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);
SQL_实验2.1.3 清华大学出版社
时间: 2024-10-10 16:10:16