Oracle Database
Factoraje de subconsultas recursivas utilizando la cláusula WITH (expresiones de tabla comunes AKA)

Observaciones

Un simple generador de enteros

Consulta :

WITH generator ( value ) AS (
  SELECT 1 FROM DUAL
UNION ALL
  SELECT value + 1
  FROM   generator
  WHERE  value < 10
)
SELECT value
FROM   generator;

Salida :

VALUE
-----
    1
    2
    3
    4
    5
    6
    7
    8
    9
   10

Dividir una cadena delimitada

Datos de muestra :

CREATE TABLE table_name ( value VARCHAR2(50) );

INSERT INTO table_name ( value ) VALUES ( 'A,B,C,D,E' );

Consulta :

WITH items ( list, item, lvl ) AS (
  SELECT value,
         REGEXP_SUBSTR( value, '[^,]+', 1, 1 ),
         1
  FROM   table_name
UNION ALL
  SELECT value,
         REGEXP_SUBSTR( value, '[^,]+', 1, lvl + 1 ),
         lvl + 1
  FROM   items
  WHERE  lvl < REGEXP_COUNT( value, '[^,]+' )
)
SELECT * FROM items;

Salida :

LIST      ITEM LVL
--------- ---- ---
A,B,C,D,E    A   1
A,B,C,D,E    B   2
A,B,C,D,E    C   3
A,B,C,D,E    D   4
A,B,C,D,E    E   5