MySql的简单使用，所有的代码基于MAC

brew install mysql

ps -ef|grep mysql

mysql --verbose --help | grep my.cnf

mysql.server start

mysql -u root -p 

select now();

select version();

show databases;

create database test_db;
create database test_db charset=utf8;//设置数据库的编码格式
create database test_db character set utf8;

show create database test_db;

drop database test_db;

use test_db;

select database();

show tables;

Create table students(id int unsigned not null auto_increment primary key,name varchar(20),age tinyint unsigned,hight decimal(5,2),gender enum(‘男‘,‘女‘,‘未知‘) default ‘未知‘);

drop table test;

alter table students add birthday datetime;

alter table students modify birthday date;

alter table students change birthday birth date default "1990-01-01";

alter table students drop birth;

desc students;

insert into students values (0,‘刘‘,18,1.72,‘男‘);
insert into students values (0,‘刘‘,18,1.72,‘男‘),(0,‘刘‘,18,1.72,‘男‘);
insert into students (name,hight) values ("测试",1.77);
insert into students (name,hight) values ("测试",1.77),("测试",1.77);

select * from students;
select name,age from students;
select name as 姓名,age as 年龄 from students;
select s.name,s.age from students as s;

update students set age = 30,name = "老王" where id = 1;

update students set hight = hight +1 where id = 30;

delete from students where id = 6

select distinct gender from students;

select * from students where gender <> "男";
select * from students where id > 3;
select * from students where id = 3;
select * from students where age>18 and age<40;
select * from students where age>30 or age < 20;
select * from students where not (age > 30);
select * from students where (not age > 30) and name="刘";
select * from students where name like "测%";//查询以测开头的
select * from students where name like "__";//查询名字有两个字符的
select * from students where name rlike "^测.*";//正则
select * from students where age in (30,18);//年龄30，或者18的
select * from students where age not in (30,18);//
select * from students where age between 18 and 30;//年龄18<=age<=30
select * from students where age not between 18 and 30;//年龄18<=age<=30
select * from students where age is NULL;
select * from students where age is not NULL;

select * from students order by age;
select * from students where age is not NULL order by age;//asc 升序
select * from students where age is not NULL order by age desc;//降序

select count(*) from students where age = 18;//统计18岁的人数
select count(*) as 18岁人数 from students where age = 18
select max(age) from students ;//最大年龄
select min(age) from students ;//最小年龄
select sum(age) from students;//所有年龄总和
select avg(age) from students;//平均年龄
select sum(age)/count(*) from students;//平均年龄
select round( sum(age)/count(*),2) from students;//2代表保留两位小数
select gender from students group by gender;//分组
select gender, count(*) from students group by gender;//分组统计
select gender,group_concat(name) from students group by gender;//根据性别统计姓名
select gender,group_concat(name,age) from students where gender="女";
select gender,group_concat(name,"_",age) from students where gender="女";
select gender,group_concat(name) from students group by gender having count(*)>3;//分组的数大于3条记录
select gender,group_concat(name) from students group by gender having avg(age)>20;//分组的平均年龄大于20

select * from students limit 2;//限制查询出来的个数，最多两条
select * from students limit 0,5;//从数据里面第0条开始查5条
select * from students limit 5,5;//从第五条开始查5条
select * from students limit 5,5;//从第五条开始查5条
select name from students where gender="未知" limit 2;

Create table cls(id int unsigned not null auto_increment primary key,name varchar(20));
alter table students add cls_id int default 0;

Create table if not exists  cls(id int unsigned not null auto_increment primary key,name varchar(20));

select * from students inner join cls ;//结果是count(students)*count(cls)
select * from students inner join cls on  students.cls_id = cls.id;//查出学生所在的班级对应的班级
select students.id,students.name,cls.name from students inner join cls on  students.cls_id = cls.id;//显示指定的列
select s.*,c.name from students as s inner join cls as c on  s.cls_id = c.id;//显示学生所有字段，给表设置别名
select s.*,c.name from students as s inner join cls as c on  s.cls_id = c.id order by c.name,s.id;//排序
select * from students as s left join cls as c on s.cls_id = c.id ;//用students为基础，查cls，等价下面
select * from cls as c right join students as s on s.cls_id = c.id ;
select * from students as s left join cls as c on s.cls_id = c.id having c.name is null;//查询没有班级的学生
select * from students as s left join cls as c on s.cls_id = c.id where c.name is null;//这个不建议
select s.id,s.name,c.name from students as s,cls as c where s.cls_id = c.id;//不建议

 create table areas(aid int primary key,atitle varchar(20),pid int);

source aaa.sql;//如果aaa.sql是一个sql文件，可以使用这个,前提是ls必须路径里面必须有aaa.sql

Select areas.atitle from areas inner join areas as city on city.pid = areas.aid having areas.atitle="北京市区";

select * from students where hight = (select max(hight) from students where gender=‘男‘);//查询最高的男生信息
select * from students inner join (select max(hight) as hight from students group by cls_id) as stu on students.hight = stu.hight;//查询每个班最高的学生信息
select * from (select cls_id,  max(hight) as hight from students group by cls_id) as stu left join students as s on stu.cls_id = s.cls_id and stu.hight = s.hight;//同上
select * from students inner join (select max(hight) as hight from students group by cls_id) as stu on students.hight = stu.hight having students.gender="女";//查询每个班最高的女学生信息
select * from students left join (select max(hight) as hight from students group by cls_id) as stu on students.hight = stu.hight having students.gender="女";//同上

select s.id,s.name,s.age,s.hight,s.gender,c.name from students as s inner join cls as c on  s.cls_id = c.id;

drop view students_cls_view;

start transaction;//或者begin
update students set name="好人3" where id = 28;
commit;//提交 rollback；//可以回滚

set profiling=1;//开启时间运行监测
select * from students;
show profiles;//查看查询的时间
set profiling=1;
create index id_index on students(id);//如果是字符串，需要把id改成name(10)  删除 drop index  id_index

show profiles;