스칼라서브쿼리로 풀었습니다. 다른건 생각이 안나네요.
SELECT empno, lpad(ename,length(ename)+level -1,'-'), mgr, level LV , (select ename from emp b where b.empno = a.mgr) mgr_ename FROM emp a
start with empno = 7839
connect by prior empno = mgr

좀더 간단한 방법은 없을까요?

오랜만에 들어왔는데 퀴즈가 있군요 ㅋㅋ

WITH emp AS
(
SELECT 7369 empno, 'SMITH' ename, 7902 mgr FROM dual
UNION ALL SELECT 7499, 'ALLEN' , 7698 FROM dual
UNION ALL SELECT 7521, 'WARD' , 7698 FROM dual
UNION ALL SELECT 7566, 'JONES' , 7839 FROM dual
UNION ALL SELECT 7654, 'MARTIN', 7698 FROM dual
UNION ALL SELECT 7698, 'BLAKE' , 7839 FROM dual
UNION ALL SELECT 7782, 'CLARK' , 7839 FROM dual
UNION ALL SELECT 7788, 'SCOTT' , 7566 FROM dual
UNION ALL SELECT 7839, 'KING' , NULL FROM dual
UNION ALL SELECT 7844, 'TURNER', 7698 FROM dual
UNION ALL SELECT 7876, 'ADAMS' , 7788 FROM dual
UNION ALL SELECT 7900, 'JAMES' , 7698 FROM dual
UNION ALL SELECT 7902, 'FORD' , 7566 FROM dual
UNION ALL SELECT 7934, 'MILLER', 7782 FROM dual
)
SELECT empno,
LPad(' ',level,'-')||ename ename,
mgr,
LEVEL lv,
SubStr(sys_connect_by_path(ename,',')||',',
InStr(sys_connect_by_path(ename,',')||',',',',1,Decode(LEVEL,1,2,level)-1)+1,
InStr(sys_connect_by_path(ename,',')||',',',',1,level)-InStr(sys_connect_by_path(ename,',')||',',',',1,Decode(LEVEL,1,2,level)-1)-1) mgr_ename
FROM emp
START WITH mgr IS NULL
CONNECT BY PRIOR empno = mgr


위와 같이 풀어봤습니다 ㅎㅎ

똑같이 정렬해서 나오는게 힘드네요 휴

WITH emp AS
(
SELECT 7369 empno, 'SMITH' ename, 7902 mgr FROM dual
UNION ALL SELECT 7499, 'ALLEN' , 7698 FROM dual
UNION ALL SELECT 7521, 'WARD' , 7698 FROM dual
UNION ALL SELECT 7566, 'JONES' , 7839 FROM dual
UNION ALL SELECT 7654, 'MARTIN', 7698 FROM dual
UNION ALL SELECT 7698, 'BLAKE' , 7839 FROM dual
UNION ALL SELECT 7782, 'CLARK' , 7839 FROM dual
UNION ALL SELECT 7788, 'SCOTT' , 7566 FROM dual
UNION ALL SELECT 7839, 'KING' , NULL FROM dual
UNION ALL SELECT 7844, 'TURNER', 7698 FROM dual
UNION ALL SELECT 7876, 'ADAMS' , 7788 FROM dual
UNION ALL SELECT 7900, 'JAMES' , 7698 FROM dual
UNION ALL SELECT 7902, 'FORD' , 7566 FROM dual
UNION ALL SELECT 7934, 'MILLER', 7782 FROM dual
)
SELECT EMP1.EMPNO,
REPLACE(LPAD(' ', (LEVEL-1)) || EMP1.ENAME, ' ', '-') ENAME,
LEVEL,
(SELECT EMP2.ENAME
FROM EMP EMP2
WHERE EMP1.MGR = EMP2.EMPNO(+) )
FROM EMP EMP1
START WITH EMP1.MGR IS NULL
CONNECT BY EMP1.MGR = PRIOR EMP1.EMPNO;

스칼라 서브쿼리나 Sys_connect_by_path를 Substr하여 잘 풀어 주셨네요.
하지만 정답은 "헉!"소리 날 정도로 상당히 간단 명료합니다.
계층쿼리만으로 풀리는 문제입니다. 잘 생각해 보세요.

prior ename 사용에 관련된 문제군요.

진짜 헉이네요.
혹시나. 컬럼옆에 prior를 붙여볼까(ㅡㅡ;) 해서 해봤더니. 먼가요 이런게 있었나요?
ㅎㄷㄷㄷ

SELECT empno, lpad(ename,length(ename)+level -1,'-'), mgr, level LV,prior ename
FROM emp a
start with empno = 7839
connect by prior empno = mgr;

그런게 있었지요. 계층쿼리를 하는 순간 이미 사용하고 계셨구요.
connect by 구문에서 잘 사용하고 있던 구문입니다.
select절에서 사용하실 생각을 못했던 것 뿐이죠.

감사합니다. 관계형데이터베이스 기본 다지기에 좋은 예문이군요