select name as '姓名', age as '年龄’ from students;
select s.name from students as s;
select distinct gender from s;
条件语句查询 where
where age <=18;
age > 18 or height >=180;
wehre age > 18 and gender=2;
不在范围
wehre not (age>18 and gender=2);
wehre name like " %小%"；
找到有2个名字的
where name like "__";
where age in (18,34);
where age between 18 and 34;
where age is null ;
order by 关键字：
where (age between 18 and 29) and gender=1 order by age asc; desc
多个排序字段
聚合函数：
select count(height) from s where gender=2; count(*)所有数据
select max(age) from s ;
sum（）
select round(avg(age),2) from s;
group by 分组
所以用group_concat(name)
select gender,avg(age) from s group by gender;
having avg(age) > 30;
写在最后
两张表from students as s inner join classes as c on s.cls_id = c.id;