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SQL_实验2.1.3 清华大学出版社

本文出自:http://blog.csdn.net/svitter

实验目标:熟悉实体完整性,参照完整性,事务的处理;

/*1.在数据库school表中建立表Stu_uion,进行主键约束,在没有违反实体完整性的前提下插入并更新一条记录*/
Use school 
create table stu_uion
(
	sno char(5) not null unique,
	sname char(8),
	ssex char(1),
	sage int,
	sdept char(20),
	constraint pk_stu_uion primary key (sno)
);
insert stu_uion values(‘10000‘, ‘Wangmin‘, ‘1‘, 23, ‘CS‘);
update stu_uion set sno = ‘ ‘ 
	where sdept = ‘CS‘;
update stu_uion set sno = ‘92002‘
	where sname = ‘Wangmin‘;
select * from stu_uion;


/*2.3.演示违反实体完整性的插入,更新操作*/
Use school
insert stu_uion values(‘10000‘, ‘Li hua‘, ‘1‘, 23, ‘CS‘);/*unique*/
update stu_uion set sno = NULL where sno = ‘10000‘; /*not null*/



/*4.演示事务的处理,包括事务的建立,处理,以及出错事务的回退*/
Use school
set xact_abort on
/*设置xact_abort 为on时,如果transaction(事务)语句出现错误,那么整个事务都会回滚
 *如果设置其为off时,只回滚出错的语句*/
begin transaction t1
insert into stu_uion values(‘95009‘, ‘Li yong‘, ‘M‘, 25, ‘EE‘);
insert into stu_uion values(‘95003‘, ‘wang hao‘, ‘0‘, 25, ‘EE‘);
insert into stu_uion values(‘95005‘, ‘wang hao‘, ‘0‘, 25, ‘EE‘);
select * from stu_uion ;
commit transaction t1


/*5.通过建立scholarship表 ,插入数据,演示当前与现有的数据环境不等时,无法建立实体完整性以及参照完整性*/
Use school
Create table Scholarship
(
	M_ID varchar(10),
	Stu_id char(10),
	R_money int
);
insert into scholarship values(‘0001‘, ‘700000‘, 5000);
insert into scholarship values(‘0001‘, ‘800000‘, 5000);
select * from scholarship;

/*constraint*/
Use school
alter table scholarship add
constraint pk_scholarship primary key(M_ID);/*pk: primary key*/
/*存在两个0001,无法建立主键约束*/

/**scholarship中的数据,不满足stu_id和students表中的sid对应性,创建参照完整性失败*/

Use school
alter table scholarship add
constraint fk_scholarship foreign key (Stu_id) references students(sid);