Grundlegende Fragen zu Java-Hochfrequenzinterviews——(9)

 student(sno,sname,sage,ssex)学生表        
 course(cno,cname,tno) 课程表        
 sc(sno,cno,score) 成绩表
 teacher(tno,tname) 教师表

select a.sno from
(select sno,score from sc where cno=1) a,
(select sno,score from sc where cno=2) b
where a.score>b.score and a.sno=b.sno

select a.sno as "学号", avg(a.score) as "平均成绩" 
from
(select sno,score from sc) a 
group by sno having avg(a.score)>60

select a.sno as 学号, b.sname as 姓名,
count(a.cno) as 选课数, sum(a.score) as 总成绩
from sc a, student b
where a.sno = b.sno
group by a.sno, b.sname

selectstudent.sno as 学号, student.sname as 姓名,
 count(sc.cno) as 选课数, sum(score) as 总成绩
from student left Outer join sc on student.sno = sc.sno
group by student.sno, sname

selectcount(distinct(tname)) from teacher where tname like '张%‘

select tname as "姓名", count(distinct(tname)) as "人数" 
from teacher 
where tname like'张%'
group by tname

select student.sno,student.sname from student
where sno not in (select distinct(sc.sno) from sc,course,teacher
where sc.cno=course.cno and teacher.tno=course.tno and teacher.tname='张三')

select sno, sname from student
where sno in (select sno from sc where sc.cno = 1)
and sno in (select sno from sc where sc.cno = 2)

selectc.sno, c.sname from
(select sno from sc where sc.cno = 1) a,
(select sno from sc where sc.cno = 2) b,
student c
where a.sno = b.sno and a.sno = c.sno

select student.sno,student.sname from student,sc where student.sno=sc.sno and sc.cno=1
and exists( select * from sc as sc_2 where sc_2.sno=sc.sno and sc_2.cno=2)

select a.sno, a.sname from student a, sc b
where a.sno = b.sno and b.cno in
(select c.cno from course c, teacher d where c.tno = d.tno and d.tname = '李四')

select a.sno, a.sname from student a, sc b,
(select c.cno from course c, teacher d where c.tno = d.tno and d.tname = '李四') e
where a.sno = b.sno and b.cno = e.cno