql;
  RETURN v_recordset;
END;
/

并从 SQL Developer 的命令行调用以查看结果集

VAR rc REFCURSOR
EXEC :rc := Get_Pivoted_Cols;
PRINT rc

【参考方案2】：

您需要旋转行，然后使用解码功能来映射您的输出，如下所示。 （希望 CASE 不是你专栏的真名）。

with your_data (SN, "CASE", CASE_VALUE ) as (
select 'A', 'AA', 1 from dual union all
select 'A', 'AB', 5 from dual union all
select 'A', 'AC', 3 from dual union all
select 'A', 'AD', 4 from dual union all
select 'B', 'BA', 5 from dual union all
select 'B', 'BB', 7 from dual union all
select 'B', 'BC', 5 from dual union all
select 'B', 'BD', 1 from dual
)
select SN
, decode(SN, 'A', CASE_1, CASE_5)CASE_1
, decode(SN, 'A', CASE_1_VALUE, CASE_5_VALUE)CASE_1_VALUE
, decode(SN, 'A', CASE_2, CASE_6)CASE_2
, decode(SN, 'A', CASE_2_VALUE, CASE_6_VALUE)CASE_2_VALUE
, decode(SN, 'A', CASE_3, CASE_7)CASE_3
, decode(SN, 'A', CASE_3_VALUE, CASE_7_VALUE)CASE_3_VALUE
, decode(SN, 'A', CASE_4, CASE_8)CASE_4
, decode(SN, 'A', CASE_4_VALUE, CASE_8_VALUE)CASE_4_VALUE
from your_data t
pivot (
max(case_value) as value, max("CASE") FOR "CASE" in (
 'AA' CASE_1
,'AB' CASE_2
,'AC' CASE_3
,'AD' CASE_4
,'BA' CASE_5
,'BB' CASE_6
,'BC' CASE_7
,'BD' CASE_8
  )
)
;

【参考方案3】：

我能得到的最接近，假设（根据您的数据）case 中的第一个字符将始终与 sn 匹配，我们可以以第二个字符为轴

select *
from (
  select
    data.*
  , substr(case, 2, 1) case_t
  from data
)
pivot (
  max(case) type
, max(case_value) value

  for case_t in (
    'A' case_1
  , 'B' case_2
  , 'C' case_3
  , 'D' case_4
  )
)