Mysql-45题回顾-已完成

1.1、 查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

# 查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
select t3.*
      ,t1.CId
      ,t1.score
      ,t2.cid
      ,t2.score
from sc t1 
join sc t2 on t1.SId=t2.SId and t1.CId='01' and t2.CId='02'
join student t3 on t1.SId=t3.SId
where t1.score>t2.score;
1.1

1.2 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 *
from sc as t1
left join sc t2 on t1.sid = t2.sid and t2.cid='02'
where t1.cid='01';
select *
from 
(select sc.sid
        ,sc.cid
 from sc where sc.cid='01') as t1
left join
(select sc.sid
     ,sc.cid
from sc where sc.cid='02'
) as t2 on t1.sid=t2.sid
1.2

1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

select *
from sc where sc.sid not in 
(select sc.sid from sc where sc.cid='01')
and sc.cid='02';
1.3

2、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
      ,st.Sname
      ,avg(sc.score)
from sc 
join student as st on sc.sid=st.SId
group by sc.SId having avg(sc.score)>60
order by avg(sc.score) desc;
2

3、查询在 SC 表存在成绩的学生信息

# 答案一
select * 
from student st where st.sid in
(select sc.sid from sc);
# 答案二
select distinct student.*
from student,sc
where student.SId=sc.sid;
# 答案三
select distinct st.*
from sc
inner join student st on sc.sid=st.sid
3

4、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 st.sid
      ,st.Sname
      ,count(distinct sc.cid) as courses
      ,sum(sc.score) as score
from student as st 
left join sc on st.SId=sc.SId
group by st.sid
        ,st.Sname;
4

5、查询「李」姓老师的数量

select count(*)
from teacher t
where t.Tname like '李%';
5

6、查询学过「张三」老师授课的同学的信息

select st.*
      ,t.Tname
from teacher as t
join course as c on t.TId=c.TId
join sc on c.CId=sc.CId
join student as st on sc.SId=st.SId
where t.Tname='张三';
6

7、查询没有学全所有课程的同学的信息

select st.*
      ,count(distinct sc.cid) as 课程数
from student st
left join sc on st.sid = sc.sid
group by st.sid having 课程数 <3;
7

8、查询至少有一门课与学号为" 01 "的同学所学相同的同学的信息

select distinct st.*
from sc 
join student as st on sc.sid=st.sid 
where sc.cid in
(select sc.CId from sc) ;
8
select student.* 
from student 
where sid in 
(select sid from sc group by sid where and sid <> "01" having count(cid) = (select count(cid) from sc where sid = '01') 
and sid not in(select sid from sc where cid not in(select cid from sc where sid = '01')));

9.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

1、选出所学课程不在(01,02,03)的同学-排除
2、剩下的同学肯定选了01,02,03中的某几门,判断所学课程数是否等于3

select st.*
from sc 
join student st on sc.sid=st.sid
where sc.sid !='01' and sc.sid  not in (select sc.sid
from sc where sc.cid not in(select sc.CId from sc where sc.sid='01'))
group by sc.sid 
having count(distinct sc.cid)=(select count(distinct sc.cid) from sc where sc.sid='01')
9

10.查询没学过"张三"老师讲授的任一门课程的学生姓名

select distinct st.*
from sc 
right join student st on sc.sid = st.sid
where st.sid not in
(select sc.sid
from  teacher as t 
join course as c on t.tid=c.tid 
join sc on c.cid = sc.cid
where t.Tname='张三');

10

11.查询两门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.score<60 or sc.score is null
group by st.sid having count(distinct sc.cid)>=2;
11
# 我的解法
select t.sid
      ,t.sname
      ,avg(t.score) as avg_score
from
(select sc.score
       ,sc.cid
       ,st.sid
       ,st.sname
from student st 
left join sc on sc.sid=st.sid) t
where t.score<60 or t.score is null
group by t.sid,t.sname having count(distinct t.cid)>=2 or avg_score is null;
11

12、检索" 01 "课程分数小于 60,按分数降序排列的学生信息

select *
from student st
left join sc on st.sid=sc.sid
where sc.cid='01' and score<60
order by sc.score desc;

13、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 t1.sid
      ,t1.sname
      ,max(case when t1.cname='语文' then t1.score else 0 end) as 语文
      ,max(case when t1.cname='数学' then t1.score else 0 end) as 数学
      ,max(case when t1.cname='英语' then t1.score else 0 end) as 英语
      ,ifnull(avg(t1.score) ,0) as avg_score
from
(
    select st.sid
          ,st.sname
          ,sc.cid
          ,c.cname
          ,sc.score
    from sc
    join course as c on sc.cid=c.cid
    right join student as st on st.sid = sc.sid) t1
    group by t1.sid
            ,t1.sname
    order by avg_score desc;
13
select sc.cid
      ,c.cname
      ,count(distinct sc.sid) as 选修人数
      ,max(sc.score) as 最高分
      ,min(sc.score) as 最低分
      ,round(avg(sc.score),2) as 平均分
      ,concat(round(sum(case when sc.score>=60 then 1 else 0 end)/count(distinct sc.sid)*100,2),'%') as 及格率
      ,concat(round(sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end)/count(distinct sc.sid)*100,2),'%') as 中等率
      ,concat(round(sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end)/count(distinct sc.sid)*100,2),'%') as 优良率
      ,concat(round(sum(case when sc.score>=90 then 1 else 0 end)/count(distinct sc.sid)*100,2),'%') as 优秀率
from sc
join course as c on sc.cid=c.cid
group by sc.cid
        ,c.cname
order by 选修人数 desc,sc.cid
14
select sc.cid
      ,c.cname
      ,count(distinct sc.sid) as 选修人数
      ,max(sc.score) as 最高分
      ,min(sc.score) as 最低分
      ,round(avg(sc.score),2) as 平均分
      ,concat(round(sum(if(sc.score>60,1,0))/count(distinct sc.sid)*100,2),'%') as 及格率
      ,concat(round(sum(if(sc.score>=70 and sc.score<80 , 1 , 0 ))/count(distinct sc.sid)*100,2),'%') as 中等率
      ,concat(round(sum(if(sc.score>=80 and sc.score<90 , 1 , 0 ))/count(distinct sc.sid)*100,2),'%') as 优良率
      ,concat(round(sum(if(sc.score>=90 , 1 , 0 ))/count(distinct sc.sid)*100,2),'%') as 优秀率
from sc
join course as c on sc.cid=c.cid
group by sc.cid
        ,c.cname
order by 选修人数 desc,sc.cid
14

15、按各科成绩进行排序,并显示排名, Score 重复时保留名次空缺

select @rank := 0;
select *
      ,@rank:=@rank+1
from sc
ORDER BY sc.score desc;

15.1 按各科成绩进行排序,并显示排名, Score 重复时合并名次

select sc.SId
      ,sc.CId
      ,@dense_rank:=if(@font=sc.score,@dense_rank,@dense_rank+1) as 排名
      ,@font:=sc.score as score
from sc
join (select @dense_rank :=0, @font:=NULL) a
ORDER BY sc.score desc;
15

16 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 a.sid
      ,@rank:=if(@sco=总成绩,'',@rank+1) as 排名
      ,@sco :=总成绩 as 总成绩
from(
SELECT sc.SId
      ,sum(sc.score) as 总成绩     
from sc
GROUP BY sc.SId
ORDER BY sum(sc.score) desc) a
join (select @rank:=0 , @sco :=null) b
16

17 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]

SELECT sc.CId
      ,c.cname
      ,case when sc.score<60 then '(60-0]'
            when sc.score>=60 and sc.score<70 then '[60-70)'
            when sc.score>=70 and sc.score<85 then '[70-85)'
            when sc.score>=85 then '[100-85]'
       end as 区间
      ,count(DISTINCT sc.SId) as 人数
FROM sc
join course c on sc.CId=c.cid
GROUP BY sc.CId
        ,c.cname
        ,case when sc.score<60 then '(60-0]'
            when sc.score>=60 and sc.score<70 then '[60-70)'
            when sc.score>=70 and sc.score<85 then '[70-85)'
            when sc.score>=85 then '[100-85]'
       end
17
# 17.1 [100-85],[85-70],[70-60],[60-0] 各区间段所占比例
SELECT sc.CId
      ,c.cname
      ,concat(round(avg(case when sc.score<60 then 1 else 0 end )*100,2),'%') as '(60-0]'
      ,concat(round(avg(case when sc.score>=60 and sc.score<70 then 1 else 0 end )*100,2),'%') as '[60-70)'
      ,concat(round(avg(case when sc.score>=70 and sc.score<85 then 1 else 0 end )*100,2),'%') as '[70-85)'
      ,concat(round(avg(case when sc.score>=85 then 1 else 0 end )*100,2),'%') as '[100-85]'
FROM sc
join course c on sc.CId=c.cid
GROUP BY sc.CId
        ,c.cname
17.1

18、查询各科成绩前三名的记录

select sc.cid
      ,c.cname
      ,st.sname
      ,sc.score
from sc
join course as c on sc.cid=c.cid
join student as st on st.sid=sc.sid
where (select count(*) from sc as a where a.cid=sc.cid and sc.score <a.score) <3
order by sc.cid ,sc.score desc;
18

21.查询男生、女生人数

select sum(case when st.Ssex='男' then 1 else 0 end) as 男生人数
      ,sum(case when st.Ssex='女' then 1 else 0 end) AS 女生人数
from student as st
21
select st.ssex
      ,count(*)
from student as st
group by st.ssex
21

22、查询名字中含有「风」字的学生信息

select *
from student as st 
where st.sname like '%风%';
22

23.查询同名同性学生名单,并统计同名人数

select sid
      ,sname
      ,ssex
      ,count(sname)
from student 
group by sname,ssex having count(sname)>1
23

24.查询 1990 年出生的学生名单

select *
from student
where year(Sage)='1990';
24

25.查询每门课程的平均成绩,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 cid
      ,avg(score)
from sc
group by cid
order by avg(score) desc,cid desc
25

26 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 sc 
join student as st on sc.sid=st.sid
group by st.sid
having avg(sc.score)>=85;
26

27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
      ,a.score
from student as st
join 
(select sc.sid
       ,sc.score
from sc 
join course as c on sc.cid=c.cid
where c.cname='数学' and sc.score<60) a
on st.sid = a.sid;
27

28、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 *
from student st
left join sc on st.sid=sc.sid;
28

29、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 sc.sid
    ,st.sname
    ,c.cname
    ,sc.score
from sc 
join course as c on c.cid=sc.cid
join student st on st.sid=sc.sid
where sc.score>70;
29

30.查询存在不及格的课程

select *
from sc 
join course as c on sc.cid=c.cid
where sc.score<60;
30

31.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 sc.sid
      ,st.sname
      ,sc.cid
      ,sc.score
from sc
join student as st on sc.sid=st.sid
where sc.cid='01' and sc.score>=80;
31

32、求每门课程的学生人数

select sc.cid
      ,count(sc.sid)
from sc
group by sc.cid;
32

33、成绩不重复,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 st.*
      ,max(sc.score)
      ,c.cname
      ,sc.score
from teacher as t
join course as c on t.tid=c.tid
join sc on sc.cid=c.cid
join student as st on st.sid=sc.sid
where t.tname='张三';
33

34、成绩有重复的情况下,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 t1.*
from 
(select    a.*
          ,@rank_num :=if(@score=a.score, @rank_num, @rank_num:=@rank_num+1) as rn
          ,@score:=a.score as r_score
from 
    (select st.sid
          ,st.sname
          ,sc.cid
          ,c.cname
          ,t.tname
          ,sc.score
    from teacher as t
    join course as c on t.tid=c.tid
    join sc on sc.cid=c.cid
    join student as st on st.sid=sc.sid
    where t.tname='张三') a
join (select @score := null, @rank_num :=0) b
order by a.score desc) t1
where t1.rn=1;
34

35、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 *
from sc 
where sc.sid 
in (select a.sid from sc as a where a.sid=sc.sid and sc.cid != a.cid and sc.score=a.score);
35

36.查询每门功成绩最好的前两名

select a.*
from 
(select *, rank() over(partition by sc.cid order by sc.score desc) as rank_num
from sc) a
where rank_num=1 or rank_num=2;
36

答案

select *
from sc as t1
where (select count(*) from sc as t2 where t1.cid=t2.cid and t2.score>t1.score)<2
order by t1.cid;
36-2

37.统计每门课程的学生选修人数(超过 5 人的课程才统计)

select sc.cid
      ,count(distinct sc.sid) as person_num
from sc
group by sc.cid having person_num>5;
37

38.检索至少选修两门课程的学生学号

select sc.sid
      ,count(distinct sc.cid) as cos_num
from sc
group by sc.sid having cos_num>=2;
38

39、查询选修了全部课程的学生信息

select st.*
      ,count(distinct sc.cid) as 课程数
from sc
join student as st on sc.sid=st.sid
group by sc.sid having count(distinct sc.cid)=(select count(*) from course);

40.查询各学生的年龄,只按年份来算

select *
    ,year(now())-year(Sage) as age
from student;
40

41、按照出生日期来算,当前月日 < 出生年月的月日则,年龄减一

select *
      ,case when substr(sage,6,5)<substr(now(),6,5) then year(now())-year(sage)
            when substr(sage,6,5)>=substr(now(),6,5) then year(now())-year(sage)-1
            end as age
from student;
41

42.查询本周过生日的学生

select *
     ,substr(yearweek(sage),5,2) as birth_week
     ,substr(yearweek(curdate()),5,2) as now_week
from student
where substr(yearweek(sage),5,2)=substr(yearweek(curdate()),5,2)
42

43、查询下周过生日的学生

select *
      ,substr(yearweek(sage),5,2) as birth_week
      ,substr(yearweek(curdate()),5,2)+1 as next_week
from student
where substr(yearweek(sage),5,2)=substr(yearweek(curdate()),5,2)+1;
43

44.查询本月过生日的学生

select *
      ,month(sage) as bir_month
      ,month(curdate()) as now_month
from student
where month(sage) = month(curdate())
44

EXTRACT() 函数用于返回日期/时间的单独部分,比如年、月、日、小时、分钟等等。EXTRACT(unit FROM date)

DATE_ADD() 函数向日期添加指定的时间间隔。DATE_ADD(date,INTERVAL expr type)-语法

# 45.查询下月过生日的学生

select *
from student
where extract(month from sage) = extract(month from date_add(curdate(), interval 1 month))
45

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