2021-11-15 MySQL 查询练习

1. 查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

   -- 思路：查询sid相等且'01'课程>'02'课程的信息，然后与student表联结
   select * from student RIGHT JOin(
      select t1.sid, score1,score2 from
         (select sid, cid, score as score1 from sc where cid='01') as t1,
         (select sid, cid, score as score2 from sc where cid='02') as t2
      where t1.sid=t2.sid and score1>score2) as r
   on student.sid=r.sid;
   
   -- 1.1 查询同时存在" 01 "课程和" 02 "课程的情况
   -- 思路：分别查询课程号='01'和'02'表，然后做内联结
   select * from
   (select * from sc where cid='01') as t1,
   (select * from sc where Cid='02') as t2
   where t1.sid=t2.sid;
   
   -- 等价于
   select * from
   (select * from sc where cid='01') as t1 inner join
   (select * from sc where Cid='02') as t2
   on t1.sid=t2.sid;
   
   
   -- 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 )
   -- 思路：分别查询课程号='01'和'02'表，然后做左联结
   select * from
   (select * from sc where cid='01') as t1 left join
   (select * from sc where Cid='02') as t2
   on t1.sid=t2.sid;
   
   
   -- 1.3 查询不存在" 01 "课程但存在" 02 "课程的情况
   -- 思路1：分别查询课程号='01'和'02'表，然后做有右联结，再用where is NULL
   select * from
      (select * from sc where cid='01') as t1 RIGHT JOin
      (select * from sc where Cid='02') as t2
   on t1.sid=t2.sid
   where t1.cid is null;
   
   -- 思路2：先查询课程号='01'的学生Id，然后用where判断学生Id不在这里面的，并且课程号='02'
   select * from sc
   where sc.sid not in (
      select sid from sc where sc.cid = '01'
   )
   and sc.cid= '02';
   

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

   -- 思路：使用group by之后再having判断平均成绩，与student表联结得到学生姓名
   select t1.sid,t2.sname,avg(score) as average  from sc as t1
   left join student as t2
   on t1.sid=t2.sid
   group by t1.sid,t2.sname
   having average>=60;
   

3. 查询在 SC 表存在成绩的学生信息

   -- 思路：查询成绩表中成绩不为0的学生sid，然后where判断学生表中的sid在其中
   select * from student
   where sid in (select distinct(sid) from sc where score is not NULL);
   

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 )

   -- 思路：先对成绩表进行group by然后用学生表left join
   select student.sid,student.sname,t1.cc,t1.ss from student left join
      (select sc.Sid,count(sc.cid) as cc,sum(sc.score) as ss from sc
      group by sc.sid) as t1
   on student.sid=t1.sid;
   
   -- 4.1 查有成绩的学生信息
   -- in()适用于student表大于score表的情况
   select * from student
   where student.sid in
      (select sid from sc where score is not NULL);
   
   -- exists()适用于score表大于student表的情况
   select * from student
   where exists (select sc.sid from sc where student.sid=sc.sid);
   

5. 查询「李」姓老师的数量

select '李老师' as name,count(tname) as num from teacher
where tname like '李%';

6. 查询学过「张三」老师授课的同学的信息

   -- 思路1：使用子查询
   select * from student
   where  sid in (
      select sid from sc where cid in(
         select cid from course,teacher where course.tid=teacher.tid and teacher.tname='张三'));
   -- 思路2：多表联查
   select student.* from student,sc,course,teacher
   where student.sid=sc.sid
            and sc.cid=course.cid
         and course.tid=teacher.tid
         and teacher.tname='张三';
   

7. 查询没有学全所有课程的同学的信息

   select * from student
   where sid not in(
      select sc.sid from sc
      group by sc.sid
      having count(sc.cid)=(select count(cid) from course)
   );
   

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

   -- 思路：先求出学号为'01'同学的课程，然后成绩表sc中的课程cid在里面
   select * from student
   where student.sid in (
      select sc.sid from sc
      where sc.cid in(
         select sc.cid from sc
         where sc.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

   -- 思路：否定之否定，等于肯定。先查询 只要学过的课程不在"01"同学课程中 的sid，那么其它的sid对应课程肯定都在01同学的课程中，最后限制数量相等
   select * from sc
   where sid in(
      select sid from sc
      where sid  not in(
         select sid from sc where cid not in (select cid from sc where sid='01')
      )
      group by sid
      having count(*)=(select count(*) from sc where sid='01')
   );
   

10. 查询没学过"张三"老师讲授的任一门课程的学生姓名

    -- 思路：查询有学过"张三"老师的课的学生sid，然后排除这些sid
    select sid,sname from student
    where sid not in(
       select sid from sc,course,teacher
       where sc.cid=course.cid
             and course.tid=teacher.tid
             and teacher.tname='张三'
    );
    

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

    -- 先查询大于2门不及格的学生sid，再联查其平均成绩和姓名
    select t1.sid,t2.avgs,student.sname from
       (select sid from sc
       where score<60
       group by sid
       having count(*)>=2) as t1
    left join(
       select sid,avg(score)as avgs from sc group by sid) as t2
    on t1.sid=T2.sid
    left join student
    on t1.sid=student.sid;
    

12. 检索" 01 "课程分数小于 60，按分数降序排列的学生信息

    select student.*,sc.score from sc,student
    where sc.sid=student.sid and sc.cid='01' and sc.score<60
    order by 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

    -- 思路：使用开窗函数over()
    select sid,cid,score,
    avg(score) over(partition by sid) as avgs
    from sc;
    

14. 查询各科成绩最高分、最低分和平均分

    -- 思路：直接使用max()/min()/avg()/sum()和case语句
    /*
    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 sc.cid,course.cname,max(score) as maxs,min(score)as mins,avg(score)as avgs,
    sum(case when score>=60 then 1
          else 0 end)/count(*) as '及格率',
    sum(case when score>=70 and score<80 then 1
          else 0 end)/count(*) as '中等率',
    sum(case when score>=80 and score<90 then 1
          else 0 end)/count(*) as '优良率',
    sum(case when score>=90 then 1
          else 0 end)/count(*) as '优秀率',
    count(*) as '选修人数'
    from sc left join course
    on sc.cid=course.cid
    group by sc.cid,course.cname
    order by '选修人数' desc,sc.cid ;
    

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 sc.*,rank() over(partition by cid order by score desc) as 'rank' from sc;
    -- 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.*,dense_rank() over(partition by cid order by score desc) as 'rank' from sc;
    -- 15.2 按各科成绩进行排序，并显示排名， Score 重复时随机排名
    select sc.*,row_number() over(partition by cid order by score desc) as 'rank' from sc;
    

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 sc.sid,sum(score) as 'sums',rank() over(order by sum(score) DESC) as 'rank'
    from sc
    group by sid;
    -- 16.1 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 sc.sid,sum(score) as 'sums',dense_rank() over(order by sum(score) DESC) as 'rank'
    from sc
    group by sid;
    

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] 及所占百分比

    select sc.cid,course.cname,
    sum(case when score<=60 then 1 else 0 end) as '[0-60]',
    concat(round(sum(case when score<60 then 1 else 0 end)/count(*)*100,2),'%') as '[0-60]%',
    sum(case when score>=60 and score<70 then 1 else 0 end) as '[70-60]',
    concat(round(sum(case when score>=60 and score<70 then 1 else 0 end)/count(*)*100,2),'%') as '[70-60]%',
    sum(case when score>=70 and score<85 then 1 else 0 end) as '[85-70]',
    concat(round(sum(case when score>=70 and score<85 then 1 else 0 end)/count(*)*100,2),'%') as '[85-70]%',
    sum(case when score>=85 then 1 else 0 end) as '[100-85]',
    concat(round(sum(case when score>=85 then 1 else 0 end)/count(*)*100,2),'%') as '[100-85]%'
    from sc left join course
    on sc.cid=course.cid
    group by sc.cid,course.cname;
    

18. 查询各科成绩前三名的记录

    select * from sc as a
    where 3>(select count(*) from sc as b where a.cid=b.cid and a.score

19. 查询每门课程被选修的学生数

    select cid,count(*) as num
    from sc
    group by cid;
    

20. 查询出只选修两门课程的学生学号和姓名

    select sc.sid, student.sname
    from sc left join student
    on sc.sid=student.sid
    group by sid,sname
    having count(*)=2;
    

21. 男生、女生人数

    select ssex,count(*) as num
    from student
    group by ssex;
    

22. 查询名字中含有「风」字的学生信息

    select * from student
    where sname LIKE '%风%';
    

23. 查询同名学生名单，并统计同名人数

    select * from student
    where sname in(
       select sname from student group by sname having count(*)>1
    );
    

24. 查询 1990 年出生的学生名单

    select * from student
    where YEAR(Sage)=1990;
    

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) as avgs
    from sc
    group by cid
    order by avg(score) DESC,cid;
    

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 sc.sid,student.sname,avg(score) as 'avgs'
    from sc left join student
    on sc.sid=student.sid
    group by sc.sid,student.sname
    having avg(score)>=85;
    

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 student.sname,sc.score,course.cname
    from student,sc,course
    where course.cname='数学'
          and course.cid=sc.cid
          and sc.sid=student.sid
          and sc.score<60;
    

28. 查询所有学生的课程及分数情况

    select * from student
    left join sc
    on student.sid=sc.sid;
    

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

    -- 思路：有分数低于70的学生sid就排除
    select student.sname,course.cname,sc.score
    from   sc left join student
    on sc.sid=student.sid
    left join course
    on sc.cid=course.cid
    where sc.sid not in(select sid from sc where score<=70)
    order by student.sname;
    

30. 查询存在不及格的课程

    select * from sc where score<60;
    

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

32. 每门课程的学生人数

    select cid,count(*) as 'Snum'
    from sc
    group by
    cid;
    

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 student.*,sc.score from sc,student,course,teacher
    where teacher.tname='张三'
             and teacher.tid=course.tid
          and course.cid=sc.cid
             and sc.sid=student.sid
    order by sc.score DESC
    limit 1;
    

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 student.*,sc.score,
    rank() over(order by score DESC) as 'rank'
    from sc,student,course,teacher
    where teacher.tname='张三'
             and teacher.tid=course.tid
          and course.cid=sc.cid
             and sc.sid=student.sid;
    

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 distinct t1.sid,t1.cid,t1.score from sc as t1
    inner join sc as t2
    on t1.sid=t2.sid
    where t1.score=t2.score and t1.cid<>t2.cid;
    

36. 查询每门功成绩最好的前两名

    -- 感觉用dense_rank()更公平一些，并列算一个人，排名连续。
    select * from(
       select *,
       dense_rank() over(partition by cid order by score DESC) as 'rank'
       from sc) as t1
    where t1.rank<3;
    
    select * from(
       select *,
       rank() over(partition by cid order by score DESC) as 'rank'
       from sc) as t1
    where t1.rank<3;
    

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 cid,count(*) as 'Snum'
    from sc
    group by cid
    having count(*)>5;
    

38. 检索至少选修两门课程的学生学号

    select sid,count(*) as 'Cnum'
    from sc
    group by sid
    having count(*)>2;
    

39. 查询选修了全部课程的学生信息

    select * from student
    where sid in(
       select sid from sc
       group by sid
       having count(sc.cid)=(select count(course.cid) from course)
       );
    

40. 查询各学生的年龄，只按年份来算

    select student.*,YEAR(NOW())-YEAR(Sage) as 'age' from student;
    

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 student.*,
    (
    CASE WHEN MonTH(NOW())

42. 查询本周过生日的学生

    select * from student
    where WEEKOFYEAR(NOW())=WEEKOFYEAR(Sage);
    

43. 查询下周过生日的学生

    select * from student
    where WEEKOFYEAR(NOW())+1=WEEKOFYEAR(Sage);
    

44. 查询本月过生日的学生

    select * from student
    where MonTH(NOW())=MonTH(Sage);
    

45. 查询下月过生日的学生

    select * from student
    where MonTH(NOW())+1=MonTH(Sage);