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标题:关于第三次作业的说明(SQL语句)

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admin 发表于:2008/11/22 7:14:12

1、查询学生姓名和年龄

select name,extract(year from sysdate)-extract(year from birthday) as age from student

2、查询平均分大于85分的学生姓名

select min(name),round(avg(grade),1) from student inner join grade on student."NUMBER"=grade."NUMBER" group by student."NUMBER" having avg(grade)>85

3、利用自动生成的序列数字来给学号主键赋值

create sequence count start with 10 increment by 2 maxvalue 2000 nocycle cache 20;

create table stu ( sid integer primary key, name varchar(8) not null, sex char(1) check(sex in('M','F')) );

insert into stu values(count.nextval,'Tom','M');

select * from stu;

[此贴子已经被作者于2010-12-11 20:04:26编辑过]
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