Oracle SQL - Write SQL to list emp at the bottom of the management hierarchy.

Requirements

Here is the table

Demo

SQL>
SQL> drop table emp;
SQL> create table emp(
  2  empno      NUMBER(4)    primary key,
  3  ename      VARCHAR2(8)  not null   ,
  4  init       VARCHAR2(5)  not null   ,
  5  job        VARCHAR2(8)             ,
  6  mgr        NUMBER(4)               ,
  7  bdate      DATE         not null   ,
  8  msal       NUMBER(6,2)  not null   ,
  9  comm       NUMBER(6,2)             ,
 10  deptno     NUMBER(2)    default 10) ;
SQL>-- from w ww . j  a  v  a  2  s  .  c  om
SQL> insert into emp values(7001,'SMITH','N',  'TRAINER', 7902,date '1975-12-17',  1800 , NULL, 20);
SQL> insert into emp values(7002,'ALLEN','JAM','SALESREP',7006,date '1971-05-20',  1600, 300,   30);
SQL> insert into emp values(7003,'WARD', 'TF' ,'SALESREP',7006,date '1972-03-02',  1250, 500,   10);
SQL> insert into emp values(7004,'JACK', 'JM', 'MANAGER', 7009,date '1977-04-02',  2975, NULL,  20);
SQL> insert into emp values(7005,'BROWN','P',  'SALESREP',7006,date '1976-09-28',  1250, 1400,  30);
SQL> insert into emp values(7006,'BLAKE','R',  'MANAGER', 7009,date '1973-11-01',  2850, NULL,  10);
SQL> insert into emp values(7007,'CLARK','AB', 'MANAGER', 7009,date '1975-06-09',  2450, NULL,  10);
SQL> insert into emp values(7008,'SCOTT','DEF','TRAINER', 7004,date '1979-11-26',  3000, NULL,  20);
SQL> insert into emp values(7009,'KING', 'CC', 'DIRECTOR',NULL,date '1972-10-17',  5000, NULL,  10);
SQL> insert into emp values(7010,'BREAD','JJ', 'SALESREP',7006,date '1978-09-28',  1500, 0,     30);
SQL> insert into emp values(7011,'ADAMS','AA', 'TRAINER', 7008,date '1976-12-30',  1100, NULL,  20);
SQL> insert into emp values(7012,'JONES','R',  'ADMIN',   7006,date '1979-10-03',  8000, NULL,  30);
SQL> insert into emp values(7902,'FORD', 'MG', 'TRAINER', 7004,date '1979-02-13',  3000, NULL,  20);
SQL> insert into emp values(7934,'MARY', 'ABC','ADMIN',   7007,date '1972-01-23',  1300, NULL,  10);
SQL>

Write SQL to list emp at the bottom of the management hierarchy.

With a third column showing the number of management levels above them.

Demo

SQL>
SQL> select     ename, init
  2  ,          (level - 1) as levels_above
  3  from       emp
  4  where      connect_by_isleaf = 1-- w  ww  . j a  v a  2  s.  co m
  5  start with mgr is null
  6  connect by prior empno = mgr;

ENAME    INIT  LEVELS_ABOVE
-------- ----- ------------
ADAMS    AA               3
SMITH    N                3
ALLEN    JAM              2
WARD     TF               2
BROWN    P                2
BREAD    JJ               2
JONES    R                2
MARY     ABC              2

8 rows selected.

SQL>

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