值得思考的sql问题

先给出关系型数据库 表结构：

Student(Sn,Sname,Sage,Ssex) 学生表       Sn：学号；Sname：学生姓名；Sage：学生年龄；Ssex：学生性别

Course(Cn,Cname,Tn) 课程表                    Cn,课程编号；Cname：课程名字；Tn：教师编号

SC(Sn,Cn,score) 成绩表                             Sn：学号；Cn,课程编号；score：成绩

Teacher(Tn,Tname) 教师表                        Tn：教师编号； Tname：教师名字

 

 

以下语句在Mysql 5.6运行通过，问题（问题编号后的*越多表示难度越大）

1、查询“001”课程比“002”课程成绩高的所有学生的学号；

  select a.Sn from (select sn,score from SC where Cn='001') a,(select sn,score

  from SC where Cn='002') b

  where a.score>b.score and a.sn=b.sn;

 

2、查询平均成绩大于60分的同学的学号和平均成绩；

    select Sn,avg(score)

    from sc

    group by Sn having avg(score) >60;

 

3、查询所有同学的学号、姓名、选课数、总成绩；

  select Student.Sn,Student.Sname,count(SC.Cn),sum(score)

  from Student left Outer join SC on Student.Sn=SC.Sn

  group by Student.Sn,Sname;

 

4、查询姓“李”的老师的个数；

  select count(distinct(Tname))

  from Teacher

  where Tname like '李%';

 

5、查询没学过“叶平”老师课的同学的学号、姓名；

    select Student.Sn,Student.Sname

    from Student 

    where Sn not in (select distinct( SC.Sn) from SC,Course,Teacher where  SC.Cn=Course.Cn and Teacher.Tn=Course.Tn and Teacher.Tname='叶平');

6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；

  select Student.Sn,Student.Sname from Student,SC where Student.Sn=SC.Sn and SC.Cn='001'and exists( Select * from SC as SC_2 where SC_2.Sn=SC.Sn and SC_2.Cn='002');

7、查询学过“叶平”老师所教的所有课的同学的学号、姓名；

  select Sn,Sname

  from Student

  where Sn in (select Sn from SC ,Course ,Teacher where SC.Cn=Course.Cn and Teacher.Tn=Course.Tn and Teacher.Tname='叶平' group by Sn having count(SC.Cn)=(select count(Cn) from Course,Teacher  where Teacher.Tn=Course.Tn and Tname='叶平'));

8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；

  Select Sn,Sname from (select Student.Sn,Student.Sname,score ,(select score from SC SC_2 where SC_2.Sn=Student.Sn and SC_2.Cn='002') score2

  from Student,SC where Student.Sn=SC.Sn and Cn='001') S_2 where score2

9、查询所有课程成绩小于60分的同学的学号、姓名；

  select Sn,Sname

  from Student

  where Sn not in (select Student.Sn from Student,SC where S.Sn=SC.Sn and score>60);

10、查询没有学全所有课的同学的学号、姓名；

    select Student.Sn,Student.Sname

    from Student,SC

    where Student.Sn=SC.Sn group by  Student.Sn,Student.Sname having count(Cn) <(select count(Cn) from Course);

11、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名；

    select Sn,Sname from Student,SC where Student.Sn=SC.Sn and Cn in select Cn from SC where Sn='1001';

12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名；

    select distinct SC.Sn,Sname

    from Student,SC

    where Student.Sn=SC.Sn and Cn in (select Cn from SC where Sn='001');

13、把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩；

    update SC set score=(select avg(SC_2.score)

    from SC SC_2

    where SC_2.Cn=SC.Cn ) from Course,Teacher where Course.Cn=SC.Cn and Course.Tn=Teacher.Tn and Teacher.Tname='叶平');

14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名；(下面语句错的,因为是先sql是先用where先筛选条件再group by的)

    select Sn from SC where Cn in (select Cn from SC where Sn='1002')

    group by Sn having count(*)=(select count(*) from SC where Sn='1002');

15**、删除学习“叶平”老师课的SC表记录；

    Delect SC

    from SC, Course ,Teacher 

    where Course.Cn=SC.Cn and Course.Tn= Teacher.Tn and Tname='叶平';

16**、向SC表中插入一些记录，这些记录要求符合以下条件：没有上过编号“003”课程的同学学号、2、此号课的平均成绩；

    Insert into SC select Sn,'002',(Select avg(score)

    from SC where Cn='002') from Student where Sn not in (Select Sn from SC where Cn='002');

17**、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩，按如下形式显示： 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分

    SELECT Sn as 学生ID

        ,(SELECT score FROM SC WHERE SC.Sn=t.Sn AND Cn='004') AS 数据库

        ,(SELECT score FROM SC WHERE SC.Sn=t.Sn AND Cn='001') AS 企业管理

        ,(SELECT score FROM SC WHERE SC.Sn=t.Sn AND Cn='006') AS 英语

        ,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩

    FROM SC AS t

    GROUP BY Sn

    ORDER BY avg(t.score) 

18、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.Cn, max(s1.score), min(s1.score) from Course c left join SC s1 on c.Cn=s1.Cn group by s1.Cn; 

与下面语句有差异么：

    SELECT L.Cn As 课程ID,L.score AS 最高分,R.score AS 最低分

    FROM SC L ,SC AS R

    WHERE L.Cn = R.Cn and

        L.score = (SELECT MAX(IL.score)

                      FROM SC AS IL,Student AS IM

                      WHERE L.Cn = IL.Cn and IM.Sn=IL.Sn

                      GROUP BY IL.Cn)

        AND

        R.Score = (SELECT MIN(IR.score)

                      FROM SC AS IR

                      WHERE R.Cn = IR.Cn

                  GROUP BY IR.Cn

                    );

 

19、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

用这个语句：

select c.Cn, avg(s1.score) pingjun, count(s1.Sn) renshu, 

(select count(1) from SC s2 where s2.Cn=s1.Cn and s2.score>=60)/count(Sn)*100 as hegelv

from Course c left join SC s1 on c.Cn=s1.Cn group by s1.Cn order by pingjun asc, hegelv desc;

 

有这么复杂么：

    SELECT t.Cn AS 课程号,max(course.Cname)AS 课程名,isnull(AVG(score),0) AS 平均成绩

        ,100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) AS 及格百分数

    FROM SC T,Course

    where t.Cn=course.Cn

    GROUP BY t.Cn

    ORDER BY 100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC

20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理（001），马克思（002），OO&UML （003），数据库（004）

 

select Cn, avg(score),(select count(1) from SC s2 where s2.Cn=s1.Cn and s2.score>=60)/count(Sn)*100 as hegelv

from SC s1 where Cn=4004 or Cn=4005 group by Cn;

 

要这么复杂么：

 

    SELECT SUM(CASE WHEN Cn ='001' THEN score ELSE 0 END)/SUM(CASE Cn WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分

        ,100 * SUM(CASE WHEN Cn = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cn = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数

        ,SUM(CASE WHEN Cn = '002' THEN score ELSE 0 END)/SUM(CASE Cn WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分

        ,100 * SUM(CASE WHEN Cn = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cn = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数

        ,SUM(CASE WHEN Cn = '003' THEN score ELSE 0 END)/SUM(CASE Cn WHEN '003' THEN 1 ELSE 0 END) AS UML平均分

        ,100 * SUM(CASE WHEN Cn = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cn = '003' THEN 1 ELSE 0 END) AS UML及格百分数

        ,SUM(CASE WHEN Cn = '004' THEN score ELSE 0 END)/SUM(CASE Cn WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分

        ,100 * SUM(CASE WHEN Cn = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN Cn = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数

  FROM SC;

 

21、查询不同老师所教不同课程平均分从高到低显示(以下语句错，因为有些老师的课没人选的话这样就不会显示出来了)

  SELECT max(Z.Tn) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.Cn AS 课程ＩＤ,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩

    FROM SC AS T,Course AS C ,Teacher AS Z

    where T.Cn=C.Cn and C.Tn=Z.Tn

  GROUP BY C.Cn

  ORDER BY AVG(Score) DESC;

 

22、查询如下课程成绩第 3 名到第 6 名的学生成绩单：企业管理（001），马克思（002），UML （003），数据库（004）(描述不请，什么课程的3到6名)

    [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩

    SELECT  DISTINCT top 3

      SC.Sn As 学生学号,

        Student.Sname AS 学生姓名 ,

      T1.score AS 企业管理,

      T2.score AS 马克思,

      T3.score AS UML,

      T4.score AS 数据库,

      ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) as 总分

      FROM Student,SC  LEFT JOIN SC AS T1

                      ON SC.Sn = T1.Sn AND T1.Cn = '001'

            LEFT JOIN SC AS T2

                      ON SC.Sn = T2.Sn AND T2.Cn = '002'

            LEFT JOIN SC AS T3

                      ON SC.Sn = T3.Sn AND T3.Cn = '003'

            LEFT JOIN SC AS T4

                      ON SC.Sn = T4.Sn AND T4.Cn = '004'

      WHERE student.Sn=SC.Sn and

      ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)

      NOT IN

      (SELECT

            DISTINCT

            TOP 15 WITH TIES

            ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0)

      FROM sc

            LEFT JOIN sc AS T1

                      ON sc.Sn = T1.Sn AND T1.Cn = 'k1'

            LEFT JOIN sc AS T2

                      ON sc.Sn = T2.Sn AND T2.Cn = 'k2'

            LEFT JOIN sc AS T3

                      ON sc.Sn = T3.Sn AND T3.Cn = 'k3'

            LEFT JOIN sc AS T4

                      ON sc.Sn = T4.Sn AND T4.Cn = 'k4'

      ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC);

 

23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]

    SELECT SC.Cn as 课程ID, Cname as 课程名称

        ,SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS [100 - 85]

        ,SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS [85 - 70]

        ,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS [70 - 60]

        ,SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -]

    FROM SC,Course

    where SC.Cn=Course.Cn

    GROUP BY SC.Cn,Cname;

24***、查询学生平均成绩及其名次

 

这句可以，但并列的情况显示并列第几没处理好:

select tb_avg.*,  (@rownum:=@rownum + 1) as paiming

from

(select s1.Sn, avg(s1.score) pingjun 

from SC s1 group by s1.Sn order by pingjun desc) tb_avg,  (select (@rownum:=0)) tb2 ;

 

下面子查询太多：

      SELECT 1+(SELECT COUNT( distinct 平均成绩)

               FROM (SELECT Sn,AVG(score) AS 平均成绩

                       FROM SC

                   GROUP BY Sn

                   ) AS T1

             WHERE 平均成绩 > T2.平均成绩) as 名次,

       Sn as 学生学号, 平均成绩

    FROM (SELECT Sn, AVG(score) 平均成绩 FROM SC GROUP BY Sn) AS T2

    ORDER BY 平均成绩 desc;

 

25、查询各科成绩前三名的记录:(不考虑成绩并列情况)

有比这更好写法么：

 

select s1.Cn, 

(select s2.Sn from SC s2 where s2.Cn=s1.Cn  order by score desc, Sn asc limit 0,1) as No1,

(select s2.Sn from SC s2 where s2.Cn=s1.Cn  order by score desc, Sn asc limit 1,1) as No2,

(select s2.Sn from SC s2 where s2.Cn=s1.Cn  order by score desc, Sn asc limit 2,1) as No3

from SC s1 group by s1.Cn;

 

 

30、查询同名同性学生名单，并统计同名人数

  select Sname,count(*) from Student group by Sname having  count(*)>1;;

31、1981年出生的学生名单

    select Sname,  2016-Sage as age from student where (2016-Sage)=1981;

 

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 Sname,SC.Sn ,avg(score)

    from Student,SC

    where Student.Sn=SC.Sn group by SC.Sn,Sname having    avg(score)>85;

 

35、查询所有学生的选课情况(包括没选课的)

    SELECT Student.Sn, Student.Sname, SC.Cn,Cname

    FROM Student left jion CS on  SC.Sn=Student.Sn 

left jion Course on SC.Cn=Course.Cn ;

 

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

（从下面语句看出，distinct是对查出结果整行而言的, 就算写成select distinct(student.Sn) ,student.Sname, SC.Cn, SC.score FROM student,Sc from ... 结果也是一样的）

    SELECT  distinct student.Sn,student.Sname,SC.Cn,SC.score

    FROM student,Sc

    WHERE SC.score>=70 AND SC.Sn=student.Sn;

 

40、查询选修“叶平”老师所授课程的学生中，成绩最高的学生姓名及其成绩

注意不能用下面的语句, 因为max函数不能关联本行的, 这样查出的姓名可能是错的：

 select Student.Sname,max(score) from Student,SC,Course C,Teacher

    where Student.Sn=SC.Sn and SC.Cn=C.Cn and C.Tn=Teacher.Tn and Teacher.Tname='叶平';

 

要用这个语句：

    select Student.Sname,score

    from Student,SC,Course C,Teacher

    where Student.Sn=SC.Sn and SC.Cn=C.Cn and C.Tn=Teacher.Tn and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where Cn=C.Cn );

 

43***、查询每门功成绩最好的前两名 

Mysql的limit在子关联查询中用不了，如何解?dong?

 

下面语句在mysql不能用：

 

    SELECT t1.Sn as 学生ID,t1.Cn as 课程ID,Score as 分数

      FROM SC t1

      WHERE score IN (SELECT TOP 2 score

              FROM SC

              WHERE t1.Cn= Cn

            ORDER BY score DESC

              )

      ORDER BY t1.Cn;

 

 

48、查询两门以上不及格课程的同学的学号及其平均成绩

 

下面哪条语句执行快：

select Sn, avg(score) from SC where Sn in ( select Sn  from SC where score<60 group by Sn having count(score)>=2) group by Sn;

 

或者：

select s1.Sn, avg(s1.score) from SC s1, 

( select Sn, count(score) as  bujigeshu from SC where score<60 group by Sn having count(score)>=2) cl60 

where s1.Sn = cl60.Sn  group by Sn;

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