mysql初学者题目及答案

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假设有下面4张表

student(sid,sname,sage,ssex) 学生表

course(cid,cname,tid) 课程表

grade(sid,cid,score) 成绩表

teacher(tid,Tname) 教师表

 

问题：

1、  查询“001”课程比“002”课程成绩高的所有学生的学号

 #方式1

select a.sid

from grade a,grade b

where a.sid=b.sidand a.cid=2 and b.cid=3 and a.score>b.score;

 #方式2

select a.sid

from (select sid,score from grade where cid=2) a,

(select sid,score from grade where cid=3) b

where a.sid=b.sidand a.score>b.score;

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) as avg

from grade

group by sid

havingavg>60;

 

3、  查询所有同学的学号、姓名、选课数、总成绩；

 #方式1 

select s.sid,s.sname,count(cid),sum(score)

from student s

left join grade g

on s.sid=g.sid

group by s.sid;

 

#方式2

select s.sid,s.sname,a.num,a.totals

from student s

left join (select sid,count(cid) as num,

sum(score) as totals from grade group by sid)as a

on s.sid=a.sid;

 

#方式3 

select s.sid,s.sname,(select count(cid)from grade where sid=s.sid) as num,

(select sum(score) from grade where sid=s.sid)as totals

from student s;

 

4、  查询姓“李”的老师的个数；

select count(tname)

from teacher

where tname like '李%';

 

5、查询没学过“ta”老师课的同学的学号、姓名；

#方式1

select sid,sname

from student

where sid not in

(select sid from grade where cid in

(select cid from course c,teacher t wherec.tid=t.tid and t.tname='ta')

);

 #方式2

select sid,sname

from student

where sid not in

(selectdistinct sid from grade g,course c,teacher t where g.cid=c.cid and c.tid=t.tidand t.tname='ta');

 

6、查询学过“002”并且也学过编号“003”课程的同学的学号、姓名；

 #方式1

select sid,sname

from student

where sid in (

select a.sid

from (select sid from grade where cid=2) a,(select sid from grade where cid=3) b

wherea.sid=b.sid);

 #方式2

select sid,sname

from student

where sid in (select sid from grade wherecid=2)

and

sid in (select sidfrom grade where cid=3);

 

#方式3

select s.sid,sname

from student s,grade g

where s.sid=g.sid and g.cid=2 and

s.sid in (select sid from grade wherecid=3);

 

 #方式4

 

select s.sid, sname

from student s,grade g

where s.sid=g.sid and g.cid=2 and

exists( Select * from grade as g2 where g2.sid=g.sidand g2.cid=3);

 

 

7、查询学过“tb”老师所教的所有课的同学的学号、姓名；

 

#方式1

#利用sum统计每个同学上tb老师课的门数，选出那些门数=tb所教课的门数

 

Select sid,sname

from student

where sid in

(select sid

from grade

group by sid

having sum(case when cid in (select cidfrom Course c,Teacher t where T.tid=C.tid and Tname='tb') then 1 else 0 end)

=(select count(cid) from Course c,Teacher twhere T.tid=C.tid and Tname='tb')

);

 

 #方式2

select sid,sname

from Student

where sid in

(select sid

from GRADE g,Course c,Teacher t

where g.cid=C.cid and c.tid=t.tid andTname='tb'

group by sid

having count(g.cid)=

(select count(cid) from Course c,Teacher twhere T.tid=C.tid and Tname='tb')

);

 

8、查询那些所有课程成绩小于60分的同学的学号、姓名；

 

#方式1

 

select s.sid,sname

from student s,grade g

where s.sid=g.sid and s.sid not in (select sidfrom grade where score>=60);

 

 #方式2

select s.sid, sname

from student s,grade g

where s.sid=g.sid

group by s.sid

havingmax(score)<60;

 

9、查询没有学全所有课的同学的学号、姓名；

 

#方式1

 

select s.sid,sname

from student s

left join grade g

on s.sid=g.sid

group by s.sid

havingcount(cid)<(select count(distinct cname) from course);

 

 #方式2

 

select sid,sname

from student

where sid not in (

select sid

from grade 

group by sid

havingcount(cid)=(select count(distinct cname) from course));

 

 

10、查询至少有一门课与学号为“1”的同学所学相同的同学的学号和姓名；

 

#方式1

 

select distinct s.sid,sname

from student s,grade g

where s.sid=g.sid and s.sid<>1 andcid in (select cid from grade where sid=1);

 

 

#方式2

 

select sid,sname 

from student

where sid in (

select sid

from grade

where sid<>1 and cid in (select cidfrom grade where sid=1)

);

 

 

11、查询至少学过学号为“005”同学所有课的其他同学学号和姓名；

 

#方式1 

Select s.sid,sname

From student s, grade g, (select cid fromgrade where sid=5) as a

Where s.sid=g.sid and g.cid=a.cid ands.sid<>5

Group by s.sid

Having count(g.cid)=

(select count(b.cid) from (select cid fromgrade where sid=5) as b);

 

 

12、把“GRADE”表中“tb”老师教的课的成绩都更改为此课程的平均成绩；

 

#方式1

 

Update grade as g1,

(select g.cid, avg(score) avg from gradeg,course c,teacher t where g.cid=c.cid and c.tid=t.tid and t.tname=’tb’ groupby g.cid) as a

Set g1.score=a.avg

Where g1.cid=a.cid

 

#方式2，虽然巧，但是效率低

 

Update grade as g

set g.score=

(select a.avg

From (select cid,avg(score) as avg  from grade group by cid) as a

Where g.cid=a.cid)

Where  g.cid in (Select cid from course c,teacher twhere c.tid=t.tid and t.tname='tb');

 

13、查询和“1”号的同学学习的课程完全相同的其他同学学号；

 #方式1

#巧妙的数学逻辑

 

select distinct sid

from grade

where sid <>1 and sid not in (

select sid from grade where cid not in (

select cid from grade where sid=1))

group by sid

havingcount(sid)= (select count(cid) from grade where sid=1);

 

14、删除学习“ta”老师课的GRADE表记录；

 #方式1

delete

from grade

where cid in (select cid from coursec,teacher t where c.cid=t.tid and t.tname='ta');

#方式2 

delete g

from grade g,course c,teacher t

whereg.cid=c.cid and c.tid=t.tid and t.tname='ta';

 

15、向GRADE表中插入一些记录（包含以下3个字段）：

没有上过编号“001”课程的同学学号，1号课，1号课的平均成绩；

 

insert into grade

select sid,1,(select avg(score) from gradewhere cid=1)

from student

where sid notin (select sid from grade where cid=1);

 

 

16、按所有课程平均成绩从高到低显示所有学生的“math”、“phy”、“che”三门的课程成绩，按如下形式显示：

 学生ID,,math,phy,che,有效课程数,有效平均分

 

 

Select s.sid,

(select score from grade,course wheregrade.sid=s.sid and grade.cid=course.cid and course.cname='math') as 'math',

(select score from grade,course wheregrade.sid=s.sid and grade.cid=course.cid and course.cname='phy') as 'phy',

(select score from grade,course wheregrade.sid=s.sid and grade.cid=course.cid and course.cname='che') as 'che',

(select count(cid) from grade where sid=s.sid)as num,

(select avg(score) from grade where sid=s.sid)as avg

From student s

order by avg desc;

 

 

 

17、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 c.cid,max(score) as max,min(score)as min

from course c

left join grade g

on c.cid=g.cid

group by c.cid;

 

18、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

 

#方式1

 

select g.cid, avg(g.score) as avg,100*(selectcount(sid) from grade where cid=g.cid and score>=60)/count(sid) as rate

from grade g

group by cid

order by avgasc, rate desc;

 

#方式2

#利用sum函数和分支语句，sum(1) 貌似与 count(*) 效果一样

 

select g.cid, avg(g.score) asavg,100*sum(case when score>=60 then 1 else 0 end)/sum(1) as rate

from grade g

group by cid

order by avgasc, rate desc;

 

19、查询如下课程平均成绩和及格率的百分数(用"1行"显示): math（001），phy（002）

 

#方式1

select  

(select avg(score) from grade where cid=1)as mathavg,

((select count(sid) from grade where cid=1and score>=60)/(select count(sid) from grade where cid=1)) as mathrate,

(select avg(score) from grade where cid=2)as phyavg,

((select count(sid)from grade where cid=2 and score>=60)/(select count(sid) from grade wherecid=2)) as phyrate;

 #方式2

#利用sum函数和分支语句进行全局统计

 

SELECT

SUM(CASE WHEN cid =1 THEN score ELSE 0END)/SUM(CASE cid WHEN 1 THEN 1 ELSE 0 END) AS mathavg,

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 mathrate,

SUM(CASE WHEN cid = 2 THEN score ELSE 0END)/SUM(CASE cid WHEN '002' THEN 1 ELSE 0 END) AS phyavg,

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 phyrate

FROM GRADE

 

 

20、查询不同老师所教不同课程平均分从高到低显示

 

#方式1

 

select c.tid, c.cid, avg(g.score) as avg

from course c,grade g

where c.cid=g.cid

group by c.tid, c.cid

order by c.tid,avg desc

 

 

21、查询各科成绩前三名的记录:(不考虑成绩并列情况)

 

#方式1（错误）

#mysql 不支持在子查询中取一定数量的记录，所以下面的语句逻辑上是对的，执行是错的

 

select g.cid,sid,score

from grade g

where sid in (select sid from grade wherecid=g.cid order by score limit 3)

order by g.cid,score desc;

 

#方式2 

#下面的语句很巧妙，一方面利用了传值计算判断，另一方面用了判断排名的技巧

 

select cid,sid,score

from grade g

where 2>=(select count(sid) from gradewhere score>g.score and cid=g.cid )

order by cid,score desc;

 

#方式3

#下面的语句也很巧妙，一方面，没有用传值判断，因为用两张大表左关联代替，另一方面，判断排名的技巧同上

 

select a.sid, a.cid,a.score

from grade a

left join grade b

on a.cid=b.cid and a.score

group by a.sid, a.cid,a.score

having count(b.sid)<=2

order bya.cid,a.score desc;

#方式4 ，下面的语句在Oracle上可运行

 

selectcid,sid ,score

from  (select cid,sid,score,

row_number()OVER(PARTITION BY cid ORDER BY score DESC) rid

FROMgrade)

where rid<=3;

 

 

23、统计各科成绩各分数段人数:课程ID,课程名称,[100-85],[84-60],[ <60]

#方式1

#利用sum函数和分支语句

 

select g.cid,cname,

sum(case when score>=85 then 1 else 0end ) as '[85-100]',

sum(case when score <85 and score>=60then 1 else 0 end) as '[60-84]',

sum(case when score<60 then 1 else 0end) as'[0-59]'

from grade g,course

where g.cid=course.cid

group by g.cid;

 

 

24、查询学生平均成绩及其名次

 

#方式1

#很巧妙，先构造（学生，平均成绩）表，然后为排名构造一个技巧

 

select

1+(select count(b.avg) from (select sid,avg(score)as avg from grade group by sid) as b where b.avg > a.avg)  as ranking,

sid,a.avg

from (select sid,avg(score) as avg fromgrade group by sid) as a

order byranking;

 

26、查询每门课程被选修的学生数

 

select cid,count(sid)

from grade

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 g.sid,s.sname

from grade g,student s 

where g.sid=s.sid

group by g.sid

having count(cid)=1;

 

 

28、查询男生、女生人数

select ssex,count(sid)

from student

group by ssex;

 

29、查询姓“张”的学生名单

select sid,sname

from student

where snamelike '张%'

 

30、查询同名同姓学生名单，并统计同名人数

 

select sname,count(sid) as num

from student

group by sname

having  num>1;

 

 

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) as avg

from grade

group by cid

order by avgdesc,cid asc;

 

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 s.sid,sname,avg(score) as avg

from student s,grade g

where s.sid=g.sid

group by s.sid

havingavg>85;

 

34、查询课程名称为“phy”，且分数低于60的学生姓名和分数

 

select sname,score

from student s,grade g,course c

where s.sid=g.sidand g.cid=c.cid and c.cname='phy' and g.score<60;

 

35、查询所有学生的选课情况；

# 方式1

select s.sid,sname,g.cid,cname

from student s,grade g,course c

where s.sid=g.sid and g.cid=c.cid

order by s.sid,g.cid;

 

#方式2 

select s.sid,s.sname,a.cid,a.cname

from student s

left join (select sid,g.cid,cname fromgrade g,course c where g.cid=c.cid) a

on s.sid=a.sid

order by s.sid,a.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 s.sid,sname,cname,score

from student s,grade g,course c

where s.sid=g.sid and g.cid=c.cid andscore>70

order by s.sid;

 

 

37、查询有不及格学生的课程，并按课程号从大到小排列

select cid

from grade

where score<60

order by ciddesc;

 

38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名；

select s.sid,s.sname

from student s,grade g

where s.sid=g.sidand score>80 and g.cid=3;

 

39、求选了课程的学生人数

select count(distinct sid)

from grade;

 

40、查询选修“tb”老师所授课程的学生中，成绩最高的学生学号、姓名、课程编号及其成绩

 

本题有两种理解，一是不区分tb所教不同课程，求出最高分；二是进行区分，分别求出不同课程的最高分

#方式1，不区分

select cid,s.sid,sname ,score

from grade g,student s

where g.sid=s.sid and g.cid in

(select cid from course c,teacher t wherec.tid=t.tid and tname='tb') 

andscore=(select max(score) from grade where cid in  (select cid from course c,teacher t where c.tid=t.tidand tname='tb'));

 

#方式2，不区分

Select g.cid,s.sid,sname,score

from student s,grade g,course c,teacher t

where s.sid=g.sid and g.cid=c.cid andc.tid=t.tid and t.tname='tb' and score=

(selectmax(score) from grade g1,course c1,teacher t1 where g1.cid=c1.cid andc1.tid=t1.tid and t1.tname='tb');

 

 #方式3，区分

Select g.cid,s.sid,sname,score

from student s,grade g,course c,teacher t

where s.sid=g.sid and g.cid=c.cid andc.tid=t.tid and t.tname='tb' and score=

(selectmax(score) from grade where cid=g.cid);

 

 

41、查询各个课程及相应的选修人数

select cid,count(sid)

from grade

group by cid;

 

 

 

 

44、统计每门课程的学生选修人数（超过4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，若人数相同，按课程号升序排列 

select cid,count(sid) as num

from grade

group by cid

having num>4

order by num desc, cid;

 

 

45、检索至少选修两门课程的学生学号

select sid

from grade

group by sid

havingcount(cid)>=2

 

46、查询全部学生都选修的课程的课程号和课程名

select c.cid,cname

from course c,grade g

where c.cid=g.cid

group by c.cid

having count(sid)=(select count(sid) fromstudent);

 

 

47、查询没学过“ta”老师讲授的任一门课程的学生姓名

 

select sname

from student

where sid not in

(select distinct sid from grade g,coursec,teacher t where g.cid=c.cid and c.tid=t.tid and t.tname='ta');

 

 

48、查询至少有一门不及格课程的同学的学号及其平均成绩

select sid,avg(score)

from grade

where sid in (select sid from grade wherescore<60 group by sid having count(cid)>0)

group by sid;

 

 

 

50、删除“002”同学的“001”课程的成绩

 

delete from Grade

where sid='002'andcid='001';