Enhanced version of sql statements that you must know about MySQl database

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This article is an enhanced version. The questions and SQL statements are as follows.

Create users table, set id, name, gender, sal fields, where id is the primary key

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);

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

One-to-one: What is AA’s identity number

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);

[Note: inner join means within Connection]

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';

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

One-to-many: Query Which employees are in the "Development Department"

Create groups table

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');

Create emps table

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);

Query which employees are in the "Development Department"

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';

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

Many-to-many: Query which students "Zhao" has taught

Create students table

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');

Create teachers table

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');

Create middles table primary key(sid,tid) represents the joint primary key, The whole of these two fields must be unique

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);

Query which students "Zhao" has taught

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";

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

Identify employees who earn more than 5,000 yuan (inclusive) It is "high salary", otherwise it is marked as "starting salary"

Mark employees whose salary is NULL as "no salary"

Mark employees whose salary is more than 5,000 yuan (inclusive) as "high salary" , otherwise marked as "starting salary"

Mark employees with 7,000 yuan as "high salary", employees with 6,000 yuan as "medium salary", 5,000 yuan as "starting salary", otherwise mark as " Probationary salary"

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

Inner join (equivalent join): Query customer name, order number, order price

[Note: customers c inner join orders o uses an alias, and o will represent orders in the future]

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+Conditions for connecting two tables. Primary key of one table, foreign key of another table

Inner join: Only records that exist in two tables according to the connection conditions can be queried, which is somewhat similar to the intersection in mathematics

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

Outer connection: Group by customer, Query the name and order number of each customer

Outer join: You can query the records that exist in both tables according to the connection conditions, or you can also force the records of the other party based on one party even if they are not satisfied with the conditions. Can be queried

Outer join can be subdivided into:

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 means that the content on the left will be displayed, for example, customers c left out join means that all the contents of a certain column in customers will be displayed Find them all

----------------------------------------- -------------
Self-connection: Find out whether AA’s boss is EE. Think of yourself as two tables. One on each side

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;

------------------------------------------ -------------------------------------------------- ------
Demonstrate the function in MySQL (query manual)

Date and time function:

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());

String function ：

select charset('哈哈'); 
select concat('你好','哈哈','吗'); 
select instr('www.baidu.com','baidu'); 
select substring('www.baidu.com',5,3);

Mathematical function:

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();//

Encryption function:

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;

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