oracle——SQL复习08

select * from emp e order by deptno ;

/**范围   分区*/
drop table  emp_part01;
CREATE TABLE emp_part01
(  "EMPNO" NUMBER(4,0),
"ENAME" VARCHAR2(10),
"JOB" VARCHAR2(9),
"MGR" NUMBER(4,0),
"HIREDATE" DATE,
"SAL" NUMBER(7,2),
"COMM" NUMBER(7,2),
"DEPTNO" NUMBER(2,0)
)
partition by range(sal)
(
partition bujige values less than(2000), --不及格
partition jige values less than(3200), --及格
partition youxiu values less than(maxvalue) --优秀
)  ;

--
insert into emp_part01
select * from emp e   ;

commit;

select * from emp_part01 partition(bujige);
select * from emp_part01 partition(jige);
select * from emp_part01 partition(youxiu);

/**列表   分区*/

CREATE TABLE emp_part02
(  "EMPNO" NUMBER(4,0),
"ENAME" VARCHAR2(10),
"JOB" VARCHAR2(9),
"MGR" NUMBER(4,0),
"HIREDATE" DATE,
"SAL" NUMBER(7,2),
"COMM" NUMBER(7,2),
"DEPTNO" NUMBER(2,0)
)
partition by list(deptno)
(
partition d10 values('10'),
partition d20 values('20'),
partition d30 values('30')
)

--
insert into emp_part02
select * from emp e   ;

commit;

select * from emp_part02 partition(d10);
select * from emp_part02 partition(d20);
select * from emp_part02 partition(d30);
select * from emp_part02 ;

/**散列  分区*/

CREATE TABLE emp_part03
(  "EMPNO" NUMBER(4,0),
"ENAME" VARCHAR2(10),
"JOB" VARCHAR2(9),
"MGR" NUMBER(4,0),
"HIREDATE" DATE,
"SAL" NUMBER(7,2),
"COMM" NUMBER(7,2),
"DEPTNO" NUMBER(2,0)
)
partition by hash(EMPNO)
(
partition p1,
partition p2,
partition p3
);

insert into emp_part03
select * from emp e   ;

commit;

select * from emp_part03 partition(p1);
select * from emp_part03 partition(p2);
select * from emp_part03 partition(p3);
select * from emp_part03 ;