• mysql基础


    - SQL语句
    - 数据库
    create database db1; ?
    drop database db1;

    - 数据表
    先创建tb2部门表

    create table tb1用户表(
    id int not null auto_increment primary key,
    name char(10),
    department_id int,
    p_id int,
    constraint fk_1 foreign key (department_id,p_id) references tb2(tid,xid)
    )engine=innodb default charset=utf8;


    补充:主键
    一个表只能有一个主键
    主键可以由多列组成


    补充:外键 ?
    CREATE TABLE t5 (
    nid int(11) NOT NULL AUTO_INCREMENT,
    pid int(11) not NULL,
    num int(11),
    primary key(nid,pid)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8;

    create table t6(
    id int auto_increment primary key,
    name char(10),
    id1 int,
    id2 int,
    CONSTRAINT fk_t5_t6 foreign key (id1,id2) REFERENCES t1(nid,pid)
    )engine=innodb default charset=utf8;



    - 数据行

    insert into tb1(name,age) values('alex',18);

    delete from tb1;
    truncate table tb1;
    delete from tb1 where id > 10


    update tb1 set name='root' where id > 10

    select * from tb;
    select id,name from tb;


    4 对于自增补充:
    desc t10;

    show create table t10;

    show create table t10 G;

    alter table t10 AUTO_INCREMENT=20;


    MySQL: 自增步长
    基于会话级别:
    show session variables like 'auto_inc%'; 查看全局变量
    set session auto_increment_increment=2; 设置会话步长
    # set session auto_increment_offset=10;
    基于全局级别:
    show global variables like 'auto_inc%'; 查看全局变量
    set global auto_increment_increment=2; 设置会话步长
    # set global auto_increment_offset=10;


    SqlServer:自增步长:
    基础表级别:
    CREATE TABLE `t5` (
    `nid` int(11) NOT NULL AUTO_INCREMENT,
    `pid` int(11) NOT NULL,
    `num` int(11) DEFAULT NULL,
    PRIMARY KEY (`nid`,`pid`)
    ) ENGINE=InnoDB AUTO_INCREMENT=4, 步长=2 DEFAULT CHARSET=utf8

    CREATE TABLE `t6` (
    `nid` int(11) NOT NULL AUTO_INCREMENT,
    `pid` int(11) NOT NULL,
    `num` int(11) DEFAULT NULL,
    PRIMARY KEY (`nid`,`pid`)
    ) ENGINE=InnoDB AUTO_INCREMENT=4, 步长=20 DEFAULT CHARSET=utf8


    今日内容:
    0. 唯一索引

    create table t1(
    id int ....,
    num int,
    xx int,
    unique 唯一索引名称 (列名,列名),
    constraint ....
    )
    #
    1 1 1
    2 1 2
    PS:
    唯一:
    约束不能重复(可以为空)
    PS: 主键不能重复(不能为空)
    加速查找

    1. 外键的变种

    a. 用户表和部门表

    用户:
    1 alex 1
    2 root 1
    3 egon 2
    4 laoyao 3

    部门:
    1 服务
    2 保安
    3 公关
    ===》 一对多
    b. 用户表和博客表
    用户表:
    1 alex
    2 root
    3 egon
    4 laoyao
    博客表:
    FK() + 唯一
    1 /yuanchenqi/ 4
    2 /alex3714/ 1
    3 /asdfasdf/ 3
    4 /ffffffff/ 2

    ===> 一对一

    create table userinfo1(
    id int auto_increment primary key,
    name char(10),
    gender char(10),
    email varchar(64)
    )engine=innodb default charset=utf8;

    create table admin(
    id int not null auto_increment primary key,
    username varchar(64) not null,
    password VARCHAR(64) not null,
    user_id int not null,
    unique uq_u1 (user_id),
    CONSTRAINT fk_admin_u1 FOREIGN key (user_id) REFERENCES userinfo1(id)
    )engine=innodb default charset=utf8;




    c. 用户表(百合网) 相亲记录表

    示例1:
    用户表
    相亲表

    示例2:
    用户表
    主机表
    用户主机关系表
    ===》多对多

    create table userinfo2(
    id int auto_increment primary key,
    name char(10),
    gender char(10),
    email varchar(64)
    )engine=innodb default charset=utf8;

    create table host(
    id int auto_increment primary key,
    hostname char(64)
    )engine=innodb default charset=utf8;


    create table user2host(
    id int auto_increment primary key,
    userid int not null,
    hostid int not null,
    unique uq_user_host (userid,hostid),
    CONSTRAINT fk_u2h_user FOREIGN key (userid) REFERENCES userinfo2(id),
    CONSTRAINT fk_u2h_host FOREIGN key (hostid) REFERENCES host(id)
    )engine=innodb default charset=utf8;


    2. SQL语句数据行操作补充
    create table tb12(
    id int auto_increment primary key,
    name varchar(32),
    age int
    )engine=innodb default charset=utf8;


    insert into tb11(name,age) values('alex',12);

    insert into tb11(name,age) values('alex',12),('root',18);

    insert into tb12(name,age) select name,age from tb11;

    delete from tb12;
    delete from tb12 where id !=2
    delete from tb12 where id =2
    delete from tb12 where id > 2
    delete from tb12 where id >=2
    delete from tb12 where id >=2 or name='alex'


    update tb12 set name='alex' where id>12 and name='xx'
    update tb12 set name='alex',age=19 where id>12 and name='xx'


    select * from tb12;

    select id,name from tb12;

    select id,name from tb12 where id > 10 or name ='xxx';

    select id,name as cname from tb12 where id > 10 or name ='xxx';

    select name,age,11 from tb12;

    其他:
    select * from tb12 where id != 1
    select * from tb12 where id in (1,5,12);
    select * from tb12 where id not in (1,5,12);
    select * from tb12 where id in (select id from tb11)
    select * from tb12 where id between 5 and 12;


    通配符:

    select * from tb12 where name like "a%"
    select * from tb12 where name like "a_"


    分页:

    select * from tb12 limit 10;

    select * from tb12 limit 0,10;
    select * from tb12 limit 10,10;
    select * from tb12 limit 20,10;

    select * from tb12 limit 10 offset 20;


    # page = input('请输入要查看的页码')
    # page = int(page)
    # (page-1) * 10
    # select * from tb12 limit 0,10; 1
    # select * from tb12 limit 10,10;2


    排序:
    select * from tb12 order by id desc; 大到小
    select * from tb12 order by id asc; 小到大
    select * from tb12 order by age desc,id desc;

    取后10条数据
    select * from tb12 order by id desc limit 10;

    分组:

    select count(id),max(id),part_id from userinfo5 group by part_id;

    count
    max
    min
    sum
    avg

    **** 如果对于聚合函数结果进行二次筛选时?必须使用having ****
    select count(id),part_id from userinfo5 group by part_id having count(id) > 1;

    select count(id),part_id from userinfo5 where id > 0 group by part_id having count(id) > 1;


    连表操作:

    select * from userinfo5,department5

    select * from userinfo5,department5 where userinfo5.part_id = department5.id

    select * from userinfo5 left join department5 on userinfo5.part_id = department5.id
    select * from department5 left join userinfo5 on userinfo5.part_id = department5.id
    # userinfo5左边全部显示


    # select * from userinfo5 right join department5 on userinfo5.part_id = department5.id
    # department5右边全部显示



    select * from userinfo5 innder join department5 on userinfo5.part_id = department5.id
    将出现null时一行隐藏






    select * from
    department5
    left join userinfo5 on userinfo5.part_id = department5.id
    left join userinfo6 on userinfo5.part_id = department5.id


    select
    score.sid,
    student.sid
    from
    score

    left join student on score.student_id = student.sid

    left join course on score.course_id = course.cid

    left join class on student.class_id = class.cid

    left join teacher on course.teacher_id=teacher.tid




    select count(id) from userinfo5;



    作业练习:
    http://www.cnblogs.com/wupeiqi/articles/5729934.html
    10-15个完成

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  • 原文地址:https://www.cnblogs.com/hlan/p/6955654.html
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