Как да премахнете дубликатите от списъка, разделен със запетая, чрез regexp_replace в Oracle?

Две решения, които използват само SQL и трето решение, което използва малка/проста PL/SQL функция, която прави много кратка окончателна SQL заявка.

Настройка на Oracle :

CREATE TABLE data ( value ) AS
SELECT 'POW,POW,POWPRO,PRO,PRO,PROUTL,TNEUTL,TNEUTL,UTL,UTLTNE,UTL,UTLTNE' FROM DUAL;

CREATE TYPE stringlist AS TABLE OF VARCHAR2(4000);
/

Запитване 1 :

SELECT LISTAGG( t.COLUMN_VALUE, ',' ) WITHIN GROUP ( ORDER BY t.COLUMN_VALUE ) AS list
FROM   data d,
       TABLE(
         SET(
           CAST(
             MULTISET(
              SELECT REGEXP_SUBSTR( d.value, '[^,]+', 1, LEVEL )
              FROM   DUAL
              CONNECT BY LEVEL <= REGEXP_COUNT( d.value, '[^,]+' )
             ) AS stringlist
           )
         )
       ) t
GROUP BY d.value;

Изходи :

LIST
---------------------------------------
POW,POWPRO,PRO,PROUTL,TNEUTL,UTL,UTLTNE

Запитване 2 :

SELECT ( SELECT LISTAGG(  COLUMN_VALUE, ',' ) WITHIN GROUP ( ORDER BY ROWNUM )
         FROM TABLE( d.uniques ) ) AS list
FROM   (
  SELECT ( SELECT CAST(
                    COLLECT(
                      DISTINCT
                      REGEXP_SUBSTR( d.value, '[^,]+', 1, LEVEL )
                    )
                    AS stringlist
                  )
            FROM  DUAL
            CONNECT BY LEVEL <= REGEXP_COUNT( d.value, '[^,]+' )
         ) uniques
  FROM   data d
) d;

Изход :

LIST
---------------------------------------
POW,POWPRO,PRO,PROUTL,TNEUTL,UTL,UTLTNE

Настройка на Oracle :

Малка помощна функция:

CREATE FUNCTION split_String(
  i_str    IN  VARCHAR2,
  i_delim  IN  VARCHAR2 DEFAULT ','
) RETURN stringlist DETERMINISTIC
AS
  p_result       stringlist := stringlist();
  p_start        NUMBER(5) := 1;
  p_end          NUMBER(5);
  c_len CONSTANT NUMBER(5) := LENGTH( i_str );
  c_ld  CONSTANT NUMBER(5) := LENGTH( i_delim );
BEGIN
  IF c_len > 0 THEN
    p_end := INSTR( i_str, i_delim, p_start );
    WHILE p_end > 0 LOOP
      p_result.EXTEND;
      p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, p_end - p_start );
      p_start := p_end + c_ld;
      p_end := INSTR( i_str, i_delim, p_start );
    END LOOP;
    IF p_start <= c_len + 1 THEN
      p_result.EXTEND;
      p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, c_len - p_start + 1 );
    END IF;
  END IF;
  RETURN p_result;
END;
/

Запитване 3 :

SELECT ( SELECT LISTAGG(  COLUMN_VALUE, ',' ) WITHIN GROUP ( ORDER BY ROWNUM )
         FROM TABLE( SET( split_String( d.value ) ) ) ) AS list
FROM   data d;

или (ако искате да предадете само една стойност):

SELECT LISTAGG(  COLUMN_VALUE, ',' ) WITHIN GROUP ( ORDER BY ROWNUM ) AS list
FROM   TABLE( SET( split_String(
          'POW,POW,POWPRO,PRO,PRO,PROUTL,TNEUTL,TNEUTL,UTL,UTLTNE,UTL,UTLTNE'
       ) ) );

Изход :

LIST
---------------------------------------
POW,POWPRO,PRO,PROUTL,TNEUTL,UTL,UTLTNE