[ Java面试题 ]数据库篇
基本表结构:
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
1、查询课程1的成绩比课程2的成绩高的所有学生的学号
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
2、查询平均成绩大于60分的同学的学号和平均成绩
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
3、查询所有同学的学号、姓名、选课数、总成绩
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
或者:
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
4、查询姓“张”的老师的个数
selectcount(distinct(tname)) from teacher where tname like '张%‘
或者:
select tname as "姓名", count(distinct(tname)) as "人数"
from teacher
where tname like'张%'
group by tname
5、查询没学过“张三”老师课的同学的学号、姓名
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')
6、查询同时学过课程1和课程2的同学的学号、姓名
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)
或者:
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
或者:
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)
7、查询学过“李四”老师所教所有课程的所有同学的学号、姓名
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')
或者:
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
8、查询课程编号1的成绩比课程编号2的成绩高的所有同学的学号、姓名
select a.sno, a.sname from student a,
(select sno, score from sc where cno = 1) b,
(select sno, score from sc where cno = 2) c
where b.score > c.score and b.sno = c.sno and a.sno = b.sno
9、查询所有课程成绩小于60分的同学的学号、姓名
select sno,sname from student
where sno not in (select distinct sno from sc where score > 60)
10、查询至少有一门课程与学号为1的同学所学课程相同的同学的学号和姓名
select distinct a.sno, a.sname
from student a, sc b
where a.sno <> 1 and a.sno=b.sno and
b.cno in (select cno from sc where sno = 1)
或者:
select s.sno,s.sname
from student s,
(select sc.sno
from sc
where sc.cno in (select sc1.cno from sc sc1 where sc1.sno=1)and sc.sno<>1
group by sc.sno)r1
where r1.sno=s.sno
11、把“sc”表中“王五”所教课的成绩都更改为此课程的平均成绩
update sc set score = (select avg(sc_2.score) from sc sc_2 wheresc_2.cno=sc.cno)
from course,teacher where course.cno=sc.cno and course.tno=teacher.tno andteacher.tname='王五'
12、查询和编号为2的同学学习的课程完全相同的其他同学学号和姓名
这一题分两步查:
第一步
select sno
from sc
where sno <> 2
group by sno
having sum(cno) = (select sum(cno) from sc where sno = 2)
第二步
select b.sno, b.sname
from sc a, student b
where b.sno <> 2 and a.sno = b.sno
group by b.sno, b.sname
having sum(cno) = (select sum(cno) from sc where sno = 2)
13、删除学习“王五”老师课的sc表记录
delete sc from course, teacher
where course.cno = sc.cno and course.tno = teacher.tno and tname = '王五'
14、向sc表中插入一些记录,这些记录要求符合以下条件:
将没有课程3成绩同学的该成绩补齐, 其成绩取所有学生的课程2的平均成绩
insert sc select sno, 3, (select avg(score) from sc where cno = 2)
from student
where sno not in (select sno from sc where cno = 3)
15、按平平均分从高到低显示所有学生的如下统计报表:
-- 学号,企业管理,马克思,UML,数据库,物理,课程数,平均分
select sno as 学号
,max(case when cno = 1 then score end) AS 企业管理
,max(case when cno = 2 then score end) AS 马克思
,max(case when cno = 3 then score end) AS UML
,max(case when cno = 4 then score end) AS 数据库
,max(case when cno = 5 then score end) AS 物理
,count(cno) AS 课程数
,avg(score) AS 平均分
FROM sc
GROUP by sno
ORDER by avg(score) DESC
16、查询各科成绩最高分和最低分:
以如下形式显示:课程号,最高分,最低分
select cno as 课程号, max(score) as 最高分, min(score) 最低分
from sc group by cno
或
select course.cno as '课程号'
,MAX(score) as '最高分'
,MIN(score) as '最低分'
from sc,course
where sc.cno=course.cno
group by course.cno
17、按各科平均成绩从低到高和及格率的百分数从高到低顺序
SELECT t.cno AS 课程号,
max(course.cname)AS 课程名,
isnull(AVG(score),0) AS 平均成绩,
100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/count(1) AS 及格率
FROM sc t, course
where t.cno = course.cno
GROUP BY t.cno
ORDER BY 及格率 desc
18、查询如下课程平均成绩和及格率的百分数(用"1行"显示):
企业管理(001),马克思(002),UML (003),数据库(004)
select
avg(case when cno = 1 then score end) as 平均分1,
avg(case when cno = 2 then score end) as 平均分2,
avg(case when cno = 3 then score end) as 平均分3,
avg(case when cno = 4 then score end) as 平均分4,
100 * sum(case when cno = 1 and score > 60 then 1 else 0 end) / sum(casewhen cno = 1 then 1 else 0 end) as 及格率1,
100 * sum(case when cno = 2 and score > 60 then 1 else 0 end) / sum(casewhen cno = 2 then 1 else 0 end) as 及格率2,
100 * sum(case when cno = 3 and score > 60 then 1 else 0 end) / sum(casewhen cno = 3 then 1 else 0 end) as 及格率3,
100 * sum(case when cno = 4 and score > 60 then 1 else 0 end) / sum(casewhen cno = 4 then 1 else 0 end) as 及格率4
from sc
19、查询不同老师所教不同课程平均分, 从高到低显示
select max(c.tname) as 教师, max(b.cname) 课程, avg(a.score) 平均分
from sc a, course b, teacher c
where a.cno = b.cno and b.tno = c.tno
group by a.cno
order by 平均分 desc
或者:
select r.tname as '教师',r.rname as '课程' , AVG(score) as '平均分'
from sc,
(select t.tname,c.cno as rcso,c.cname as rname
from teacher t ,course c
where t.tno=c.tno)r
where sc.cno=r.rcso
group by sc.cno,r.tname,r.rname
order by AVG(score) desc
20、查询如下课程成绩均在第3名到第6名之间的学生的成绩:
-- [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩
select top 6 max(a.sno) 学号, max(b.sname) 姓名,
max(case when cno = 1 then score end) as 企业管理,
max(case when cno = 2 then score end) as 马克思,
max(case when cno = 3 then score end) as UML,
max(case when cno = 4 then score end) as 数据库,
avg(score) as 平均分
from sc a, student b
where a.sno not in (select top 2 sno from sc where cno = 1 order by score desc)
and a.sno not in (select top 2 sno from sc where cno = 2 order by scoredesc)
and a.sno not in (select top 2 sno from sc where cno = 3 order by scoredesc)
and a.sno not in (select top 2 sno from sc where cno = 4 order by scoredesc)
and a.sno = b.sno
group by a.sno
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