sql练习题

sql练习题

1.查询" 01 “课程⽐” 02 "课程成绩⾼的学⽣的信息及课程分数

SELECT s1.sid,s1.score from 
(select sid,score FROM sc where cid='01') as s1 
join
(SELECT sid,score FROM sc where cid='02') as s2
on s1.sid=s2.sid
WHERE s1.score>s2.score;						

2.查询同时存在" 01 “课程和” 02 "课程的情况

SELECT s1.sid,s1.cid,s1.score,s2.cid,s2.score from 
(select sid,cid,score FROM sc where cid='01') as s1 
join
(SELECT sid,cid,score FROM sc where cid='02') as s2
on s1.sid=s2.sid

3.查询存在" 01 “课程但可能不存在” 02 "课程的情况(不存在时显示为 null )

SELECT s1.*,s2.* from 
(select sid,cid,score FROM sc where cid='01') as s1 
left join
(SELECT sid,cid,score FROM sc where cid='02') as s2
on s1.sid=s2.sid

4.查询不存在" 01 “课程但存在” 02 "课程的情况

SELECT * from sc
where sid  not in(select sid FROM sc where cid='01')
and cid='02';

5.查询平均成绩⼤于等于 60 分的同学的学⽣编号和学⽣姓名和平均成绩

SELECT sc.SId, sname,ROUND(AVG(score))
from sc,student
WHERE sc.SId=student.SId
GROUP BY sid,sname HAVING AVG(score)>=60

6.查询在 SC 表存在成绩的学⽣信息

SELECT DISTINCT student.*  FROM student JOIN sc on sc.SId=student.SId

7.查询所有同学的学⽣编号、学⽣姓名、选课总数、所有课程的总成绩(没成绩的显示为 null )

SELECT student.SId,student.sname,count(sc.cId) ,SUM(sc.score)
from student left JOIN sc 
on student.SId=sc.sid
GROUP BY student.SId,student.Sname

8.查询「李」姓⽼师的数量

SELECT COUNT(Tname) FROM teacher WHERE Tname LIKE '李%'

9.查询学过「张三」⽼师授课的同学的信息

select st.* from student st 
left join score sc on sc.sid=st.sid
left join course c on c.cid=sc.cid
left join teacher t on t.tid=c.tid
 where t.tname="张三"

10.查询没有学全所有课程的同学的信息

select SId, count(CId) as Cn from sc GROUP BY SId HAVING Cn<(select count(CId) as Cnum  from sc)

11.查询⾄少有⼀⻔课与学号为" 01 "的同学所学相同的同学的信息

select distinct st.* from student st 
left join  sc on sc.sid=st.sid
where sc.cid in (
select sc2.cid from student st2
left join sc sc2 on sc2.sid=st2.sid
where st2.sid ='01'
)

12.查询和" 01 "号的同学学习的课程 完全相同的其他同学的信息

select  st.* from student st 
left join sc sc on sc.sid=st.sid
group by st.sid
having group_concat(sc.cid) = 
(
select  group_concat(sc2.cid) from student st2
left join sc sc2 on sc2.sid=st2.sid
where st2.sid ='01'
)

13.查询没学过"张三"⽼师讲授的任⼀⻔课程的学⽣姓名

select st.s_name from student st 
where st.s_id not in (
select sc.s_id from score sc 
inner join course c on c.c_id=sc.c_id
inner join teacher t on t.t_id=c.t_id and t.t_name="张三"

14.查询两⻔及其以上不及格课程的同学的学号，姓名及其平均成绩

select st.sid,st.sname,avg(sc.score) from student st
left join  sc on sc.sid=st.sid
where sc.sid in (
select sc.sid from  sc 
where sc.score<60 or sc.score is NULL
group by sc.sid having COUNT(1)>=2
)
group by st.sid

15.检索" 01 "课程分数⼩于 60，按分数降序排列的学⽣信息

select st.*,sc.score from student st 
inner join  sc on sc.sid=st.sid and sc.cid="01" and sc.score<60
order by sc.score desc

16.按平均成绩从⾼到低显示所有学⽣的所有课程的成绩以及平均成绩

SELECT sc.*, a FROM sc LEFT JOIN
   ( SELECT sid, AVG(score) AS a FROM sc GROUP BY SId )r
ON sc.SId = r.sid ORDER BY a DESC

17.查询各科成绩最⾼分、最低分和平均分： 以如下形式显示：课程 ID，课程 name，最⾼分，最低分，平均分，及格率，中等率，优良率，优秀率 及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90 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 c.cid,c.cname,max(sc.score) "最高分",MIN(sc2.score) "最低分",avg(sc3.score) "平均分" 
,((select count(sid) from sc where score>=60 and cid=c.cid )/(select count(sid) from sc where cid=c.cid)) "及格率"
,((select count(sid) from sc where score>=70 and score<80 and cid=c.cid )/(select count(sid) from sc where cid=c.cid)) "中等率"
,((select count(sid) from sc where score>=80 and score<90 and cid=c.cid )/(select count(sid) from sc where cid=c.cid)) "优良率"
,((select count(sid) from sc where score>=90 and cid=c.cid )/(select count(sid) from sc where cid=c.cid)) "优秀率"
from course c
left join  sc on sc.cid=c.cid 
left join sc sc2 on sc2.cid=c.cid 
left join sc sc3 on sc3.cid=c.cid 
group by c.cid

18.按各科平均成绩进⾏排序，并显示排名， Score 重复时保留名次空缺

SELECT s.SId, s.Sname, 
avg(score) as 平均分
from student s LEFT JOIN sc  ON  s.sid = sc.sid GROUP BY s.sid 
ORDER BY 平均分

select s2.CId,s2.avg_score,count(DISTINCT s1.avg_score) as rank
FROM
(select CId,round(avg(score),2) as avg_score from 
sc group by SId) as s1
join
(select CId,round(avg(score),2) as  avg_score from 
sc group by SId) as s2
on s1.avg_score >= s2.avg_score
GROUP BY s2.cid,s2.avg_score
ORDER BY rank

19.按各科平均成绩进⾏排序，并显示排名， Score 重复时不保留名次空缺

select a.*,AVG(a.score) as 平均分 from sc a 
left join sc b on a.cid = b.cid and a.score < b.score
group by a.sid
order by a.cid,平均分

select b.CId,b.avg_score,@i:=@i+1 as rank from (SELECT @i:=0) as a,
(select CId,round(avg(score),2) as avg_score from sc GROUP BY cid ORDER BY avg_score desc) as b

20.查询学⽣的总成绩，并进⾏排名，总分重复时保留名次空缺

SELECT s.SId, s.Sname, 
sum(score) as 总分
from student s LEFT JOIN  sc ON  s.SId = sc.SId GROUP BY s.SId

21.查询学⽣的总成绩，并进⾏排名，总分重复时不保留名次空缺

select st.SId,st.sname
,(case when sum(sc.score) is null then 0 else sum(sc.score) end)
 from student as st
left join sc on sc.sid=st.sid
group by st.sid order by sum(sc.score) desc

22.统计各科成绩各分数段⼈数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0]及所占百分⽐

select c.cid,c.cname 
,((select count(1) from sc where sc.cid=c.cid and sc.score<=100 and sc.score>80)/(select count(1) from sc where sc.cid=c.cid )) "100-85"
,((select count(1) from sc where sc.cid=c.cid and sc.score<=85 and sc.score>70)/(select count(1) from  sc where sc.cid=c.cid )) "85-70"
,((select count(1) from sc where sc.cid=c.cid and sc.score<=70 and sc.score>60)/(select count(1) from  sc where sc.cid=c.cid )) "70-60"
,((select count(1) from sc where sc.cid=c.cid and sc.score<=60 and sc.score>=0)/(select count(1) from sc where sc.cid=c.cid )) "60-0"
from course c order by c.cid

23.查询各科成绩前三名的记录

select a.* from (
 select st.sid,st.sname,c.cid,c.cname,sc.score from student st
 left join sc on sc.sid=st.sid
 inner join course c on c.cid=sc.cid and c.cid='01'
 order by sc.score desc LIMIT 0,3) a
union all 
select b.* from (
 select st.sid,st.sname,c.cid,c.cname,sc.score from student st
 left join  sc on sc.sid=st.sid
 inner join course c on c.cid=sc.cid and c.cid='02'
 order by sc.score desc LIMIT 0,3) b
union all
select c.* from (
 select st.sid,st.sname,c.cid,c.cname,sc.score from student st
 left join  sc on sc.sid=st.sid
 inner join course c on c.cid=sc.cid and c.cid='03'
 order by sc.score desc LIMIT 0,3) c

24.查询每⻔课程被选修的学⽣数

select c.cid,c.cname,count(1) from course c 
left join  sc on sc.cid=c.cid
inner join student st on st.sid=c.cid
group by st.sid

25.查询出只选修两⻔课程的学⽣学号和姓名

select st.sid,st.sname from student st 
left join  sc on sc.sid=st.sid
inner join course c on c.cid=sc.cid 
group by st.sid having count(1)=2

26.查询男⽣、⼥⽣⼈数

select st.ssex,count(1) from student st group by st.ssex

27.查询名字中含有「⻛」字的学⽣信息

select st.* from student st where st.sname like "%风%";

28.查询同名同性学⽣名单，并统计同名⼈数

select st.*,count(1) from student st group by st.s_name,st.s_sex having count(1)>1

29.查询 1990 年出⽣的学⽣名单

select st.* from student st where st.sage like "1990%";

30.查询每⻔课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列

select c.cid,c.cname,avg(sc.score) from course c
inner join sc on sc.cid=c.cid  
group by c.cid order by avg(sc.score) desc,c.cid asc

31.查询平均成绩⼤于等于 85 的所有学⽣的学号、姓名和平均成绩

select st.sid,st.sname,avg(sc.score) from student st
left join sc on sc.sid=st.sid
group by st.sid having avg(sc.score)>=85

32.查询课程名称为「数学」，且分数低于 60 的学⽣姓名和分数

select st.sid,st.sname,sc.score from student st
inner join  sc on sc.sid=st.sid and sc.score<60
inner join course c on c.cid=sc.cid and c.cname ="数学" 

33.查询所有学⽣的课程及分数情况（存在学⽣没成绩，没选课的情况）

select st.sid,st.sname,c.cname,sc.score from student st
left join  sc on sc.sid=st.sid
left join course c on c.cid =sc.cid
order by st.sid,c.cname

34.查询任何⼀⻔课程成绩在 70 分以上的姓名、课程名称和分数

select st2.sid,st2.sname,c2.cname,sc2.score from student st2
left join sc sc2 on sc2.sid=st2.sid
left join course c2 on c2.cid=sc2.cid 
where st2.sid in(
select st.sid from student st 
left join sc on sc.sid=st.sid 
group by st.sid having min(sc.score)>=70)
order by sid

35.查询不及格的课程

select st.sid,c.cname,st.sname,sc.score from student st
inner join sc on sc.sid=st.sid and  sc.score<60
inner join course c on c.cid=sc.cid 

36.查询课程编号为 01 且课程成绩在 80 分以上的学⽣的学号和姓名

select st.sid,st.sname,sc.score from student st
inner join sc on sc.sid=st.sid and sc.cid="01" and sc.score>=80

37.求每⻔课程的学⽣⼈数

select c.cid,c.cname,count(1) from course c
inner join score sc on sc.cid=c.cid
group by c.cid

38.成绩有重复，查询选修「张三」⽼师所授课程的学⽣中，成绩最⾼的学⽣信息及其成绩

select sId, Score, Rank from (
               select sId, Score, @tempRank := @tempRank + if(@preScore := Score, 0, 1) as Rank,                           
							 @preScore := Score  from Sc, 
                               (select @tempRank:= 0) r, (select @preScore:= null) p 
                                        order by Score desc, SId asc
                                            )  t
limit 0,1

39.成绩不重复的情况下，查询选修「张三」⽼师所授课程的学⽣中，成绩最⾼的学⽣信息及其成绩

select st.*,c.cname,sc.score,t.tname from student st
inner join  sc on sc.sid=st.sid
inner join course c on c.cid=sc.cid 
inner join teacher t on t.tid=c.tid and  t.tname="张三"
order by sc.score desc
limit 0,1

40.查询不同课程成绩相同的学⽣的学⽣编号、课程编号、学⽣成绩

select st.sid,st.sname,sc.cid,sc.score from student st 
left join  sc on sc.sid=st.sid
left join course c on c.cid=sc.cid
where (
select count(1) from student st2 
left join sc sc2 on sc2.sid=st2.sid
left join course c2 on c2.cid=sc2.cid
where sc.score=sc2.score and c.cid!=c2.cid 
)>1

41.查询每⻔课程成绩最好的前两名

 select a.sid,a.cid,a.score from sc a
 where (select COUNT(1) from sc b where b.cid=a.cid and b.score>=a.score)<=2 order by a.cid

42.统计每⻔课程的学⽣选修⼈数（超过 5 ⼈的课程才统计）

select sc.cid,count(1) from  sc
left join course c on c.cid=sc.cid
group by c.cid having count(1)>5
order by count(1) desc,sc.cid asc

43.检索⾄少选修两⻔课程的学⽣学号

select st.sid from student st 
left join  sc on sc.sid=st.sid
group by st.sid having count(1)>=2

44.查询选修了全部课程的学⽣信息

select st.* from student st 
left join  sc on sc.sid=st.sid
group by st.sid having count(1)=(select count(1) from course)

45.查询各学⽣的年龄，只按年份来算

select st.*,timestampdiff(year,st.sage,now()) from student st

46.按照出⽣⽇期来算，当前⽉⽇ < 出⽣年⽉的⽉⽇则，年龄减⼀

47.查询本周过⽣⽇的学⽣

select st.* from student st 
where week(now())=week(date_format(st.sage,'%Y%m%d'))

48.查询下周过⽣⽇的学⽣

select st.* from student st 
where week(now())+1=week(date_format(st.sage,'%Y%m%d'))

49.查询本⽉过⽣⽇的学⽣

select st.* from student st 
where month(now())=month(date_format(st.sage,'%Y%m%d'))

50.查询下⽉过⽣⽇的学⽣

select st.* from student st 
where month(timestampadd(month,1,now()))=month(date_format(st.sage,'%Y%m%d'))
-- 或


select st.* from student st where (month(now()) + 1) mod 12 = month(date_format(st.sage,'%Y%m%d'))