MySQL单表查询

单表语法：
    select
         distinct
         字段1,字段2,...
    from 表名
    where 约束条件
    group by 分组条件
    having 过滤条件
    order by 排序字段
    limit n;

#简单查询 
     select * from emp;
     select id,name from emp;
# 去除重复
     select distinct post from emp;
# 四则运算
     select name,salary*12 as annual_salary from emp;
     select name,salary*12 annual_salary from emp;
# 定义显示格式
   concat() 函数用于连接字符串
     select concat(‘姓名: ‘,name,‘  年薪: ‘, salary*12)  as annual_salary from employee;
     concat_ws() 第一个参数为分隔符
     select concat_ws(‘:‘,name,salary*12)  as annual_salary from employee;  

结合CASE语句：
      select
          (
            case
            when name = ‘egon‘ then
                name
            when name = ‘alex‘ then
                concat(name,‘_BIGSB‘)
            else
                concat(name,‘_SB‘)
            end
        ) as new_name
        from emp;

#1:单条件查询
    SELECT name FROM employee
        WHERE post=‘sale‘;
        
#2:多条件查询
    SELECT name,salary FROM employee
        WHERE post=‘teacher‘ AND salary>10000;

#3:关键字BETWEEN AND
    SELECT name,salary FROM employee 
        WHERE salary BETWEEN 10000 AND 20000;

    SELECT name,salary FROM employee 
        WHERE salary NOT BETWEEN 10000 AND 20000;
    
#4:关键字IS NULL(判断某个字段是否为NULL不能用等号，需要用IS)
    SELECT name,post_comment FROM employee 
        WHERE post_comment IS NULL;

    SELECT name,post_comment FROM employee 
        WHERE post_comment IS NOT NULL;
        
    SELECT name,post_comment FROM employee 
        WHERE post_comment=‘‘; 注意‘‘是空字符串，不是null
    ps：
        执行
        update employee set post_comment=‘‘ where id=2;
        再用上条查看，就会有结果了

#5:关键字IN集合查询
    SELECT name,salary FROM employee 
        WHERE salary=3000 OR salary=3500 OR salary=4000 OR salary=9000 ;
    
    SELECT name,salary FROM employee 
        WHERE salary IN (3000,3500,4000,9000) ;

    SELECT name,salary FROM employee 
        WHERE salary NOT IN (3000,3500,4000,9000) ;

#6:关键字LIKE模糊查询
    通配符’%’
    SELECT * FROM employee 
            WHERE name LIKE ‘eg%‘;

    通配符’_’
    SELECT * FROM employee 
            WHERE name LIKE ‘al__‘;

单独使用GROUP BY关键字分组
    SELECT post FROM employee GROUP BY post;
    注意：我们按照post字段分组，那么select查询的字段只能是post，想要获取组内的其他相关信息，需要借助函数

GROUP BY关键字和GROUP_CONCAT()函数一起使用
    SELECT post,GROUP_CONCAT(name) FROM employee GROUP BY post;#按照岗位分组，并查看组内成员名
    SELECT post,GROUP_CONCAT(name) as emp_members FROM employee GROUP BY post;

GROUP BY与聚合函数一起使用
    select post,count(id) as count from employee group by post;#按照岗位分组，并查看每个组有多少人

#强调：聚合函数聚合的是组的内容，若是没有分组，则默认一组

示例：
    SELECT COUNT(*) FROM employee;
    SELECT COUNT(*) FROM employee WHERE depart_id=1;
    SELECT MAX(salary) FROM employee;
    SELECT MIN(salary) FROM employee;
    SELECT AVG(salary) FROM employee;
    SELECT SUM(salary) FROM employee;
    SELECT SUM(salary) FROM employee WHERE depart_id=3;

#执行优先级从高到低：where > group by > having 
#1. Where 发生在分组group by之前，因而Where中可以有任意字段，但是绝对不能使用聚合函数。

#2. Having发生在分组group by之后，因而Having中可以使用分组的字段，无法直接取到其他字段,可以使用聚合函数

select post,avg(salary) from emp group by post having avg(salary) > 20000;

select *　from emp order by age asc; # 默认升序，从小到大
select *　from emp order by age desc; #从大到小
#按多列排序:先按照age排序，如果年纪相同，则按照id排序
select * from emp order by age asc,id desc;
select post,avg(salary) from emp group by post order by avg(salary);

#默认初始位置为0
SELECT * FROM employee ORDER BY salary DESC LIMIT 3;                   
#从第0开始，即先查询出第一条，然后包含这一条在内往后查5条
SELECT * FROM employee ORDER BY salary DESC LIMIT 0,5; 
#从第5开始，即先查询出第6条，然后包含这一条在内往后查5条
SELECT * FROM employee ORDER BY salary DESC LIMIT 5,5; 

SELECT * FROM employee WHERE name REGEXP ‘^ale‘;
SELECT * FROM employee WHERE name REGEXP ‘on$‘;
SELECT * FROM employee WHERE name REGEXP ‘m{2}‘;

小结：对字符串匹配的方式
WHERE name = ‘egon‘;
WHERE name LIKE ‘yua%‘;
WHERE name REGEXP ‘on$‘;

#查看所有员工中名字是jin开头，n或者g结果的员工信息
select * from employee where name regexp ‘^jin.*[gn]$‘;