Oracle分析函数PERCENTILE_CONT

查询各部门中薪水分布处于25%、50%、75%位置的人的薪水，percent_rank()是确定排行中的相对位置。

create table EMP
(
  EMPNO    NUMBER(4) not null,
  ENAME    VARCHAR2(10),
  JOB      VARCHAR2(9),
  MGR      NUMBER(4),
  HIREDATE DATE,
  SAL      NUMBER(7,2),
  COMM     NUMBER(7,2),
  DEPTNO   NUMBER(2)
);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7369, ‘SMITH‘, ‘CLERK‘, 7902, to_date(‘17-12-1980‘, ‘dd-mm-yyyy‘), 800.00, null, 20);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7499, ‘ALLEN‘, ‘SALESMAN‘, 7698, to_date(‘20-02-1981‘, ‘dd-mm-yyyy‘), 1600.00, 300.00, 30);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7521, ‘WARD‘, ‘SALESMAN‘, 7698, to_date(‘22-02-1981‘, ‘dd-mm-yyyy‘), 1250.00, 500.00, 30);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7566, ‘JONES‘, ‘MANAGER‘, 7839, to_date(‘02-04-1981‘, ‘dd-mm-yyyy‘), 2975.00, null, 20);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7654, ‘MARTIN‘, ‘SALESMAN‘, 7698, to_date(‘28-09-1981‘, ‘dd-mm-yyyy‘), 1250.00, 1400.00, 30);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7698, ‘BLAKE‘, ‘MANAGER‘, 7839, to_date(‘01-05-1981‘, ‘dd-mm-yyyy‘), 2850.00, null, 30);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7782, ‘CLARK‘, ‘MANAGER‘, 7839, to_date(‘09-06-1981‘, ‘dd-mm-yyyy‘), 2450.00, null, 10);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7788, ‘SCOTT‘, ‘ANALYST‘, 7566, to_date(‘19-04-1987‘, ‘dd-mm-yyyy‘), 3000.00, null, 20);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7839, ‘KING‘, ‘PRESIDENT‘, null, to_date(‘17-11-1981‘, ‘dd-mm-yyyy‘), 5000.00, null, 10);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7844, ‘TURNER‘, ‘SALESMAN‘, 7698, to_date(‘08-09-1981‘, ‘dd-mm-yyyy‘), 1500.00, 0.00, 30);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7876, ‘ADAMS‘, ‘CLERK‘, 7788, to_date(‘23-05-1987‘, ‘dd-mm-yyyy‘), 1100.00, null, 20);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7900, ‘JAMES‘, ‘CLERK‘, 7698, to_date(‘03-12-1981‘, ‘dd-mm-yyyy‘), 950.00, null, 30);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7902, ‘FORD‘, ‘ANALYST‘, 7566, to_date(‘03-12-1981‘, ‘dd-mm-yyyy‘), 3000.00, null, 20);
insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7934, ‘MILLER‘, ‘CLERK‘, 7782, to_date(‘23-01-1982‘, ‘dd-mm-yyyy‘), 1300.00, null, 10);
commit;

SQL> select e.ename,e.sal,e.deptno,