admin管理员组

文章数量:1530518

2024年3月27日发(作者:)

建库:

create database db_name character set =utf8;

/*建立库db_name,默认字符集为utf8*/

建表:

create table tb_1 (id int auto_increment primary key, remark varchar(20))

engine=innodb default character set=utf8 auto_increment =100;

/*建立表tb_1,存储引擎为innodb,默认字符集为utf8,自增列开始值为100*/

create table tb_2 (id2 int auto_increment primary key, remark varchar(20))

engine=innodb default character set=utf8 auto_increment =100;

内联接:

select * from t1 inner join t2 where =2;

左联接:

select * from t1 left join t2 on =2;

右联接:

select * from t1 right join t2 on =2;

全联接:

select * from t1,t2

create table tb_student (

Student_ID int auto_increment primary key,

Student_Name varchar(20),

Sex char(1),

Birthday varchar(10)

) engine=innodb default character set =utf8;

create table tb_subject (

Subject_ID int auto_increment primary key,

Subject_Name varchar(50)

) engine=innodb default character set =utf8;

create table tb_score (

ID int auto_increment primary key,

Student_ID int,

Subject_ID int,

Score decimal(10,2),

foreign key (Student_ID) references tb_student(Student_ID),

foreign key (Subject_ID) references tb_subject(Subject_ID)

) engine=innodb default character set =utf8

Data 数据 Database 数据库 RDBMS 关系数据库管理系统 GRANT 授权

REVOKE 取消权限 DENY 拒绝权限 DECLARE 定义变量 PROCEDURE存储过程

事务 Transaction 触发器 TRIGGER 继续 continue 唯一 unqiue

主键 primary key 标识列 identity 外键 foreign key 检查 check

约束 constraint

--------------------------------------------------------------------

1) 创建一张学生表,包含以下信息,学号,姓名,年龄,性别,家庭住址,联系电话

create table student

(

学号 int,

姓名 varchar(10),

年龄 int,

性别 varchar(4),

家庭住址 varchar(50),

联系电话 varchar(11)

);

--------------------------------------------------------------------

2) 修改学生表的结构,添加一列信息,学历

alter table student add column 学历 varchar(6);

--------------------------------------------------------------------

3) 修改学生表的结构,删除一列信息,家庭住址

alter table student drop column 家庭住址;//注意此处用drop而非delete

--------------------------------------------------------------------

4) 向学生表添加如下信息:

学号 姓名年龄性别联系电话学历

1A22男123456小学

2B21男119中学

3C23男110高中

4D18女114大学

insert into student (学号,姓名,年龄,性别,联系电话,学历) values(1,"A",22,"男

","123456","小学");

insert into student (学号,姓名,年龄,性别,联系电话,学历) values(1,"B",21,"男

","119","中学");

insert into student (学号,姓名,年龄,性别,联系电话,学历) values(1,"C",23,"男

","123456","高中");

insert into student (学号,姓名,年龄,性别,联系电话,学历) values(1,"D",23,"女

","114","大学");

--------------------------------------------------------------------

5) 修改学生表的数据,将电话号码以11开头的学员的学历改为“大专”

update student set 学历="大专" where 联系电话 like "11%";

--------------------------------------------------------------------

6) 删除学生表的数据,姓名以C开头,性别为‘男'的记录删除

delete from student where 姓名 like "C" and 性别="男";

--------------------------------------------------------------------

7) 查询学生表的数据,将所有年龄小于22岁的,学历为“大专”的,学生的姓名和学号示

出来

select 姓名,学号 from student where 年龄<22 and 学历="大专";

--------------------------------------------------------------------

8) 查询学生表的数据,查询所有信息,列出前25%的记录

select top 25 percent * from student ;

select * from student limit 25%;

这条有问题,在sql 2000中应该是select top 25 percent * from student ;

--------------------------------------------------------------------

9) 查询出所有学生的姓名,性别,年龄降序排列

select 姓名,性别,年龄 from student order by 年龄 desc;

--------------------------------------------------------------------

10) 按照性别分组查询所有的平均年龄

select avg(年龄) as 平均年龄 from student group by 性别;

select avg(年龄) from student group by 性别;

select avg(年龄) 平均年龄 from student group by 性别;

--------------------------------------------------------------------

3) 说出以下聚合数的含义:avg ,sum ,max ,min , count ,count(*)

AVG:求平均值

SUM:求和

MAX:求最大值

MIN:求最小值

COUNT(*):返回所有行数

COUNT返回满足指定条件的记录值

本文标签: 性别学生年龄姓名