我的表是:
mysql> select * from professor;
+-------+--------+--------+--------+------+
| empid | name | status | salary | age |
+-------+--------+--------+--------+------+
| 1 | Arun | 1 | 2000 | 23 |
| 2 | Benoy | 0 | 3000 | 25 |
| 3 | Chacko | 1 | 1000 | 36 |
| 4 | Divin | 0 | 5000 | 32 |
| 5 | Edwin | 1 | 2500 | 55 |
| 7 | George | 0 | 1500 | 46 |
+-------+--------+--------+--------+------+
6 rows in set (0.00 sec)
mysql> select * from works;
+----------+-------+---------+
| courseid | empid | classid |
+----------+-------+---------+
| 1 | 1 | 10 |
| 2 | 2 | 9 |
| 3 | 3 | 8 |
| 4 | 4 | 10 |
| 5 | 5 | 9 |
| 6 | 1 | 9 |
| 2 | 3 | 10 |
| 2 | 1 | 7 |
| 4 | 2 | 6 |
| 2 | 4 | 6 |
| 2 | 5 | 2 |
| 7 | 5 | 6 |
| 3 | 5 | 2 |
| 6 | 4 | 10 |
+----------+-------+---------+
14 rows in set (0.00 sec)
mysql> select * from course;
+----------+------------+--------+
| courseid | coursename | points |
+----------+------------+--------+
| 1 | Maths | 5 |
| 2 | Science | 1 |
| 3 | English | 6 |
| 4 | Social | 4 |
| 5 | Malayalam | 20 |
| 6 | Arts | 25 |
| 7 | Biology | 20 |
+----------+------------+--------+
7 rows in set (0.00 sec)
问题是:
列出那些没有教过三门课程的教授 班级7,8,9
我的查询是:
select professor.name from professor
inner join works
on professor.empid=works.empid
group by works.empid
having works.classid not in (7,8,9);
我得到的输出是:
Arun
Divin
我实际需要获得的输出:
Divin住宿
请帮助我。
答案 0 :(得分:1)
添加
group by works.empid, works.classid
答案 1 :(得分:0)
select professor.name from professor
inner join works
on professor.empid=works.empid
inner join course on work.courseid=course.courseid
group by works.empid
having works.classid not in (7,8,9);
答案 2 :(得分:0)
试试这个
select name from professor where name not in (
select name from professor
inner join works
on professor.empid=works.empid
where works.classid in (7,8,9)
group by professor.name );
你会得到
divin
george
乔治也没有7,8,9