oracle——笔记——存储过程简介

create table emp2 ( empno number(4),
                    ename varchar2(10),
                    job varchar2(9),
                    mgr number(4),
                    hiredate date,
                    sal number(7,2),
                    comm number(7,2),
                    deptno number(2) 
                 );
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insert into emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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 emp2 (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);

=============================================================================================================================

create table dept2 ( deptno number(2),
                     dname varchar2(14),
                     loc varchar2(13)
                   );

----------------------------------------------------

insert into dept2 (DEPTNO, DNAME, LOC)
values (10, ‘ACCOUNTING‘, ‘NEW YORK‘);

insert into dept2 (DEPTNO, DNAME, LOC)
values (20, ‘RESEARCH‘, ‘DALLAS‘);

insert into dept2 (DEPTNO, DNAME, LOC)
values (30, ‘SALES‘, ‘CHICAGO‘);

insert into dept2 (DEPTNO, DNAME, LOC)
values (40, ‘OPERATIONS‘, ‘BOSTON‘);

============================================================

create table salgrade2 ( grade number,
                         losal number,
                         hisal number
                       );

--------------------------------------------------

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (1, 700, 1200);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (2, 1201, 1400);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (3, 1401, 2000);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (4, 2001, 3000);

insert into salgrade2 (GRADE, LOSAL, HISAL)
values (5, 3001, 9999);


========================================================

select * from emp2;

select * from dept2;

select * from salgrade2;

=====================================
select * from emp2;        --employee2 员工表
select * from dept2;       --department2 部门表
select * from salgrade2;   -- salary grade2 工资等级表

-----------------------------------------------------
emp2

empno    员工编号
ename    员工姓名
job      工作/工种
mgr      manager上级编号
hiredate 入职日期
sal      salary 工资
comm     奖金/津贴
deptno   部门编号
---------------------

dept2

deptno  部门号
dname   部门名称
loc     所在地
---------------------------
salgrade2

grade                   等级
losal lowest salary     最低工资
hisal high salary       最高工资

=============================================

 

存储过程：

存储过程，sql语句的集合
一次编译，多次运行
select * from emp;
语法语义检查
权限检查
生成执行计划



create or replace procedure insertUser(uName in varchar2)
is
I integer;
begin
  for  I in 1..100 loop
       insert into db_emp2(empno,ename)  values(I,uName||I);
  end loop;
  commit;
end;
  ---调用：
begin
  insertUser(‘user‘);
end;