mysql实战

这是老男孩的练习题，做一做增加熟练度。

数据库文件

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-- Dump completed on 2019-05-15 20:47:52

　　1.查询学过“叶平”老师所教的所有课的同学的学号、姓名；

select student.sname,B.student_id from (select student_id from score where course_id in (select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.tname="李平老师") group by student_id having count(student_id)=(select count(cid) from course left join teacher on course.teacher_id = teacher.tid where teacher.tname="李平老师"))as B left join student on student.sid=B.student_id;

2.查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；

select student_id,sname from
 
    (select student_id,course_id from score where course_id = 1 or course_id = 2) as B
      
    left join student on B.student_id = student.sid group by student_id HAVING count(student_id) > 1

3.查询没有学全所有课的同学的学号、姓名；

 select student.sname,B.student_id from(select student_id,count(sid) from score group by student_id having count(sid)< (select count(cid) from course)) as B left join student on student.sid=B.student_id;

4.查询有课程成绩小于60分的同学的学号、姓名；

查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名；

查询至少学过学号为“001”同学所有课的其他同学学号和姓名；

查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名；

select student_id from score where student_id in (select student_id from score where student_id !=1 group by student_id having count(course_id)=(select count(course_id) from score where student_id =1)) and course_id in (select course_id from score where student_id =1) group by student_id having count(course_id)=(select count(course_id) from score where student_id =1);