面经 - 数据库基础面试题

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作为一枚Java后端开发者,数据库知识必不可少,对数据库的掌握熟悉度的考察也是对这个人是否有扎 实基本功的考察。特别对于初级开发者,面试可能不会去问框架相关知识,但是绝对不会不去考察数据 库知识,这里收集一些常见类型的SQL语句,无论对于平常开发还是准备面试,都会有助益。 

基础表结构

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

select student.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、查询姓的老师的个数

select count(tno)
from teacher
where tname like '张%'

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 sno
-- 子查询结果如果要当临时表来使用的话需要起个别名(比如这里面的t)
from (
	-- 学过该老师课程的同学学号(包含只学一门)
	select sno,count(cno) num
	from sc
	where cno in (
		-- 该老师教了哪些课程
		select cno
		from course
		where tno in (
			select tno 
			from teacher
			where tname = '李四'
		)
	) group by sno
) t
where t.num = (
	-- 统计该老师总课程数
	select count(cno)
	from course
	where tno = (
		select tno
		from teacher
		where tname = '李四'
	)
)

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、 “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  
where sc_2.cno = sc.cno)
where cno in 
(select c.cno from course c 
left join teacher t on t.tno = c.tno
where t.tname = '王五');

12、查询和编号为1002的同学学习的课程完全相同的其他同学学号和姓名

select sc_1.sno
from (select cno from sc where sno='1002')a
left join sc sc_1 on a.cno = sc_1.cno
where sc_1.sno<>'1002' 
group by sc_1.sno 
having count(sc_1.cno) = 
(select count(cno) from sc where sno='1002');

select a.sno,s.sname from 
(select sno,GROUP_CONCAT(cno order by cno separator ',') as cid_str 
from sc where sno='1002')b,
(select sno,GROUP_CONCAT(cno order by cno separator ',') as cid_str 
from sc group by sno)a
left join student s 
on a.sno = s.sno
where a.cid_str = b.cid_str and a.sno<>'1002';

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、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

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,
c.cname as 课程名,
COALESCE(avg(score),0) as 平均成绩,
100*sum(case 
when COALESCE(score,0)>=60 
then 1 else 0 END)/count(*) as 及格百分数
from sc t
left join course c 
on t.cno = c.cno
group by t.cno
order by 100*sum(case 
when COALESCE(score,0)>=60 
then 1 else 0 END)/count(*);

18、查询如下课程平均成绩和及格率的百分数("1"显示):企业管理(001),马克思(002),UML003),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 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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