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# A. concat :  String splicing 
select concat('Hello' , ' MySQL');
# B. lower :  All lowercase 
select lower('Hello');
# C. upper :  Capitalize all 
select upper('Hello');
# D. lpad :  padding-left 
select lpad('01', 5, '-');
# E. rpad :  Right fill 
select rpad('01', 5, '-');
# F. trim :  Remove the space 
select trim('   Hello MySQL ');
# G. substring :  Intercept substring 
select substring('Hello MySQL',1,5);

Due to changes in business requirements , Employee's job number , Uniform for 5 digit , Not enough at present 5 All digits are filled in the front 0. such as ： 1 The job number of employee No. should be 00001

update emp set workno = lpad(workno, 5, '0');

Numerical function

# A. ceil： Rounding up 
select ceil(1.1);
# B. floor： Rounding down 
select floor(1.9);
# C. mod： modulus 
select mod(7,4);
# D. rand： Get random numbers 
select rand();
# E. round： rounding 
select round(2.344,2);

Through the functions of the database , Generate a six digit random verification code

select lpad(round(rand()*1000000 , 0), 6, '0');

Ideas ： The random number can be obtained by rand() function , But the random number obtained is in 0-1 Between , So it can be based on

Multiply up 1000000, Then discard the decimal part , If the length is not enough 6 position , repair 0

Date function

# A. curdate： The current date 
select curdate();

# B. curtime： current time 
select curtime();

# C. now： Current date and time 
select now();

# D. YEAR , MONTH , DAY： The current year 、 month 、 Japan 
select YEAR(now());
select MONTH(now());
select DAY(now());

# E. date_add： Increase the specified time interval 
select date_add(now(), INTERVAL 70 YEAR );

# F. datediff： Get the number of days between the two dates 
select datediff('2021-10-01', '2021-12-01');

Query the enrollment days of all employees , And sort them in reverse order according to the number of days of employment

Ideas ： Days in position , Through the current date - Date of entry , So we need to use datediff Function to complete

select name, datediff(curdate(), entrydate) as 'entrydays' from emp order by entrydays 
desc;

Flow function

# A. if
select if(false, 'Ok', 'Error');

# B. ifnull
select ifnull('Ok','Default');
select ifnull('','Default');
select ifnull(null,'Default');

# C. case when then else end
select
name,
( case workaddress when ' Beijing ' or ' Shanghai '  then ' first-tier cities 'else
' Second-tier cities ' end ) as ' Work address '
from emp;

constraint

Concept ： Constraints are rules that act on fields in a table , Used to limit the data stored in a table

Purpose ： Ensure that the data in the database is correct 、 Effectiveness and integrity

Primary key constraint

Provide constraints when creating tables

CREATE TABLE tb_user(
id int AUTO_INCREMENT PRIMARY KEY COMMENT 'ID Unique identification ',
name varchar(10) NOT NULL UNIQUE COMMENT ' full name ' ,
age int check (age > 0 && age <= 120) COMMENT ' Age ' ,
status char(1) default '1' COMMENT ' state ',
gender char(1) COMMENT ' Gender '
);

Foreign key constraints

Create two tables

create table dept(
id int auto_increment comment 'ID' primary key,
name varchar(50) not null comment ' Department name '
)comment ' Departmental table ';
INSERT INTO dept (id, name) VALUES (1, ' R & D department '), (2, ' The Marketing Department '),(3, ' Finance Department '), (4,
' The sales department '), (5, ' General manager office ');

create table empp(
id int auto_increment comment 'ID' primary key,
name varchar(50) not null comment ' full name ',
age int comment ' Age ',
job varchar(20) comment ' Position ',
salary int comment ' Salary ',
entrydate date comment ' Entry time ',
managerid int comment ' Direct leadership ID',
dept_id int comment ' department ID'
)comment ' The employee table ';
INSERT INTO empp (id, name, age, job,salary, entrydate, managerid, dept_id) VALUES (1, ' Jin yong ', 66, ' President ',20000, '2000-01-01', null,5),(2, ' zhang wuji ', 20, ' project manager ',12500, '2005-12-05', 1,1), (3, ' Yang Xiao ', 33, ' Development ', 8400,'2000-11-03', 2,1),(4, ' Xiangr ', 48, ' open   Hair ',11000, '2002-02-05', 2,1), (5, ' chang yuchun ', 43, ' Development ',10500, '2004-09-07', 3,1),(6, ' Small zhao ', 19, ' cheng   The foreman encouraged the teacher ',6600, '2004-10-12', 2,1);

1). Add foreign keys

CREATE TABLE Table name (
Field name data type ,
...
[CONSTRAINT] [ Name of the foreign key ] FOREIGN KEY ( Foreign key field name ) REFERENCES Main table ( Name of main table )
);
ALTER TABLE Table name ADD CONSTRAINT Name of the foreign key FOREIGN KEY ( Foreign key field name )
REFERENCES Main table ( Name of main table ) ;

by empp Tabular dept_id Field add foreign key constraint , relation dept Primary Key id

alter table empp add constraint fk_emp_dept_id foreign key (dept_id) references dept(id);

2). Delete foreign key

ALTER TABLE Table name DROP FOREIGN KEY Name of the foreign key ;

alter table emp drop foreign key fk_emp_dept_id;

Delete / Update behavior

grammar

ALTER TABLE Table name ADD CONSTRAINT Name of the foreign key FOREIGN KEY ( Foreign key field ) REFERENCES
Main table name ( Main table field name ) ON UPDATE CASCADE ON DELETE CASCADE;

because NO ACTION It's the default behavior , When we demonstrated the grammar earlier , It's been tested , I don't want to show it anymore , Here we demonstrate two other behaviors ：CASCADE、SET NULL.

1.CASCADE

alter table emp add constraint fk_emp_dept_id foreign key (dept_id) references dept(id) on 
update cascade on delete cascade ;

The phenomenon is that the primary and secondary tables are synchronized

2.SET NULL

alter table emp add constraint fk_emp_dept_id foreign key (dept_id) references dept(id) on 
update set null on delete set null ;

We delete id by 1 The data of , It is found that the records of the parent table can be deleted normally , After deleting the data of the parent table , Then open the sub table empp, We found that the sub table empp Of dept_id Field , original dept_id by 1 The data of , Now it's all set to NULL 了 .

over