1、将如下表中的每门成绩都大于80分的人名？

select * from tb_stu2;
select distinct stu2_name from tb_stu2 where stu2_name not in
(select distinct stu2_name from tb_stu2 where stu2_score<80);

2、输出老爹对应的大儿子的信息？

create table student(
stu_id int,
stu_age int,
stu_name varchar(30),
stu_f int
);
create table father(
c_id int,
c_master varchar(30)
);
insert into student values(1,28,‘王思聪‘,1);
insert into student values(2,20,‘王五‘,1);
insert into student values(3,24,‘李嘉‘,2);
insert into student values(4,25,‘李泽‘,2);

insert into father values(1,‘王健林‘);
insert into father values(2,‘李嘉诚‘);

#先建立父亲对儿子的一对多的关系，然后选出各自的大儿子显示：
select * from student join father on student.stu_f=father.c_id where (student.stu_age=(select max(stu_age) from student where stu_f=1))
or (student.stu_age=(select max(stu_age) from student where stu_f=2)) ;

mySQL 的 2个分类