MySQL经典50题

mysq经典50题拿来练手·····

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创建表及信息录入

# 学生表
create table student
(
    sid   varchar(10),
    sname varchar(32),
    saged datetime,
    ssex  varchar(10)
) comment '学生表';
# 学生表插入数据
insert into student(sid,sname,saged,ssex)values('01' , '赵雷' , '1990-01-01' , '男');
insert into student(sid,sname,saged,ssex)values('02' , '钱电' , '1990-12-21' , '男');
insert into student(sid,sname,saged,ssex)values('03' , '孙风' , '1990-05-20' , '男');
insert into student(sid,sname,saged,ssex)values('04' , '李云' , '1990-08-06' , '男');
insert into student(sid,sname,saged,ssex)values('05' , '周梅' , '1991-12-01' , '女');
insert into student(sid,sname,saged,ssex)values('06' , '吴兰' , '1992-03-01' , '女');
insert into student(sid,sname,saged,ssex)values('07' , '郑竹' , '1989-07-01' , '女');
insert into student(sid,sname,saged,ssex)values('08' , '王菊' , '1990-01-20' , '女');

-- -------------------------------------------------------------
# 课程表
create table course
(
    cid   varchar(10),
    cname varchar(32),
    tid   varchar(10)
) comment '课程表';
# 插入课程数据
insert into course(cid,cname,tid)values('01' , '语文' , '02');
insert into course(cid,cname,tid)values('02' , '数学' , '01');
insert into course(cid,cname,tid)values('03' , '英语' , '03');

-- --------------------------------------------------------
# 教师表
create table teacher
(
    tid   varchar(10),
    tname varchar(32)
)comment '教师表';
# 插入教师表信息
insert into teacher(tid,tname)values('01' , '张三');
insert into teacher(tid,tname)values('02' , '李四');
insert into teacher(tid,tname)values('03' , '王五');

-- --------------------------------------------------
# 成绩表
create table sc
(
    sid   varchar(10),
    cid   varchar(10),
    score int(30)
)comment '成绩表';
# 插入成绩信息

insert into SC values('01','01',80);
insert into SC values('01','02',90);
insert into SC values('01','03',99);
insert into SC values('02','01',70);
insert into SC values('02','02',60);
insert into SC values('02','03',80);
insert into SC values('03','01',80);
insert into SC values('03','02',80);
insert into SC values('03','03',80);
insert into SC values('04','01',50);
insert into SC values('04','02',30);
insert into SC values('04','03',20);
insert into SC values('05','01',76);
insert into SC values('05','02',87);
insert into SC values('06','01',31);
insert into SC values('06','03',34);
insert into SC values('07','02',89);
insert into SC values('07','03',98);

1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数

# 涉及表 student ,sc
select stu.*, s1.score, s2.score
from student stu
         left join sc s1 on stu.sid = s1.sid and s1.cid = 01
         left join sc s2 on stu.sid = s2.sid and s2.cid = 02
where s1.score > s2.score;

2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数

select stu.*,s1.score,s2.score
from student stu
         left join sc s1 on stu.sid = s1.sid and s1.cid = 01
         left join sc s2 on stu.sid = s2.sid and s2.cid = 02
where s1.score < s2.score;

3、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

# 涉及表：student ，sc
# 聚合函数 avg---分组
# having---分组之后过滤平均分大于等于60
select stu.sid, stu.sname, avg(s.score) avgscore
from student stu
         left join sc s on stu.sid = s.sid
group by stu.sid, stu.sname
having avgscore >= 60;

4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩

情况一：参加考试的学生

# 涉及表：student ，sc
# 聚合函数 avg---分组
# having---分组之后过滤平均分小于60
select stu.sid, stu.sname, ifnull(avg(s.score),0 )avgscore
from student stu
         left join sc s on stu.sid = s.sid
group by stu.sid, stu.sname
having avgscore < 60;

情况二：
注意：这里有个同学没有参加选课和考试，所以显示成绩为null,如果判定没参加考试的为0分，那么算平均分之前加上ifnull函数，将null的值判定为0

# 涉及表：student ，sc
# 聚合函数 avg---分组
# having---分组之后过滤平均分小于60
select stu.sid, stu.sname, ifnull(avg(s.score),0) avgscore
from student stu
         left join sc s on stu.sid = s.sid
group by stu.sid, stu.sname
having avgscore < 60;

5、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

这里将没有选课和考试的学生原为null值，用ifnull判定为null值

# 涉及表:student,sc
# 聚合函数count,sum---分组
select stu.sid, stu.sname, count(s.cid), ifnull(sum(s.score), 0)
from student stu
         left join sc s on stu.sid = s.sid
group by stu.sid, stu.sname;

6、查询"李"姓老师的数量

#涉及表：teacher
#聚合函数：count
#模糊查询 like %
select count(*)
from teacher
where tname like '李%'

7、查询学过"张三"老师授课的同学的信息

#涉及表：student，sc,course,teacher
select stu.*
from student stu
         left join sc s on stu.sid = s.sid
         left join course c on s.cid = c.cid
         left join teacher t on c.tid = t.tid
where t.tname = '张三';

8、查询没学过"张三"老师授课的同学的信息

#涉及表：student，sc,course,teacher
# 联系第7题，可以根据学号 not in，将上题范围排除
select stu.*
from student stu
where stu.sid not in (
    select stu.sid
    from student stu
             left join sc s on stu.sid = s.sid
             left join course c on s.cid = c.cid
             left join teacher t on c.tid = t.tid
    where t.tname = '张三'
);

9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息

# 涉及表 student ,sc
select stu.*
from student stu
         left join sc s1 on stu.sid = s1.sid and s1.cid = 01
         left join sc s2 on stu.sid = s2.sid and s2.cid = 02
where s1.sid = s2.sid;

10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息

# 涉及表 student ,sc
select *
from student stu
where stu.sid in (
    select s1.sid
    from sc s1
    where s1.cid = 01
)
  and stu.sid not in (
    select s2.sid
    from sc s2
    where s2.cid = 02
);

11、查询没有学全所有课程的同学的信息

# 涉及表：student,sc,course
# 用聚合函数统计course的课程数
# 根据课程数来判断
select stu.sid, stu.sname, stu.saged, stu.ssex
from student stu
         left join sc s on stu.sid = s.sid
group by stu.sid, stu.sname, stu.saged, stu.ssex
having count(s.cid) < (select count(cid) from course)

12、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

# 涉及表 student,sc
# 先找出学号01同学的所学课程
# 用in在其中筛选，并且排除自己（排除01）
select distinct stu.*
from student stu
         left join sc s on stu.sid = s.sid
where stu.sid != 01
  and s.cid in (
    select s2.cid
    from student stu2
             left join sc s2 on stu2.sid = s2.sid
    where stu2.sid = 01
);

13、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

# 涉及表：student,sc
select s1.sid, stu.*, count(s1.cid) num
from sc s1
         left join sc s2 on s1.cid = s2.cid
    and s1.sid != 01 and s2.sid = 01
         left join student stu on s1.sid = stu.sid
where stu.sid != 01
  and s2.sid is not null
group by s1.sid, stu.sid, stu.sname, stu.saged, stu.ssex
having num = (select count(cid) from sc where sid = 01);

14、查询没学过"张三"老师讲授的任一门课程的学生姓名

# 涉及表：student,sc,course,teacher
# 先找到张三老师的课程号--02
# 再找到学生选修02的学生信息
select stu.*
from student stu
where stu.sid in (
    select stu.sid
    from student stu
             left join sc s on stu.sid = s.sid
    where s.cid in (
        select c.cid
        from course c
                 left join teacher t on c.tid = t.tid
        where t.tname = '张三')
    );

15、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

# 涉及表：student,sc
# 条件：where <60
# 聚合函数 avg---分组
select stu.sid, stu.sname
from student stu
         left join sc s on stu.sid = s.sid
where s.sid in (
    select s1.sid
    from sc s1
    where s1.score < 60
    group by s1.sid
    having count(*) >= 2
)
group by stu.sid, stu.sname;

16、检索"01"课程分数小于60，按分数降序排列的学生信息

# 涉及表 student,sc
# 条件 where cid=01 and score<60
# 排序 order by desc
select stu.*
from sc s
         left join student stu on s.sid = stu.sid
where s.cid = 01
  and s.score < 60
order by s.score desc;

17、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

# 涉及表 student,sc
# 聚合函数 sum,avg
# case when then else end
select stu.sid,
       stu.sname,
       ifnull(t.avgscore, 0) avgscore,
       sum(case when s2.cid = 01 then s2.score else 0 end) '语文',
       sum(case when s2.cid = 02 then s2.score else 0 end) '数学',
       sum(case when s2.cid = 03 then s2.score else 0 end) '英语'
from (select s1.sid, avg(s1.score) avgscore
      from sc s1
      group by s1.sid
      order by avgscore desc) t
         left join sc s2 on t.sid = s2.sid
         right join student stu on t.sid = stu.sid
group by stu.sid, stu.sname, t.avgscore

18、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

# 涉及表：sc
# 聚合函数 max,min,avg,sum,count
select c.cid,
       c.cname,
       max(s.score)'最高分',
       min(s.score)'最低分',
       avg(s.score)'平均分',
       sum(case when score >= 60 then 1 else 0 end) / count(*) '及格率',
       sum(case when score between 70 and 80 then 1 else 0 end) / count(*) '中等率',
       sum(case when score between 80 and 90 then 1 else 0 end) / count(*) '优良率',
       sum(case when score >=90 then 1 else 0 end) / count(*) '优秀率'
from sc s
         left join course c on s.cid = c.cid
group by c.cid, c.cname

19、按各科成绩进行排序，并显示排名

#涉及表sc
#窗口函数 row_number()
select s.*, row_number() over (partition by s.cid order by s.score desc ) 排名
from sc s;

20、查询学生的总成绩并进行排名

# 涉及表 student ,sc
# 聚合函数 sum
# 排序 order by desc
select stu.sid, ifnull(sum(s.score), 0) sumscore
from student stu
         left join sc s on stu.sid = s.sid
group by stu.sid
order by sumscore desc

21、查询不同老师所教不同课程平均分从高到低显示

# 涉及表 teacher, course,sc
# 聚合函数 avg
# 排序 order by desc
select t.tid, t.tname, avg(s.score) avgscore
from teacher t
         left join course c on t.tid = c.tid
         left join sc s on c.cid = s.cid
group by t.tid, t.tname
order by avgscore desc;

22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩

#涉及表 sc
SELECT a.* , (SELECT COUNT(DISTINCT score) FROM SC WHERE CID = a.CID AND score > a.score) + 1 AS 排名
FROM SC AS a HAVING 排名 BETWEEN 2 AND 3 ORDER BY a.CID,排名;

23、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 , 0-60 及所占百分比

# 涉及表 sc,course
# 聚合函数 sum count round
# case when then else end
select c.cid  '课程编号',
       c.cname '课程名称',
       round(sum(case when s.score between 0 and 60 then 1 else 0 end) / count(*) * 100) '0-60占比%',
       round(sum(case when s.score between 60 and 70 then 1 else 0 end) / count(*) * 100) '70-60占比%',
       round(sum(case when s.score between 70 and 85 then 1 else 0 end) / count(*) * 100) '85-70占比%',
       round(sum(case when s.score between 85 and 100 then 1 else 0 end) / count(*) * 100) '100-85占比%'
from sc s
         left join course c on s.cid = c.cid
group by c.cid, c.cname

24、查询学生平均成绩及其名次

# 涉及表 sc student
# 聚合函数 avg
select stu.sname,avg(s.score)
from sc s
left join student stu on s.sid = stu.sid
group by stu.sname,stu.sid

25、查询各科成绩前三名的记录

#涉及表 sc
select s1.sid, s1.score
from sc s1
where (
          select count(*)
          from sc s2
          where s1.cid = s2.cid
            and s1.score < s2.score
      ) < 3
order by s1.cid, s1.score desc;

26、查询每门课程被选修的学生数

# 涉及表 course sc
select c.cid, c.cname, count(*)
from course c
         left join sc s on c.cid = s.cid
group by c.cid, c.cname;

27、查询出只有两门课程的全部学生的学号和姓名

# 涉及表 student sc
select stu.sid, stu.sname
from student stu
         left join sc s on stu.sid = s.sid
group by stu.sid, stu.sname
having count(s.cid) = 2

28、查询男生、女生人数

# 涉及表 student
# 聚合函数 count--分组
select stu.ssex, count(*)
from student stu
group by stu.ssex

29、查询名字中含有"风"字的学生信息

# 涉及表 student
# 模糊查询 like '%风%'
select stu.*
from student stu
where stu.sname like '%风%';

30、查询同名同姓学生名单，并统计同名人数

# 涉及表 student
# 内连接，不排除自身情况下，若出现同名同姓的至少大于1
select stu1.sid, stu2.sname
from student stu1
         join student stu2 on stu1.sname = stu2.sname
group by stu1.sid, stu2.sname
having count(*) > 1

31、查询1990年出生的学生名单(注：Student表中Sage列的类型是datetime)

# 涉及表 student
# 模糊查询 like '1990%'
select stu.*
from student stu
where stu.saged like '1990%'

32、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

# 涉及表 sc
# 聚合函数 avg---分组
select s.cid,avg(s.score) avgscore
from sc s
group by s.cid
order by avgscore desc ,s.cid asc ;

33、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

# 涉及表 student sc
# 聚合函数 avg---分组
# having 过滤
select stu.sid, stu.sname, avg(s.score) avgscore
from student stu
         left join sc s on stu.sid = s.sid
group by stu.sid, stu.sname
having avgscore >= 85

35、查询所有学生的课程及分数情况

# 涉及表：student sc,course 
select stu.sid, stu.sname, s.cid, c.cname, s.score
from student stu
         left join sc s on stu.sid = s.sid
         left join course c on s.cid = c.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

# 涉及表 student ,sc,course
select stu.sname, c.cname, s.score
from student stu
         left join sc s on stu.sid = s.sid
         left join course c on s.cid = c.cid
where s.score >= 70;

37、查询不及格的课程

# 涉及表 sc course
select distinct s.sid, s.cid, c.cname
from sc s
         left join course c on s.cid = c.cid
where s.score < 60;

38、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

# 涉及表 student sc
select stu.sid, stu.sname
from student stu
         left join sc s on stu.sid = s.sid
where s.cid = 01
  and s.score >= 80

39、求每门课程的学生人数

# 涉及表 sc course
# 聚合函数 count
select c.cname, count(s.cid)
from sc s
         left join course c on s.cid = c.cid
group by c.cid, c.cname

40、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

# 涉及表 student sc course teacher
# 分页查询 1 ---显示第一个
select stu.*, s.score
from student stu
         left join sc s on stu.sid = s.sid
         left join course c on s.cid = c.cid
         left join teacher t on c.tid = t.tid
where t.tname = '张三'
order by s.score desc
limit 1;

41、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 s.*
from sc s,
     (
         select cid, score
         from sc
         group by cid, score
         having count(*) > 1
     ) s2
where s.cid = s2.cid
  and s.score = s2.score
order by s.cid, s.sid, s.score;

42、查询每门功课成绩最好的前两名

# 涉及表 sc
select s.cid,s.score
from sc s
where (
    select count(*)
    from sc s2
    where s.cid=s2.cid and s.score<s2.score
          )<2
order by s.cid,s.score desc;

43、统计每门课程的学生选修人数（超过5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

# 涉及表 sc 
# 聚合函数 count
select count(s.cid) countpeople
from sc s
group by s.cid
having countpeople>5
order by countpeople desc ,s.cid asc;

44、检索至少选修两门课程的学生学号

# 涉及表 student
select stu.sid
from student stu
         join sc s on stu.sid = s.sid
group by stu.sid
having count(s.cid) >= 2;

45、查询选修了全部课程的学生信息

# 涉及表 student sc
select stu.*
from student stu
         join (
    select s.sid
    from sc s
    group by s.sid
    having count(s.sid) = 3
) t on stu.sid = t.sid;

46、查询各学生的年龄 只按年份来算

# 涉及表 student
# 函数 year(),now(）
select stu.sname, (year(now()) - year(stu.saged)) age
from student stu;

47、查询本周过生日的学生

# 涉及表 student
# 函数 week(),now()
select *
from student stu
where week(saged) = week(now());

48、查询下周过生日的学生

# 涉及表 student
# 函数 week(),now()
select *
from student stu
where week(saged)=week(now())+1;

49、查询本月过生日的学生

# 涉及表 student
# 函数 month(),now()
select *
from student stu
where month(now())=month(stu.saged);

50、查询下月过生日的学生

# 涉及表 student
# 函数 month(),date_sub(),curdate()
select
    stu.*
from student stu
where month(date_sub(saged,interval 1 month)) = month(curdate());