面经 - 数据库基础面试题

Posted 2021-10-03 程序员牧码

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“张”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、 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  
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、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

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(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 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