MySQL必会的50个常见面试练习题

下面的SQL题目都是比较基础,比较常见的数据库SQL面试题,在技术面试环节虽然碰到相同题目的机会比较少,但解题的基本思路都是差
不多的。下面是SQL面试题描述:

 

Student(Sid,Sname,Sage,Ssex) 学生表 Sid:学号;Sname:学生姓名;Sage:学生年龄;Ssex:学生性别
Course(Cid,Cname,Tid) 课程表 Cid,课程编号;Cname:课程名字;Tid:教师编号
SC(Sid,Cid,score) 成绩表 Sid:学号;Cid,课程编号;score:成绩
Teacher(Tid,Tname) 教师表 Tid:教师编号; Tname:教师名字

 

下面是数据库表:

/*==============================================================*/

/* DBMS name:      MySQL 5.0                                    */

/* Created on:     2018/6/29 16:05:44                           */

/*==============================================================*/

 

 

drop table if exists Course;

 

drop table if exists SC;

 

drop table if exists Student;

 

drop table if exists Teacher;

 

/*==============================================================*/

/* Table: Course                                                */

/*==============================================================*/

create table Course

(

   Cid                   bigint not null comment '课程编号',

   Cname                varchar(32) comment '课程名字',

   Tid                   bigint comment '教师编号',

   primary key (Cid)

)

ENGINE = InnoDB

DEFAULT CHARACTER SET = utf8;

 

/*==============================================================*/

/* Table: SC                                                    */

/*==============================================================*/

create table SC

(

   Sid                   bigint not null comment '学号',

   Cid                   bigint comment '课程编号',

   score                int comment '成绩',

   primary key (Sid)

)

ENGINE = InnoDB

DEFAULT CHARACTER SET = utf8;

 

/*==============================================================*/

/* Table: Student                                               */

/*==============================================================*/

create table Student

(

   Sid                   bigint not null comment '学号',

   Sname                varchar(32) comment '学生姓名',

   Sage                 int comment '学生年龄',

   Ssex                 varchar(10) comment '学生性别',

   primary key (Sid)

)

ENGINE = InnoDB

DEFAULT CHARACTER SET = utf8;

 

/*==============================================================*/

/* Table: Teacher                                               */

/*==============================================================*/

create table Teacher

(

   Tid                   bigint not null comment '教师编号',

   Tname                varchar(32) comment '教师名字',

   primary key (Tid)

)

ENGINE = InnoDB

DEFAULT CHARACTER SET = utf8;

 

 


问题及答案:
1、查询1课程比2课程成绩高的所有学生的学号;
select a.Sid from (select sid,score from SC where Cid=1) a,(select sid,score
from SC where Cid=2) b
where a.score>b.score and a.sid=b.sid;


2、查询平均成绩大于60分的同学的学号和平均成绩;
select Sid,avg(score)
from sc
group by Sid having avg(score) >60;


3、查询所有同学的学号、姓名、选课数、总成绩;
select Student.Sid,Student.Sname,count(SC.Cid),sum(score)
from Student left Outer join SC on Student.Sid=SC.Sid
group by Student.Sid,Sname


4、查询姓“李”的老师的个数;
select count(distinct(Tname))
from Teacher
where Tname like ‘李%';


5、查询没学过“叶平”老师课的同学的学号、姓名;
select Student.Sid,Student.Sname
from Student
where Sid not in (select distinct( SC.Sid) from SC,Course,Teacher where SC.Cid=Course.Cid and
Teacher.Tid=Course.Tid and Teacher.Tname=’叶平’);


6、查询学过1并且也学过编号2课程的同学的学号、姓名;
select Student.Sid,Student.Sname from Student,SC where Student.Sid=SC.Sid and SC.Cid=1and exists( Select * from SC as
SC_2 where SC_2.Sid=SC.Sid and SC_2.Cid=2);


7、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
select Sid,Sname
from Student
where Sid in (select Sid from SC ,Course ,Teacher where SC.Cid=Course.Cid and Teacher.Tid=Course.Tid and
Teacher.Tname=’叶平’ group by Sid having count(SC.Cid)=(select count(Cid) from Course,Teacher where
Teacher.Tid=Course.Tid and Tname=’叶平’));


8、查询课程编号2的成绩比课程编号1课程低的所有同学的学号、姓名;
Select Sid,Sname from (select Student.Sid,Student.Sname,score ,(select score from SC SC_2 where SC_2.Sid=Student.Sid
and SC_2.Cid=2) score2
from Student,SC where Student.Sid=SC.Sid and Cid=1) S_2 where score2 <score;


9、查询所有课程成绩小于60分的同学的学号、姓名;
select Sid,Sname
from Student
where Sid not in (select Student.Sid from Student,SC where S.Sid=SC.Sid and score>60);


10、查询没有学全所有课的同学的学号、姓名;
select Student.Sid,Student.Sname
from Student,SC
where Student.Sid=SC.Sid group by Student.Sid,Student.Sname having count(Cid) <(select count(Cid) from Course);

11、查询至少有一门课与学号为“11”的同学所学相同的同学的学号和姓名;
select Sid,Sname from Student,SC where Student.Sid=SC.Sid and Cid in select Cid from SC where Sid=’11′;


12、查询至少学过学号为1同学所有一门课的其他同学学号和姓名;
select distinct SC.Sid,Sname
from Student,SC
where Student.Sid=SC.Sid and Cid in (select Cid from SC where Sid=1);


13、把“SC”表中“叶平”老师教的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.Cid=SC.Cid ) from Course,Teacher where Course.Cid=SC.Cid and Course.Tid=Teacher.Tid and Teacher.Tname=’
叶平’);


14、查询和“12”号的同学学习的课程完全相同的其他同学学号和姓名;
select Sid from SC where Cid in (select Cid from SC where Sid=’12′)
group by Sid having count(*)=(select count(*) from SC where Sid=’12′);


15、删除学习“叶平”老师课的SC表记录;
Delect SC
from course ,Teacher
where Course.Cid=SC.Cid and Course.Tid= Teacher.Tid and Tname=’叶平';


16、向SC表中插入一些记录,这些记录要求符合以下条件:没有上过编号“3”课程的同学学号、2、
号课的平均成绩;
Insert SC select Sid,2,(Select avg(score)
from SC where Cid=2) from Student where Sid not in (Select Sid from SC where Cid=2);


17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩,按如下形式显示: 学生ID,,数据库,企
业管理,英语,有效课程数,有效平均分
SELECT Sid as 学生ID
,(SELECT score FROM SC WHERE SC.Sid=t.Sid AND Cid=’4′) AS 数据库
,(SELECT score FROM SC WHERE SC.Sid=t.Sid AND Cid=1) AS 企业管理
,(SELECT score FROM SC WHERE SC.Sid=t.Sid AND Cid=’006′) AS 英语
,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩
FROM SC AS t
GROUP BY Sid
ORDER BY avg(t.score)


18、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分
SELECT L.Cid As 课程ID,L.score AS 最高分,R.score AS 最低分
FROM SC L ,SC AS R
WHERE L.Cid = R.Cid and
L.score = (SELECT MAX(IL.score)
FROM SC AS IL,Student AS IM
WHERE L.Cid = IL.Cid and IM.Sid=IL.Sid
GROUP BY IL.Cid)
AND
R.Score = (SELECT MIN(IR.score)
FROM SC AS IR
WHERE R.Cid = IR.Cid
GROUP BY IR.Cid
);
19、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.Cid 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(*) AS 及格百分数
FROM SC T,Course
where t.Cid=course.Cid
GROUP BY t.Cid
ORDER BY 100 * SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC


20、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(1),马克思(2),OO&UML (3),数据库(4)
SELECT SUM(CASE WHEN Cid =1 THEN score ELSE 0 END)/SUM(CASE Cid WHEN ‘1’ THEN 1 ELSE 0 END) AS 企业管理平均分
,100 * SUM(CASE WHEN Cid = ‘1’ AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cid = ‘1’ THEN 1 ELSE 0 END)
AS 企业管理及格百分数
,SUM(CASE WHEN Cid = ‘2’ THEN score ELSE 0 END)/SUM(CASE Cid WHEN ‘2’ THEN 1 ELSE 0 END) AS 马克思平均分
,100 * SUM(CASE WHEN Cid = ‘2’ AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cid = ‘2’ THEN 1 ELSE 0 END)
AS 马克思及格百分数
,SUM(CASE WHEN Cid = ‘3’ THEN score ELSE 0 END)/SUM(CASE Cid WHEN ‘3’ THEN 1 ELSE 0 END) AS UML平均分
,100 * SUM(CASE WHEN Cid = ‘3’ AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cid = ‘3’ THEN 1 ELSE 0 END)
AS UML及格百分数
,SUM(CASE WHEN Cid = ‘4’ THEN score ELSE 0 END)/SUM(CASE Cid WHEN ‘4’ THEN 1 ELSE 0 END) AS 数据库平均分
,100 * SUM(CASE WHEN Cid = ‘4’ AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cid = ‘4’ THEN 1 ELSE 0 END)
AS 数据库及格百分数
FROM SC


21、查询不同老师所教不同课程平均分从高到低显示
SELECT max(Z.Tid) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.Cid AS 课程ID,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩
FROM SC AS T,Course AS C ,Teacher AS Z
where T.Cid=C.Cid and C.Tid=Z.Tid
GROUP BY C.Cid
ORDER BY AVG(Score) DESC


22、查询如下课程成绩第 3 名到第 6 名的学生成绩单:企业管理(1),马克思(2),UML (3),数据库(4)
[学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩
SELECT DISTINCT top 3
SC.Sid As 学生学号,
Student.Sname AS 学生姓名 ,
T1.score AS 企业管理,
T2.score AS 马克思,
T3.score AS UML,
T4.score AS 数据库,
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分
FROM Student,SC LEFT JOIN SC AS T1
ON SC.Sid = T1.Sid AND T1.Cid = ‘1’
LEFT JOIN SC AS T2
ON SC.Sid = T2.Sid AND T2.Cid = ‘2’
LEFT JOIN SC AS T3
ON SC.Sid = T3.Sid AND T3.Cid = ‘3’
LEFT JOIN SC AS T4
ON SC.Sid = T4.Sid AND T4.Cid = ‘4’
WHERE student.Sid=SC.Sid and
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
NOT IN
(SELECT
DISTINCT
TOP 15 WITH TIES
ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)
FROM sc
LEFT JOIN sc AS T1
ON sc.Sid = T1.Sid AND T1.Cid = ‘k1′
LEFT JOIN sc AS T2
ON sc.Sid = T2.Sid AND T2.Cid = ‘k2′
LEFT JOIN sc AS T3
ON sc.Sid = T3.Sid AND T3.Cid = ‘k3′
LEFT JOIN sc AS T4
ON sc.Sid = T4.Sid AND T4.Cid = ‘k4′
ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC);

23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
SELECT SC.Cid as 课程ID, Cname as 课程名称
,SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 – 85]
,SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 – 70]
,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 – 60]
,SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -]
FROM SC,Course
where SC.Cid=Course.Cid
GROUP BY SC.Cid,Cname;

24、查询学生平均成绩及其名次
SELECT 1+(SELECT COUNT( distinct 平均成绩)
FROM (SELECT Sid,AVG(score) AS 平均成绩
FROM SC
GROUP BY Sid
) AS T1
WHERE 平均成绩 > T2.平均成绩) as 名次,
Sid as 学生学号,平均成绩
FROM (SELECT Sid,AVG(score) 平均成绩
FROM SC
GROUP BY Sid
) AS T2
ORDER BY 平均成绩 desc;

25、查询各科成绩前三名的记录:(不考虑成绩并列情况)
SELECT t1.Sid as 学生ID,t1.Cid as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 3 score
FROM SC
WHERE t1.Cid= Cid
ORDER BY score DESC
)
ORDER BY t1.Cid;
26、查询每门课程被选修的学生数
select cid,count(Sid) from sc group by Cid;


27、查询出只选修了一门课程的全部学生的学号和姓名
select SC.Sid,Student.Sname,count(Cid) AS 选课数
from SC ,Student
where SC.Sid=Student.Sid group by SC.Sid ,Student.Sname having count(Cid)=1;


28、查询男生、女生人数
Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex=’男';
Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex=’女’;


29、查询姓“张”的学生名单
SELECT Sname FROM Student WHERE Sname like ‘张%';


30、查询同名同性学生名单,并统计同名人数
select Sname,count(*) from Student group by Sname having count(*)>1;;


31、1981年出生的学生名单(注:Student表中Sage列的类型是datetime)
select Sname, CONVERT(char (11),DATEPART(year,Sage)) as age
from student
where CONVERT(char(11),DATEPART(year,Sage))=’1981′;


32、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列
Select Cid,Avg(score) from SC group by Cid order by Avg(score),Cid DESC ;


33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩
select Sname,SC.Sid ,avg(score)
from Student,SC
where Student.Sid=SC.Sid group by SC.Sid,Sname having avg(score)>85;


34、查询课程名称为“数据库”,且分数低于60的学生姓名和分数
Select Sname,isnull(score,0)
from Student,SC,Course
where SC.Sid=Student.Sid and SC.Cid=Course.Cid and Course.Cname=’数据库’and score <60;
35、查询所有学生的选课情况;


SELECT SC.Sid,SC.Cid,Sname,Cname
FROM SC,Student,Course
where SC.Sid=Student.Sid and SC.Cid=Course.Cid ;


36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数;
SELECT distinct student.Sid,student.Sname,SC.Cid,SC.score


FROM student,Sc
WHERE SC.score>=70 AND SC.Sid=student.Sid;


37、查询不及格的课程,并按课程号从大到小排列
select cid from sc where scor e <60 order by Cid ;


38、查询课程编号为3且课程成绩在80分以上的学生的学号和姓名;
select SC.Sid,Student.Sname from SC,Student where SC.Sid=Student.Sid and Score>80 and Cid=’3′;


39、求选了课程的学生人数
select count(*) from sc;


40、查询选修“叶平”老师所授课程的学生中,成绩最高的学生姓名及其成绩
select Student.Sname,score
from Student,SC,Course C,Teacher
where Student.Sid=SC.Sid and SC.Cid=C.Cid and C.Tid=Teacher.Tid and Teacher.Tname=’叶平’ and SC.score=(select max
(score)from SC where Cid=C.Cid );


41、查询各个课程及相应的选修人数
select count(*) from sc group by Cid;


42、查询不同课程成绩相同的学生的学号、课程号、学生成绩
select distinct A.Sid,B.score from SC A ,SC B where A.Score=B.Score and A.Cid <>B.Cid ;


43、查询每门功成绩最好的前两名
SELECT t1.Sid as 学生ID,t1.Cid as 课程ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 2 score
FROM SC
WHERE t1.Cid= Cid
ORDER BY score DESC
)
ORDER BY t1.Cid;


44、统计每门课程的学生选修人数(超过10人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,查询结果按人数
降序排列,若人数相同,按课程号升序排列
select Cid as 课程号,count(*) as 人数
from sc
group by Cid
order by count(*) desc,cid


45、检索至少选修两门课程的学生学号
select Sid
from sc
group by sid
having count(*) > = 2


46、查询全部学生都选修的课程的课程号和课程名
select Cid,Cname
from Course
where Cid in (select cid from sc group by cid)


47、查询没学过“叶平”老师讲授的任一门课程的学生姓名
select Sname from Student where Sid not in (select Sid from Course,Teacher,SC where Course.Tid=Teacher.Tid and
SC.Cid=course.Cid and Tname=’叶平’);


48、查询两门以上不及格课程的同学的学号及其平均成绩
select Sid,avg(isnull(score,0)) from SC where Sid in (select Sid from SC where score <60 group by Sid having count
(*)>2)group by Sid;
49、检索“4”课程分数小于60,按分数降序排列的同学学号
select Sid from SC where Cid=4and score <60 order by score desc;


50、删除2同学的1课程的成绩
delete from Sc where Sid=2and Cid=1;

 

posted @ 2018-07-02 10:21  胡成长  阅读(10900)  评论(2编辑  收藏  举报