WITH t AS
(
SELECT 'AA' nm, 1 seq, 0 grp, 100 amt FROM dual
UNION ALL SELECT 'AA', 2, 0, 100 FROM dual
UNION ALL SELECT 'AA', 3, 0, 100 FROM dual
UNION ALL SELECT 'AA', 2, 1, 200 FROM dual
UNION ALL SELECT 'AA', 1, 2, 110 FROM dual
UNION ALL SELECT 'AA', 2, 2, 110 FROM dual
UNION ALL SELECT 'AA', 1, 3, 150 FROM dual
UNION ALL SELECT 'AA', 2, 3, 150 FROM dual
UNION ALL SELECT 'AA', 3, 3, 150 FROM dual
)
SELECT b.nm
     , a.seq
     , b.grp
     , LAST_VALUE(b.amt IGNORE NULLS)
       OVER(PARTITION BY b.nm, a.seq ORDER BY b.grp) amt
  FROM (SELECT LEVEL seq FROM dual CONNECT BY LEVEL <= 3) a
  LEFT OUTER JOIN t b
  PARTITION BY (b.nm, b.grp)
  ON a.seq = b.seq
 ORDER BY nm, grp, seq
;

마농님 감사합니다~~^^

근데 한가지만 더 여쭤봐도 될까요?

이름에 대해서 저런 데이타들이 쭉있고

순번이 3으로 고정이 아니라 가변적입니다. 각 이름들에 대해서..

그리고 가져오는 값도 가격 하나뿐만 아니라 모든컬럼들에 대해서 가격처럼 값을 가져와야하는데

모든컬럼을 last _value 이 방법으로 하자니 좀 먹먹 하네요

혹 다른방법도 있을까요???

WITH t AS
(
SELECT 'AA' nm, 1 seq, 0 grp, 100 amt, 1 cnt FROM dual
UNION ALL SELECT 'AA', 2, 0, 100, 2 FROM dual
UNION ALL SELECT 'AA', 3, 0, 100, 3 FROM dual
UNION ALL SELECT 'AA', 2, 1, 200, 4 FROM dual
UNION ALL SELECT 'AA', 1, 2, 110, 5 FROM dual
UNION ALL SELECT 'AA', 2, 2, 110, 6 FROM dual
UNION ALL SELECT 'AA', 1, 3, 150, 7 FROM dual
UNION ALL SELECT 'AA', 2, 3, 150, 8 FROM dual
UNION ALL SELECT 'AA', 3, 3, 150, 9 FROM dual
UNION ALL SELECT 'BB', 2, 0, 100, 8 FROM dual
UNION ALL SELECT 'BB', 3, 0, 100, 7 FROM dual
UNION ALL SELECT 'BB', 1, 1, 200, 6 FROM dual
UNION ALL SELECT 'BB', 4, 1, 110, 5 FROM dual
UNION ALL SELECT 'BB', 1, 2, 110, 4 FROM dual
UNION ALL SELECT 'BB', 2, 2, 150, 3 FROM dual
UNION ALL SELECT 'BB', 3, 2, 150, 2 FROM dual
UNION ALL SELECT 'BB', 4, 2, 150, 1 FROM dual
)
SELECT a.nm
     , a.seq
     , b.grp
     , LAST_VALUE(b.amt IGNORE NULLS)
       OVER(PARTITION BY a.nm, a.seq ORDER BY b.grp) amt
     , LAST_VALUE(b.cnt IGNORE NULLS)
       OVER(PARTITION BY a.nm, a.seq ORDER BY b.grp) cnt
  FROM (SELECT DISTINCT nm, seq FROM t) a
  LEFT OUTER JOIN t b
  PARTITION BY (b.grp)
    ON a.nm = b.nm
   AND a.seq = b.seq
 ORDER BY nm, grp, seq
;

마농님 감사합니다^^

데이타 새로 올려드릴려고 했는데 이미 답을 달아주셨네요~

정말 많은 도움되었습니다^^

수고하세요~~