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  • 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) 
                     );
    -----------------------------------------------------------------------------------------------------------
    
    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;
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  • 原文地址:https://www.cnblogs.com/xiaobaibailongma/p/12258121.html
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