MySQL的建表及查询

一。建立表

mysql> 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));

mysql> 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));

二。插入数据

1.对于student表：

mysql> insert  student values(901,'张三丰','男',2002,'计算机系','北京市海淀区');
mysql> insert  student values(902,'周全有','男',2000,'中文系','北京市昌平区');
mysql> insert  student values(903,'张思维','女',2003,'中文系','湖南省永州市');
mysql> insert  student values(904,'李广昌','男',1999,'英语系','辽宁省皋新市');
mysql> insert  student values(905,'王翰','男',2004,'英语系','福建省厦门市');
mysql> insert  student values(906,'王心凌','女',1998,'计算机系','湖南省衡阳市');

2.对于score表：

mysql> insert into score values(null,901,'计算机',98);
mysql> insert into score values(null,901,'英语',80);
mysql> insert into score values(null,902,'计算机',65);
mysql> insert into score values(null,902,'中文',88);
mysql> insert into score values(null,903,'中文',95);
mysql> insert into score values(null,904,'计算机',70);
mysql> insert into score values(null,904,'英语',92);
mysql> insert into score values(null,905,'英语',94);
mysql> insert into score values(null,906,'计算机',49);
mysql> insert into score values(null,906,'英语',83);

三。数据的查询：

1.分别查询student表和score表的所有记录

上述已显示

2.查询student表的第二条到第五条记录

select * from student limit 1,5;

3.从student表中查询计算机系和英语系的学生信息

mysql> select * from student where department = "计算机系" or department = "英语系";

4.从student表中查询年龄小于22岁的学生信息

mysql> select * from student where year(now())-birth < 22;

5.从student表中查询每个院系有多少人

mysql> select count(department) as "人数",department "院系" from student group by department;

6.从score表中查询每个科目的最高分

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

7.查询李广昌的考试科目和考试成绩

mysql> select c_name,grade from score join student on student.id = score.stu_id where student.name="李广昌";

8.用连接查询的方式查询所有学生的信息id和考试信息id

select id  from student union select stu_id from score;

9.计算每个学生的总成绩

mysql> select name,sum(grade) from student join score on student.id = score.stu_id  group by name;

10.查询每个科目的平均成绩

mysql> select c_name,round(avg(grade),2) "平均成绩" from score group by c_name;

11.查询计算机成绩低于95的学生信息

mysql> select student.* from student join score on student.id = score.stu_id where score.grade < 95 and score.c_name = "计算机";

12.将计算机考试成绩按从高到低排序

mysql> select grade from score where c_name = "计算机" order by grade desc;

13.从student和score表中查询查询出学生的学号，然后合并查询结果

select id  from student union select stu_id from score;

14.查询姓张或者姓王的学生的姓名，院系和考试科目和成绩

mysql> select name,department,c_name,grade from student join score on score.stu_id = student.id where student.name like "张%" or student.name like "王%";

15.查询都是湖南的学生的姓名，年龄，院系和考试科目及成绩

mysql> select name,(year(now())-birth) as "年龄" ,department,c_name,grade from student
    -> join score on student.id=score.stu_id where address like "湖南%";

原文地址：https://blog.csdn.net/2301_78530830/article/details/140605873

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