This article shares with you an article about the enhanced version of the MySQL database must know SQL statements. It is very good and has reference value. Friends who need it can refer to it.
This article belongs to the enhanced version. Questions and The sql statement is 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);
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One-on-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 inner join]
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';
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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);
Check 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';
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Many-to-many: Query which students "Zhao" has taught
Create students table
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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 the middles table primary key(sid,tid) represents the joint primary key, and the two fields must be unique as a whole
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);Check 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";------- -------------------------------------------------- --------------------------------------------------
Mark employees with a salary of more than 5,000 yuan (inclusive) as "high salary", otherwise mark them as "starting salary"
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 one table Inner join: Only records that exist in two tables according to the connection conditions can be queried, which is a bit Similar to intersection in mathematics --------------------------------------------- ------------- Outer connection: Group by customer, query the name and order number of each customer Outer connection: You can query based on the connection conditions Records that exist in both tables can also be queried based on one side, even if the other side's records are not satisfied with the conditions. Outer joins can be subdivided into:
<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 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 found
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 functions in MySQL (query manual)
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 functions:
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 functions:
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();//
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);Identify employees with more than 5,000 yuan (inclusive) It is "high salary", otherwise it is marked as "starting salary"
select name "姓名",sal "薪水", if(sal>=5000,"高薪","起薪") "描述" from users;
select name "姓名",ifnull(sal,"无薪") "薪水" from users;
select name "姓名",sal "薪水", case when sal>=5000 then "高薪" else "起薪" end "描述" from users;identifies employees earning 7,000 yuan as "high salary", employees earning 6,000 yuan as "medium salary", and employees earning 5,000 yuan as "starting salary" Salary", otherwise marked as "Trial Salary"
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;
The above is the detailed content of SQL statements that MySQl database must know. For more information, please follow other related articles on the PHP Chinese website!