MySQL基础练习

一、创建student和score表

CREATE  TABLE  student (id  INT(10)  NOT NULL  PRIMARY KEY  ,name  VARCHAR(20)  NOT NULL ,sex  VARCHAR(4)  ,birth  YEAR,department  VARCHAR(20) ,address  VARCHAR(50) ); 

二、创建score表，SQL代码如下：

CREATE  TABLE  score (id  INT(10)  NOT NULL  PRIMARY KEY  AUTO_INCREMENT ,stu_id  INT(10)  NOT NULL ,c_name  VARCHAR(20) ,grade  INT(10));

三、为student表和score表增加记录

 （1）向student表插入记录的INSERT语句如下：

INSERT INTO student VALUES( 901,'张老大', '男',1984,'计算机系', '北京市海淀区');

INSERT INTO student VALUES( 902,'张老二', '男',1987,'中文系', '北京市昌平区');

INSERT INTO student VALUES( 903,'张三', '女',1991,'中文系', '湖南省永州市');

INSERT INTO student VALUES( 904,'李四', '男',1993,'英语系', '辽宁省阜新市');

INSERT INTO student VALUES( 905,'王五', '女',1990,'英语系', '福建省厦门市');

INSERT INTO student VALUES( 906,'王六', '男',1989,'计算机系', '湖南省衡阳市');

INSERT INTO student VALUES( 907,'老七', '男',1991,'计算机系', '广东省深圳市');

INSERT INTO student VALUES( 908,'老八', '女',1990,'英语系', '山东省青岛市');

（2）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);

INSERT INTO score VALUES(NULL,907, '计算机',98);

四、基础练习题

1.查询student表的第2条到4条记录

select * from student limit 1,3;

2.从student表查询所有学生的学号（id）、姓名（name）和院系（department）的信息

mysql> select id,name,department from student;

3.从student表中查询计算机系和英语系的学生的信息

select * from student where department in ('计算机系' ,'英语系');

4.从student表中查询年龄23~26岁的学生信息

select * from student  where birth between 1990 and 1993;    2016-23=1993  2016-26=1990

select id,name,sex,2016-birth as age,department,address from student where 2016-birth;

5.从student表中查询每个院系有多少人

select department,count(id)  from student group by department;

6.从score表中查询每个科目的最高分。

select c_name,max(grade) from score group by c_name;

7.查询李四的考试科目（c_name）和考试成绩（grade）

select c_name,grade from score,student  where score. stu_id=student.id and name='李四';

select c_name,grade from score where stu_id=(select id from student where name='李四'）;

8.用连接的方式查询所有学生的信息和考试信息

select stu.*,sc.*  from student stu left join score sc on stu.id=sc.id;

9.计算每个学生的总成绩

select stu_id,sum(grade) from score group by stu_id;

10.计算每个考试科目的平均成绩

select c_name,avg(grade) from score group by c_name;

11.查询计算机成绩低于95分的学生信息

select student.*, grade from score,student where   student.id=score.stu_id and c_name like '计算机' and grade<95;

12.查询同时参加计算机和英语考试的学生的信息

select student.*,c_name  from student,score where student.id=score.stu_id and student.id =any( select stu_id from score where stu_id in (select stu_id from score where c_name= '计算机') and c_name= '英语' );

select * from student where id in(select stu_id from score where stu_id in (select stu_id from score where c_name='计算机' )and c_name='英语');

select student.* from student,(select stu_id from score where stu_id in (select stu_id from score where c_name='计算机' )and c_name='英语') t1 where student.id=t1.stu_id;

select * from student where id in (select stu_id from score sc where sc.c_name='计算机') and id in (select stu_id from score sc where sc.c_name='英语');

13.将计算机考试成绩按从高到低进行排序

select c_name,grade from score where c_name='计算机' order by grade;

14.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 id  from score;

15.查询姓张或者姓王的同学的姓名、院系和考试科目及成绩

select name,department,c_name,grade from score sc,student st where st.id=sc.stu_id and (name like'张%'or name like '王%');

16.查询都是湖南的学生的姓名、年龄、院系和考试科目及成绩中文系

select name,2016-birth age,department,address,c_name,grade from student,score where student.id=score.stu_id and address like'湖南%';

17.查询每个科目的最高分的学生信息.                                                    

分解： score=t1,    t2=select c_name,max(grade) as grade from score group by c_name，    t1.stu_id注解

分解： select * from student where id in (select t1.stu_id from score t1,t2 t2 where t1.c_name=t2.c_name and t1.grade=t2.grade);

select * from student where id in (select t1.stu_id from score t1,(select c_name,max(grade) as grade from score group by c_name) t2 where t1.c_name=t2.c_name and t1.grade=t2.grade);

select student.* from student,(select score.* from score,(select max(grade) grade,c_name from score group by c_name) t1 where score.c_name=t1.c_name and score.grade=t1.grade) t2 where student.id=t2.stu_id;