No está claro si las relaciones son reflexivas (es decir, si B
es un "hermano" de A
entonces A
es un "hermano" de B
) ya que tiene algunas filas duplicadas con relaciones inversas en sus datos y algunas en las que esta propiedad no es evidente.
Asumiendo que sus relaciones no son reflexivas entonces:
Configuración del esquema de Oracle 11g R2 :
CREATE TABLE A ( ID, SIBS ) AS
SELECT 'A', 'B' FROM DUAL UNION ALL
SELECT 'A', 'C' FROM DUAL UNION ALL
SELECT 'B', 'A' FROM DUAL UNION ALL
SELECT 'C', 'A' FROM DUAL UNION ALL
SELECT 'C', 'D' FROM DUAL UNION ALL
SELECT 'D', 'C' FROM DUAL UNION ALL
SELECT 'E', 'F' FROM DUAL UNION ALL
SELECT 'F', 'G' FROM DUAL UNION ALL
SELECT 'G', 'H' FROM DUAL;
Consulta 1 :
SELECT DISTINCT
CONNECT_BY_ROOT( ID ) AS ID,
SIBS
FROM A
WHERE CONNECT_BY_ROOT( ID ) <> SIBS
CONNECT BY NOCYCLE
PRIOR SIBS = ID
ORDER BY ID, SIBS
| ID | SIBS |
|----|------|
| A | B |
| A | C |
| A | D |
| B | A |
| B | C |
| B | D |
| C | A |
| C | B |
| C | D |
| D | A |
| D | B |
| D | C |
| E | F |
| E | G |
| E | H |
| F | G |
| F | H |
| G | H |
Consulta 2 :Si son reflexivos, puede usar UNION [ALL]
para duplicar la tabla con las relaciones en sentido inverso y luego usar la técnica anterior:
SELECT DISTINCT
CONNECT_BY_ROOT( ID ) AS ID,
SIBS
FROM (
SELECT ID, SIBS FROM A
UNION
SELECT SIBS, ID FROM A
)
WHERE CONNECT_BY_ROOT( ID ) <> SIBS
CONNECT BY NOCYCLE
PRIOR SIBS = ID
ORDER BY ID, SIBS
| ID | SIBS |
|----|------|
| A | B |
| A | C |
| A | D |
| B | A |
| B | C |
| B | D |
| C | A |
| C | B |
| C | D |
| D | A |
| D | B |
| D | C |
| E | F |
| E | G |
| E | H |
| F | E |
| F | G |
| F | H |
| G | E |
| G | F |
| G | H |
| H | E |
| H | F |
| H | G |