Introduction to MySQL (I) grammar

1.MySQL Introduce
Data storage problem ：
Variable , Array , object , aggregate
disadvantages ： Temporary storage , The magnitude of stored data is too small
IO flow
disadvantages ： The stored data is not type specific , No security measures , No backup and recovery

Introduce the concept of database ：
Store and manage data warehouse

Database classification ：
1. Relational database ：
Oracle、DB2、MySQL( a key )、SQL Server, characteristic ： Tables are associated with tables
2. Non relational database ：
ElastecSearch、MongoDB、Redis, Hash tables are commonly used , Save data through key value pairs

Database installation ---- Text interface installation mode

SQL Language Overview
Structured query language , Simply put, adding, deleting, modifying and checking are all SQL Category –CRUD

1. Operate the database from the command prompt

2. Operate the database graphically

2. Data query （ a key ）

The basic query
ctrl+： Zoom in
The query syntax ：SELECT Name FROM Table name Query a single field

select first_name from t_employees;

Query multiple fields

select first_name,salary from t_employees;

Query all fields

select * from t_employees;

The query field can be operated

SELECT employee_id,first_name,salary*12 from t_employees;

Set the alias of the query field ： as

SELECT employee_id as ' Employee number ',first_name as ' surname ',salary*12 as ' Annual salary ' from t_employees;

De duplication of field contents ：distinct

select DISTINCT manager_id from t_employees;

Sort query ： SELECT Name FROM Table name ORDER BY Sort columns [ Sort rule ]
Check the employee's number , name , Salary . Sort by salary in descending order .

select employee_id,first_name,salary from t_employees ORDER BY salary desc;

Multi column sorting ：
Check the employee's number , name , Salary . Sort by salary in ascending order （ When the salary is the same , Sort by number in ascending order ）.

select employee_id,first_name,salary from t_employees ORDER BY salary asc,employee_id asc;

Conditions of the query
Conditions of the query ：
grammar ：SELECT Name FROM Table name WHERE Conditions
Equivalent judgment （=）
Query salary is 11000 Employee information （ Number 、 name 、 Salary ）

select employee_id,first_name,salary from t_employees where salary=11000;

Logical conditions （and,or,not）
Query salary is 11000 And the Commission is 0.30 Employee information （ Number 、 name 、 Salary ）

select employee_id,first_name,salary from t_employees where salary=11000 and commission_pct=0.3;

Unequal value judgment （> 、< 、>= 、<= 、!= 、<>）
Query employee's salary in 6000~10000 Employee information between （ Number , name , Salary ）

select employee_id,first_name,salary from t_employees where salary>=6000 and salary<=10000;

Interval judgment （between and）
Query employee's salary in 6000~10000 Employee information between （ Number , name , Salary ）

select employee_id,first_name,salary from t_employees where salary BETWEEN 6000 and 10000;

null Value judgement
Search for employee information without commission （ Number , name , Salary , Royalty ）

select employee_id,first_name,salary,commission_pct from t_employees where commission_pct is NULL;

where in Enumerate queries
Inquiry Department No 70、80、90 Employee information （ Number , name , Salary , Department number ）

select employee_id,first_name,salary,department_id from t_employees where department_id in (70,80,90);

Fuzzy query ： like % _
Look up the name with "L" Employee information at the beginning （ Number , name , Salary , Department number ）

select employee_id,first_name,salary,department_id from t_employees where first_name like 'L%'

Look up the name with "L" Starts with a length of 4 Employee information （ Number , name , Salary , Department number ）

select employee_id,first_name,salary,department_id from t_employees where first_name like 'L___';

Branch structure query
Query employee information number , name , Salary , Salary level The corresponding conditional expression is generated

select employee_id,first_name,salary,
 (case
    WHEN salary>=10000 THEN 'A'
    WHEN salary>=8000 and salary<10000 THEN 'B'
    when salary>=6000 and salary<8000  then 'C'
    WHEN salary>=4000 and salary<6000 THEN  'D'
    else 'E' 
 END) as ' Salary level '
from t_employees;

System function query

The query time ： grammar ：SELECT Time function

select SYSDATE();     Query the current system date and time 
SELECT NOW();         Query the current date and time 
SELECT CURDATE();     Query date 
select CURTIME();     Query time 

select YEAR(NOW());     Query year 
SELECT DATEDIFF('2021-08-09','2021-08-01');    The difference between two time days 

SELECT ADDDATE('2021-08-09',10);    At a given time , Plus days 

String query ：

select CONCAT(CURDATE(),' ',CURTIME());    String splicing 
SELECT INSERT('helloworld',2,3,"xx");  hxxoworld   Add a substring to the specified position of the original string    
                                        The subscript of the database often starts from 1 Start 

polymerization & grouping & Filter & Restrict queries

Aggregate functions
grammar ：SELECT Aggregate functions ( Name ) FROM Table name ;
Count the total monthly salary of all employees

select sum(salary) from t_employees;

Find the total number –COUNT( Field )： barring null Number of values

SELECT count(commission_pct) from t_employees;

Find the total number , contain null The record of value （ Commonly used ）

select count(*) from t_employees;

Group query ：
grammar ：SELECT Name FROM Table name WHERE Conditions GROUP BY Group by （ Column ）
Case study ： Query the total number of people in each department
Ideas ：
1. Group by department number （ The grouping is based on department_id）
2. Then make statistics on the number of people in each department （count）

select department_id,count(*) from t_employees GROUP BY department_id;

Case study ： Query each department 、 Number of people in each position
Ideas ：
1. Group by department number （ Group by department_id）.
2. Group by position name （ Group by job_id）.
3. Count the number of people for each position in each department （count）.

select department_id,job_id,count(*) from t_employees GROUP BY department_id,job_id;

Query each department id、 The total number of 、first_name ---- Wrong query
Be careful ： The field matching the grouping query must be , Aggregate function or group by column

select department_id,count(*),first_name from t_employees GROUP BY department_id;

Group filter query
grammar ：SELECT Name FROM Table name WHERE Conditions GROUP BY Group columns HAVING Filtering rules
Case study ： Statistics 60、70、90 The highest salary in department No
Ideas ：
1). Determine the grouping basis （department_id）
2). For grouped data , The filtered department number is 60、70、90 Information
3). max() Function processing

select department_id,max(salary) from t_employees GROUP BY department_id HAVING department_id in (60,70,90);

Limited query - Limit the number of queries
SELECT Name FROM Table name [LIMIT Start line , Number of query lines ]
Case study ： Query all the information of the top five employees in the table

select * from t_employees limit 0,5;    front 5 strip    Subscript from 0 Start 
select * from t_employees limit 5,5;    the second 5 strip