SQL查询有两门以上不及格的学生及查询出全部及格的学生

1、表结构：

/*学生*/
create table student(
sno int not null primary key,
sname varchar(10)
);
/*课程*/
create table center(
cno int not null primary key,
cname varchar(10)
);
/*分数*/
create table sgrade(
sno int ,
cno int ,
sgrade int
);

2、插入数据：

insert into student values(11,‘a‘);
insert into student values(12,‘b‘);
insert into student values(13,‘c‘)

insert into center values(21,‘aa‘);
insert into center values(22,‘bb‘);
insert into center values(23,‘cc‘)

insert into sgrade values(11,21,54);
insert into sgrade values(12,22,57);
insert into sgrade values(11,21,51);
insert into sgrade values(12,21,36);
insert into sgrade values(11,21,28);
insert into sgrade values(12,22,42);
insert into sgrade values(11,21,59)
insert into sgrade values(11,21,79);
insert into sgrade values(12,22,85);
insert into sgrade values(11,21,90);
insert into sgrade values(12,21,96);
insert into sgrade values(11,21,98);
insert into sgrade values(12,22,94);
insert into sgrade values(11,21,99)

3、查询出有两门以上不及格的学生：

SELECT s.sno,s.sname
    from student s
    LEFT JOIN sgrade g on s.sno=g.sno
WHERE g.sgrade<60
GROUP BY s.sno HAVING COUNT(s.sno)>2

4、查询出全部及格的学生:

SELECT * FROM student WHERE sno not in (
    SELECT s.sno
        from student s
        LEFT JOIN sgrade g on s.sno=g.sno
    WHERE g.sgrade<60
)

原文地址：https://www.cnblogs.com/java-zzl/p/9739976.html