MySQL中having关键字详解以及与where的区别

create table if not exists student
(
	id int null,
	name varchar(50) null,
	age int null,
	sex varchar(2) null,
	score double null
)
comment '学生表';

INSERT INTO student (id, name, age, sex, score) VALUES (1, '张三', 18, '男', 70);
INSERT INTO student (id, name, age, sex, score) VALUES (2, '李四', 17, '男', 60);
INSERT INTO student (id, name, age, sex, score) VALUES (3, '王五', 19, '男', 80);
INSERT INTO student (id, name, age, sex, score) VALUES (4, '赵六', 16, '男', 90);
INSERT INTO student (id, name, age, sex, score) VALUES (5, '七七', 16, '女', 95);
INSERT INTO student (id, name, age, sex, score) VALUES (6, '九九', 17, '女', 85);
INSERT INTO student (id, name, age, sex, score) VALUES (7, '十一', 18, '女', 80);
INSERT INTO student (id, name, age, sex, score) VALUES (8, '小明', 19, '男', 90);
INSERT INTO student (id, name, age, sex, score) VALUES (9, '小军', 17, '男', 55);
INSERT INTO student (id, name, age, sex, score) VALUES (10, '小雷', 19, '女', 60);

select * from student having score >= 80

select * from student where score >= 80

select 
    id
    ,name
    ,age 
from student 
having score >= 80 -- 报错，score筛选条件没有出现在select中

select
       id
     ,name
     ,age
     ,score as fenshu
from student
where fenshu >= 80 -- 报错，where子句中不能使用字段别名

select age,avg(score) from student group by age

select
       age
     ,avg(score)
from student
group by age
having avg(score) > 80
-- 结果为16岁

select
    name
    ,sex
    ,age
    ,score
from student
where sex = '男'
group by name,sex,age,score
having avg(score) > 80