Oracle scott模式不存在的解决方法

scott 模式/用户在oracle 12c 中是没有的，需要自己手动添加

cmd命令窗口–>用sqlplus–>system–>密码–>连接到数据库

创建用户，12c里面创建用户要以这种形式（c#/c##+scott)

创建用户

create user c##scott indentified by tiger //(tiger是默认口令)

为用户授权

GRANT CONNECT ,RESOURCE,UNLIMITED TABLESPACE TO c##scott CONTAINER= ALL ;

使用c##scott登录

conn c##scott/tiger;

接下来就是需要的各种表的数据插入

DROP TABLE emp PURGE ;

DROP TABLE dept PURGE ;

DROP TABLE bonus PURGE ;

DROP TABLE salgrade PURGE ;

CREATE TABLE dept (

deptno NUMBER(2) CONSTRAINT PK_DEPT PRIMARY KEY ,

dname VARCHAR2(14) ,

loc VARCHAR2(13) );

CREATE TABLE emp (

empno NUMBER(4) CONSTRAINT PK_EMP PRIMARY KEY ,

ename VARCHAR2(10),

job VARCHAR2(9),

mgr NUMBER(4),

hiredate DATE ,

sal NUMBER(7,2),

comm NUMBER(7,2),

deptno NUMBER(2) CONSTRAINT FK_DEPTNO REFERENCES DEPT );

CREATE TABLE bonus (

enamE VARCHAR2(10) ,

job VARCHAR2(9) ,

sal NUMBER,

comm NUMBER ) ;

CREATE TABLE salgrade (

grade NUMBER,

losal NUMBER,

hisal NUMBER );

INSERT INTO emp VALUES (7369, ‘SMITH’ , ‘CLERK’ ,7902,to_date

( ‘17-12-1980’ , ‘dd-mm-yyyy’ ),800, NULL ,20);

INSERT INTO emp VALUES (7499, ‘ALLEN’ , ‘SALESMAN’ ,7698,to_date

( ‘20-2-1981’ , ‘dd-mm-yyyy’ ),1600,300,30);

INSERT INTO emp VALUES (7521, ‘WARD’ , ‘SALESMAN’ ,7698,to_date

( ‘22-2-1981’ , ‘dd-mm-yyyy’ ),1250,500,30);

INSERT INTO emp VALUES (7566, ‘JONES’ , ‘MANAGER’ ,7839,to_date

( ‘2-4-1981’ , ‘dd-mm-yyyy’ ),2975, NULL ,20);

INSERT INTO emp VALUES (7654, ‘MARTIN’ , ‘SALESMAN’ ,7698,to_date

( ‘28-9-1981’ , ‘dd-mm-yyyy’ ),1250,1400,30);

INSERT INTO emp VALUES (7698, ‘BLAKE’ , ‘MANAGER’ ,7839,to_date

( ‘1-5-1981’ , ‘dd-mm-yyyy’ ),2850, NULL ,30);

INSERT INTO emp VALUES (7782, ‘CLARK’ , ‘MANAGER’ ,7839,to_date

( ‘9-6-1981’ , ‘dd-mm-yyyy’ ),2450, NULL ,10);

INSERT INTO emp VALUES (7788, ‘SCOTT’ , ‘ANALYST’ ,7566,to_date

( ‘19-04-1987’ , ‘dd-mm-yyyy’ )-85,3000, NULL ,20);

INSERT INTO emp VALUES (7839, ‘KING’ , ‘PRESIDENT’ , NULL ,to_date

( ‘17-11-1981’ , ‘dd-mm-yyyy’ ),5000, NULL ,10);

INSERT INTO emp VALUES (7844, ‘TURNER’ , ‘SALESMAN’ ,7698,to_date

( ‘8-9-1981’ , ‘dd-mm-yyyy’ ),1500,0,30);

INSERT INTO emp VALUES (7876, ‘ADAMS’ , ‘CLERK’ ,7788,to_date

( ‘23-05-1987’ , ‘dd-mm-yyyy’ )-51,1100, NULL ,20);

INSERT INTO emp VALUES (7900, ‘JAMES’ , ‘CLERK’ ,7698,to_date

( ‘3-12-1981’ , ‘dd-mm-yyyy’ ),950, NULL ,30);

INSERT INTO emp VALUES (7902, ‘FORD’ , ‘ANALYST’ ,7566,to_date

( ‘3-12-1981’ , ‘dd-mm-yyyy’ ),3000, NULL ,20);

INSERT INTO emp VALUES (7934, ‘MILLER’ , ‘CLERK’ ,7782,to_date

( ‘23-1-1982’ , ‘dd-mm-yyyy’ ),1300, NULL ,10);

INSERT INTO salgrade VALUES (1,700,1200);

INSERT INTO salgrade VALUES (2,1201,1400);

INSERT INTO salgrade VALUES (3,1401,2000);

INSERT INTO salgrade VALUES (4,2001,3000);

INSERT INTO salgrade VALUES (5,3001,9999);

还有一种方法：不过我没试过，想试试的话

找到F:\app\ord\product\12.2.0\dbhome_1\rdbms\admin\scott.sql(我的为例)这个sql语句，

将里面的scott全改成c##scott

再用@F:\app\ord\product\12.2.0\dbhome_1\rdbms\admin\scott.sql执行语句，行不行就不知道了

纯属自己做笔记

原文转载自http://blog.csdn.net/cfchengfei002/article/details/45582259