作业要求:
作业第一条第二条代码已给
1.创建student和score表
CREATE TABLE student (
id INT(10) NOT NULL UNIQUE PRIMARY KEY ,
name VARCHAR(20) NOT NULL ,
sex VARCHAR(4) ,
birth YEAR,
department VARCHAR(20) ,
address VARCHAR(50)
);
CREATE TABLE score (
id INT(10) NOT NULL UNIQUE PRIMARY KEY AUTO_INCREMENT ,
stu_id INT(10) NOT NULL ,
c_name VARCHAR(20) ,
grade INT(10)
);
2.为student表和score表增加记录
向student表插入记录的INSERT语句如下:
INSERT INTO student VALUES( 901,'张老大', '男',1985,'计算机系', '北京市海淀区');
INSERT INTO student VALUES( 902,'张老二', '男',1986,'中文系', '北京市昌平区');
INSERT INTO student VALUES( 903,'张三', '女',1990,'中文系', '湖南省永州市');
INSERT INTO student VALUES( 904,'李四', '男',1990,'英语系', '辽宁省阜新市');
INSERT INTO student VALUES( 905,'王五', '女',1991,'英语系', '福建省厦门市');
INSERT INTO student VALUES( 906,'王六', '男',1988,'计算机系', '湖南省衡阳市');
向score表插入记录的INSERT语句如下:
INSERT INTO score VALUES(NULL,901, '计算机',98);
INSERT INTO score VALUES(NULL,901, '英语', 80);
INSERT INTO score VALUES(NULL,902, '计算机',65);
INSERT INTO score VALUES(NULL,902, '中文',88);
INSERT INTO score VALUES(NULL,903, '中文',95);
INSERT INTO score VALUES(NULL,904, '计算机',70);
INSERT INTO score VALUES(NULL,904, '英语',92);
INSERT INTO score VALUES(NULL,905, '英语',94);
INSERT INTO score VALUES(NULL,906, '计算机',90);
INSERT INTO score VALUES(NULL,906, '英语',85);
3.查询student表的所有记录
select * from student;
select * from student limit 1, 3;
5.从student表查询所有学生的学号(id)、姓名(name)和院系(department)的信息
select id, name, department from student;
select * from student where department in ('计算机系', '英语系');
7.从student表中查询年龄18~22岁的学生信息(根据已有年份随便写了年份区间)
select * from student where birth between 1987 and 1992;
select department, count(*) as number from student group by(department);
select c_name, max(grade) from score group by c_name;
10.查询李四的考试科目(c_name)和考试成绩(grade)
select c_name, grade from score inner join student on score.stu_id = student.id where student.name = '李四';
select * from score inner join student on score.stu_id=student.id;
select stu.name, sum(sc.grade) from score as sc right join student as stu on stu.id=sc.stu_id group by stu.id;
select c_name, avg(grade) as '平均成绩' from score group by c_name;
select stu.*, sc.* from score as sc inner join student as stu on stu.id = sc.stu_id where c_name = '计算机' and grade < 95;
select a.* from score as sc inner join (select stu.* from student as stu inner join score as sc on sc.stu_id=stu.id where sc.c_name='计算机') as a on a.id = sc.stu_id where sc.c_name='英语';
select stu.*, sc.* from score sc inner JOIN student stu on sc.stu_id = stu.id where sc.c_name='计算机' order by grade desc;
17.从student表和score表中查询出学生的学号,然后合并查询结果
select id from student union select stu_id from score;
select stu.name, stu.department, sc.c_name, sc.grade from student as stu inner join score as sc on sc.stu_id = stu.id where stu.name like '张%' or stu.name like'王%';
select name, (year(now())-birth) as age, department,c_name, grade from score, (select * from student where address like '湖南%') as stu where stu.id = stu_id;