MySQl数据库必会sql语句加强版

这篇承接上一篇《mysql必会sql语句》：http://blog.csdn.net/qq_32059827/article/details/51763950

这一篇属于加强版，问题和sql语句如下。

创建users表，设置id,name,gender,sal字段，其中id为主键
drop table if exists users;
create table if not exists users(
    id int(5) primary key auto_increment,
    name varchar(10) unique not null,
    gender varchar(1) not null,
    sal int(5) not null
);
insert into users(name,gender,sal) values('AA','男',1000);
insert into users(name,gender,sal) values('BB','女',1200);
--------------------------------------------------------------------------------------
一对一：AA的身份号是多少
drop table if exists users;
create table if not exists users(
    id int(5) primary key auto_increment,
    name varchar(10) unique not null,
    gender varchar(1) not null,
    sal int(5) not null
);
insert into users(name,gender,sal) values('AA','男',1000);
insert into users(name,gender,sal) values('BB','女',1200);

drop table if exists cards;
create table if not exists cards(
    id int(5) primary key auto_increment,
    num int(3) not null unique,
    loc varchar(10) not null,
    uid int(5) not null unique,
	constraint uid_fk foreign key(uid) references users(id)
);
insert into cards(num,loc,uid) values(111,'北京',1);
insert into cards(num,loc,uid) values(222,'上海',2);
【注：inner join表示内连接】
select u.name "姓名",c.num "身份证号"
from users u inner join cards c
on u.id = c.uid
where u.name = 'AA';
--
select u.name "姓名",c.num "身份证号"
from users u inner join cards c
on u.id = c.uid
where name = 'AA';

---------------------------------------------
一对多：查询"开发部"有哪些员工
创建groups表
drop table if exists groups;
create table if not exists groups(
    id int(5) primary key auto_increment,
    name varchar(10) not null
);
insert into groups(name) values('开发部');
insert into groups(name) values('销售部');

创建emps表
drop table if exists emps;
create table if not exists emps(
    id int(5) primary key auto_increment,
    name varchar(10) not null,
    gid int(5) not null,
    constraint gid_fk foreign key(gid) references groups(id)
);
insert into emps(name,gid) values('哈哈',1);
insert into emps(name,gid) values('呵呵',1);
insert into emps(name,gid) values('嘻嘻',2);
insert into emps(name,gid) values('笨笨',2);

查询"开发部"有哪些员工
select g.name "部门",e.name "员工"
from groups g inner join emps e
on g.id = e.gid
where g.name = '开发部';
--
select g.name "部门",e.name "员工"
from groups g  inner join emps e
on g.id = e.gid
where g.name = '开发部';
------------------------------------------------------
多对多：查询"赵"教过哪些学生
创建students表
drop table if exists students;
create table if not exists students(
    id int(5) primary key auto_increment,
    name varchar(10) not null
);
insert into students(name) values('哈哈');
insert into students(name) values('嘻嘻');

创建teachers表
drop table if exists teachers;
create table if not exists teachers(
    id int(5) primary key auto_increment,
    name varchar(10) not null
);
insert into teachers(name) values('赵');
insert into teachers(name) values('刘');

创建middles表   primary key(sid,tid)  表示联合主键，这两个字段的整体要唯一
drop table if exists middles;
create table if not exists middles(
   sid int(5),
   constraint sid_fk foreign key(sid) references students(id),
   tid int(5),
   constraint tid_fk foreign key(tid) references teachers(id),
   primary key(sid,tid)
);
insert into middles(sid,tid) values(1,1);
insert into middles(sid,tid) values(1,2);
insert into middles(sid,tid) values(2,1);
insert into middles(sid,tid) values(2,2);

查询"赵"教过哪些学生
select t.name "老师",s.name "学生"
from students s inner join middles m inner join teachers t
on (s.id=m.sid) and (m.tid=t.id)
where t.name = '赵';
--
select t.name "老师",s.name "学生"
from students s inner join middles m inner join teachers t
on (s.id=m.sid) and (t.id=m.tid)
where t.name = "赵";

--------------------------------------------------------------------------------------------------------
将5000元（含）以上的员工标识为"高薪"，否则标识为"起薪"

将薪水为NULL的员工标识为"无薪"

将5000元（含）以上的员工标识为"高薪"，否则标识为"起薪"

将7000元的员工标识为"高薪",6000元的员工标识为"中薪",5000元则标识为"起薪",否则标识为"试用薪"
---------------------------------------------------------------------------------------------------------
内连接（等值连接）：查询客户姓名，订单编号，订单价格
【注：customers c inner join orders o使用了别名，以后o就代表orders】
select c.name "客户姓名",o.isbn "订单编号",o.price "订单价格"
from customers c inner join orders o
on c.id = o.customers_id;
--
select c.name "客户姓名",o.isbn "订单编号",o.price "订单价格"
from customers c inner join orsers o
on c.id = o.customers_id;
on+两张表连接的条件.一张表的主键，一张表的外键
内连接：只能查询出二张表中根据连接条件都存在的记录，有点类似于数学中交集
----------------------------------------------------
外连接：按客户分组，查询每个客户的姓名和订单数
外连接：既可以根据连接条件查询出二张表中都存在的记录，也能根据一方，强行将另一方就算不满兄条件的记录也能查询出来
外连接可以细分为：
<左外连接 : 以左侧为参照，left outer join表示
select c.name,count(o.isbn)
from customers c left outer join orders o
on c.id = o.customers_id
group by c.name;
--
>右外连接 : 以右侧为参照，right outer join表示
select c.name,count(o.isbn)
from orders o right outer join customers c
on c.id = o.customers_id
group by c.name;

left outer join表示左边的内容都会显现出来，例如customers c left out join 表示会把customers中的某列所有内容都找出来

------------------------------------------------------
自连接：求出AA的老板是EE。把自己想象成两张表。左右各一张
select users.ename,bosss.ename
from emps users inner join emps bosss
on users.mgr = bosss.empno;

select users.ename,bosss.ename
from emps users left outer join emps bosss
on users.mgr = bosss.empno;

-----------------------------------------------------------------------------------------------
演示MySQL中的函数（查询手册）

日期时间函数：
select addtime('2016-8-7 23:23:23','1:1:1');   时间相加
select current_date();
select current_time();
select now();
select year( now() );
select month( now() );
select day( now() );
select datediff('2016-12-31',now());

字符串函数：
select charset('哈哈');
select concat('你好','哈哈','吗');
select instr('www.baidu.com','baidu');
select substring('www.baidu.com',5,3);

数学函数：
select bin(10);
select floor(3.14);//比3.14小的最大整数---正3
select floor(-3.14);//比-3.14小的最大整数---负4
select ceiling(3.14);//比3.14大的最小整数---正4
select ceiling(-3.14);//比-3.14大的最小整数---负3，一定是整数值
select format(3.1415926,3);保留小数点后3位，四舍五入
select mod(10,3);//取余数
select rand();//

加密函数：
select md5('123456');
返回32位16进制数 e10adc3949ba59abbe56e057f20f883e

演示MySQL中流程控制语句
use json;
drop table if exists users;
create table if not exists users(
   id int(5) primary key auto_increment,
   name varchar(10) not null unique,
   sal int(5)
);
insert into users(name,sal) values('哈哈',3000);
insert into users(name,sal) values('呵呵',4000);
insert into users(name,sal) values('嘻嘻',5000);
insert into users(name,sal) values('笨笨',6000);
insert into users(name,sal) values('明明',7000);
insert into users(name,sal) values('丝丝',8000);
insert into users(name,sal) values('君君',9000);
insert into users(name,sal) values('赵赵',10000);
insert into users(name,sal) values('无名',NULL);

将5000元（含）以上的员工标识为"高薪"，否则标识为"起薪"
select name "姓名",sal "薪水",
       if(sal>=5000,"高薪","起薪") "描述"
from users;

将薪水为NULL的员工标识为"无薪"
select name "姓名",ifnull(sal,"无薪") "薪水"
from users;

将5000元（含）以上的员工标识为"高薪"，否则标识为"起薪"
select name "姓名",sal "薪水",
       case when sal>=5000 then "高薪"
       else "起薪" end "描述"
from users;

将7000元的员工标识为"高薪"，6000元的员工标识为"中薪",5000元则标识为"起薪"，否则标识为"试用薪"
select name "姓名",sal "薪水",
       case sal
            when 3000 then "低薪"
            when 4000 then "起薪"
            when 5000 then "试用薪"
            when 6000 then "中薪"
            when 7000 then "较好薪"
            when 8000 then "不错薪"
            when 9000 then "高薪"
            else "重薪"
       end "描述"
from users;