50个常用的SQL语句练习(数据库考试复习大招)

大家对数据库等有什么问题，欢迎私聊和底下留言

基本信息Student(`S#`,Sname,Sage,Ssex) 学生表
Course(`C#`,Cname,`T#`) 课程表
SC(`S#`,`C#`,score) 成绩表
Teacher(`T#`,Tname) 教师表

问题：
1、查询“001”课程比“002”课程成绩高的所有学生的学号；
select a.`S#` from (select `S#`,score from SC where `C#`='001') a,(select `S#`,score
from SC where `C#`='002') b where a.score>b.score and a.`S#`=b.`S#`;

↑一张表中存在多对多情况的

2、查询平均成绩大于60分的同学的学号和平均成绩； 
答案一：select `S#`,avg(score) from sc  group by `S#` having avg(score) >60;

↑一对多，对组进行筛选

答案二：SELECT s ,scr

FROM (SELECT sc.`S#` s,AVG(sc.`score`) scr FROM sc GROUP BY sc.`S#`) rs

WHERE rs.scr>60 ORDER BY rs.scr DESC      

↑嵌套查询可能影响效率

3、查询所有同学的学号、姓名、选课数、总成绩；
答案一：select Student.`S#`,Student.Sname,count(`C#`),sum(score) from Student left Outer join SC on Student.`S#`=SC.`S#` group by Student.`S#`,Sname

↑如果学生没有选课，仍然能查出，显示总分null（边界情况）

答案二：SELECT student.`S#`,student.`Sname`,COUNT(sc.`score`) 选课数,SUM(sc.`score`) 总分

FROM Student,sc

WHERE student.`S#`=sc.`S#` GROUP BY sc.`S#`

↑如果学生没有选课，sc表中没有他的学号，就查不出该学生，有缺陷！

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.`S#`,Student.Sname  from Student  where `S#` not in (select distinct(SC.`S#`) from SC,Course,Teacher where SC.`C#`=Course.`C#` and Teacher.`T#`=Course.`T#` and Teacher.Tname='叶平');

↑反面思考Step1：先找学过叶平老师课的学生学号，三表联合查询

Step2：在用not in 选出没学过的

Step3：distinct以防叶平老师教多节课；否则若某同学的几节课都由叶平教，学号就会出现重复

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.`S#`,Student.Sname from Student,SC where Student.`S#`=SC.`S#` and SC.`C#`='001'and exists( Select * from SC as SC_2 where SC_2.`S#`=SC.`S#` and SC_2.`C#`='002' );

↑注意目标字段`S#`关联

exists subquery 可以用in subquery代替，如下

select Student.`S#`,Student.Sname from Student,Sc where Student.`S#`=SC.`S#` and SC.`C#`='001'and sc.`s#` in ( select sc_2.`s#` from sc as sc_2 where   sc_2.`c#`='002' );

↑不同之处，in subquery此处就不需要关联了

1. 查询学过“叶平”老师所教的所有课的同学的学号、姓名； 
   select `S#`,Sname from Student

where `S#` in

(select `S#` from SC ,Course ,Teacher where SC.`C#`=Course.`C#` and Teacher.`T#`=Course.`T#` and Teacher.Tname='叶平' group by `S#` 

having count(SC.`C#`)=(select count(`C#`) from Course,Teacher where Teacher.`T#`=Course.`T#` 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； 
（太混乱）Select `S#`,Sname from (select Student.`S#`,Student.Sname,score ,(select score from SC SC_2 where SC_2.`S#`=Student.`S#` and SC_2.`C#`='002') score2 from Student,SC where Student.`S#`=SC.`S#` and `C#`='001') S_2 where score2

自己写的另一种方法：

SELECT student.`S#`,student.Sname FROM student

WHERE `S#` IN

( SELECT a.`S#` FROM

(SELECT * FROM sc WHERE `C#`='001') a , 

(SELECT * FROM sc WHERE `C#`='002') b

  WHERE a.score>b.score AND a.`S#`=b.`S#`

) ;

↑子查询的应用方式与第1题类似，

在一对多关系表中，如果多组之间需要比较，可以将不同组抽出为几个子查询，再比较。

这里的“一”指课程编号。

9、查询所有课程成绩小于60分的同学的学号、姓名； 

↓初始答案（效率最低）：
select `S#`,Sname from Student where `S#` not in (select Student.`S#` from Student,SC where Student.`S#`=SC.`S#` and score>60); （第二个select根本不需要联合查询）

↓改进简化版（效率更高）：

select `S#`,Sname from Student

where `S#` not in

(select distinct `S#` from SC where score>60);  （从反面思考更简化）

↓自己写的另一种方法（效率其次，但有缺陷。边界情况：没有学任何课程的人，查不出来）：

SELECT Student.`S#`,Student.Sname FROM Student

WHERE `S#` IN

(SELECT `S#` FROM sc GROUP BY `S#` HAVING MAX(score)<60);

In 和not in 去构造，有时候查出来的结果并不一样，需要考虑目标字段`S#`是否在几个表中都有

10、查询没有学全所有课的同学的学号、姓名； 
select Student.`S#`,Student.Sname from Student,SC  where Student.`S#`=SC.`S#` group by Student.`S#`,Student.Sname having count(`C#`) <(select count(`C#`) from Course);

↑有缺陷，没有选任何课的人查不出来。因为使用了关联查询，若存在关联不上的（一张表有，另一张表没有），就会遗漏。

 

select student.`s#`,student.sname from student

where student.`s#` not in

(select `s#` from sc group by `s#` having count(`c#`) = (select count(`c#`) from course));

↑可以查出没有选任何课的人，单表查询操作，不涉及关联。

11、查询至少有一门课与学号为“P1001”的同学所学相同的同学的学号和姓名； 
select DISTINCT Student.`S#`,Sname from Student,SC where Student.`S#`=SC.`S#` and `C#` in (select `C#` from SC where `S#`='P1001') ; (存在性用in即可）

↑没有排除自身，↓把结果中的P1001自己去掉

12、查询至少学过学号为“P1001”同学所有一门课的其他同学学号和姓名；
select distinct SC.`S#`,Sname  from Student,SC where Student.`S#`=SC.`S#` and `C#` in (select `C#` from SC where `S#`='001') AND Student.`s#` != 'P1001'; ←(绿色为补充，排除P1001本身) 

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；
 （有错误，很混乱）update SC set score=(select avg(SC_2.score)  from SC SC_2  where SC_2.`C#`=SC.`C#` ) from Course,Teacher where Course.`C#`=SC.`C#` and Course.`T#`=Teacher.`T#` and Teacher.Tname='叶平'); mysql报错，可能其他数据库能这么写

题目特点，把一张表的值查出来再插到这张表中，但实际不允许，会报错。

 

自己写：

UPDATE SC SET score=

(SELECT AVG(R.score) FROM

(SELECT score,`C#` FROM SC)  R

 WHERE R.`C#`=sc.`C#` GROUP BY R.`C#`)

WHERE sc.`C#` IN

(SELECT `C#` FROM Course,Teacher WHERE Course.`T#`=Teacher.`T#` 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； 
（错误,条件不够）select `S#` from SC where `C#` in (select `C#` from SC where `S#`='1002') group by `S#` having count(*)=(select count(*) from SC where `S#`='1002');

自己写：

15、删除学习“叶平”老师课的SC表记录； 
(delete后能加表名吗？？)Delete SC from course ,Teacher where Course.`C#`=SC.`C#` and Course.`T#`= Teacher.`T#` and Tname='叶平';

自己写：delete from sc where `C#` in (select `C#` from Course,Teacher where Course.`T#`=Teacher.`T#` 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号课的平均成绩； 
 (省略不看了)Insert SC select `S#`,'002',(Select avg(score) from SC where `C#`='002') from Student where `S#` not in (Select `S#` from SC where `C#`='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 `S#` AS 学生ID
 ,(SELECT score FROM SC WHERE SC.`S#`=t.`S#` AND `C#`='004') AS 数据库
 ,(SELECT score FROM SC WHERE SC.`S#`=t.`S#` AND `C#`='001') AS 企业管理
 ,(SELECT score FROM SC WHERE SC.`S#`=t.`S#` AND `C#`='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 `S#` ORDER BY avg(t.score)

 

自己写：SELECT `S#` AS 学号ID,

(SELECT score FROM sc WHERE sc.`S#`=t.`S#` AND `C#`='001') AS 语文,

(SELECT score FROM sc WHERE sc.`S#`=t.`S#` AND `C#`='002') AS 数学,

(SELECT score FROM sc WHERE sc.`S#`=t.`S#` AND `C#`='003') AS 英语,

(SELECT score FROM sc WHERE sc.`S#`=t.`S#` AND `C#`='004') AS 政治,

COUNT(`C#`) AS 有效课程数,

AVG(score) AS 课程平均分

FROM SC AS t GROUP BY `S#` ORDER BY AVG(t.score) DESC;

必须要有自关联，否则返回不止1行

 

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  DISTINCT L.`C#` AS 课程ID,

        Cname  AS 课程名,

        L.score  AS 最高分,

        R.score  AS 最低分

FROM  SC AS L ,SC AS R ,course

WHERE  L.`C#`= R.`C#`  AND  course.`C#`=L.`C#`

AND  L.score = (SELECT MAX(IL.score)  FROM SC AS IL  WHERE L.`C#` = IL.`C#`  GROUP BY IL.`C#`)

AND  R.Score = (SELECT MIN(IR.score)  FROM SC AS IR  WHERE R.`C#` = IR.`C#`  GROUP BY IR.`C#`); 

自己写：

SELECT `C#` AS课程号,

(SELECT MAX(score) FROM sc WHERE sc.`C#`=t.`C#` GROUP BY `C#`) AS 最高分,

(SELECT MIN(score) FROM sc WHERE sc.`C#`=t.`C#` GROUP BY `C#`) AS 最低分

FROM sc t GROUP BY t.`C#` ;

19、按各科平均成绩从低到高和及格率的百分数从高到低 
 （问题不清，且运行不出  ）SELECT t.`C#` 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.`C#`=course.`C#` GROUP BY t.`C#` 
 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 SUM(CASE WHEN `C#` ='001' THEN score ELSE 0 END)/SUM(CASE `C#` WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分  ,100 * SUM(CASE WHEN `C#` = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数 ,SUM(CASE WHEN `C#` = '002' THEN score ELSE 0 END)/SUM(CASE `C#` WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分,100 * SUM(CASE WHEN `C#` = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数 ,SUM(CASE WHEN `C#` = '003' THEN score ELSE 0 END)/SUM(CASE `C#` WHEN '003' THEN 1 ELSE 0 END) AS UML平均分 ,100 * SUM(CASE WHEN `C#` = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = '003' THEN 1 ELSE 0 END) AS UML及格百分数 ,SUM(CASE WHEN `C#` = '004' THEN score ELSE 0 END)/SUM(CASE `C#` WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分 ,100 * SUM(CASE WHEN `C#` = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数 FROM SC

 

修改后：

SELECT SUM(CASE WHEN `C#` ='001' THEN score ELSE 0 END)/SUM(CASE `C#` WHEN '001' THEN 1 ELSE 0 END) AS 语文平均分,

       100 * SUM(CASE WHEN `C#` = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = '001' THEN 1 ELSE 0 END) AS 语文及格率,

       SUM(CASE WHEN `C#` ='002' THEN score ELSE 0 END)/SUM(CASE `C#` WHEN '002' THEN 1 ELSE 0 END) AS 数学平均分,

       100 * SUM(CASE WHEN `C#` = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = '002' THEN 1 ELSE 0 END) AS 数学及格率,

       SUM(CASE WHEN `C#` ='003' THEN score ELSE 0 END)/SUM(CASE `C#` WHEN '003' THEN 1 ELSE 0 END) AS 英语平均分,

       100 * SUM(CASE WHEN `C#` = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = '003' THEN 1 ELSE 0 END) AS 英语及格率,

       SUM(CASE WHEN `C#` ='004' THEN score ELSE 0 END)/SUM(CASE `C#` WHEN '004' THEN 1 ELSE 0 END) AS 政治平均分,

       100 * SUM(CASE WHEN `C#` = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN `C#` = '004' THEN 1 ELSE 0 END) AS 政治及格率 

FROM SC ;

21、查询不同老师所教不同课程平均分，按课程分数从高到低显示。包括教师ID、教师姓名、课程ID、课程名、平均成绩
SELECT max(Z.`T#`) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.`C#` 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.`C#`=C.`C#` and C.`T#`=Z.`T#` GROUP BY C.`C#` ORDER BY AVG(Score) DESC

Max没必要加！

修改后：

SELECT Z.`T#` AS 教师ID,

       Z.Tname AS 教师姓名,

       C.`C#` AS 课程ID,

       C.Cname AS 课程名,

       AVG(Score) AS 平均成绩

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

WHERE T.`C#`=C.`C#` AND C.`T#`=Z.`T#`   GROUP BY C.`C#`   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）  [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩

Mysql 不支持top语句

SELECT DISTINCT top 3 SC.`S#` 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.`S#` = T1.`S#`

     AND T1.`C#` = '001'  LEFT JOIN SC AS T2 ON SC.`S#` = T2.`S#`

     AND T2.`C#` = '002'  LEFT JOIN SC AS T3  ON SC.`S#` = T3.`S#`

     AND T3.`C#` = '003'  LEFT JOIN SC AS T4   ON SC.`S#` = T4.`S#`

     AND T4.`C#` = '004'WHERE student.`S#`=SC.`S#`

     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.`S#` = T1.`S#`

            AND T1.`C#` = 'k1'  LEFT JOIN sc AS T2  ON sc.`S#` = T2.`S#`

            AND T2.`C#` = 'k2'  LEFT JOIN sc AS T3  ON sc.`S#` = T3.`S#`

            AND T3.`C#` = 'k3'  LEFT JOIN sc AS T4  ON sc.`S#` = T4.`S#`

            AND T4.`C#` = '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.`C#` 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.`C#`=Course.`C#`  GROUP BY SC.`C#`,Cname;

修改后：[90,100]指定要加单引号变成一个字符串！

SELECT c.`C#` AS 课程号,

       c.`Cname` AS 课程名称,

       SUM(CASE WHEN score BETWEEN 90 AND 100 THEN 1 ELSE 0 END) AS '[90,100]',

       SUM(CASE WHEN score BETWEEN 75 AND 89 THEN 1 ELSE 0 END) AS '[89,75]',

       SUM(CASE WHEN score BETWEEN 60 AND 74 THEN 1 ELSE 0 END) AS '[74,60]',

       SUM(CASE WHEN score <60 THEN 1 ELSE 0 END) AS '[59,-]'

FROM sc s,course c

WHERE s.`C#`=c.`C#` GROUP BY s.`C#`;

 

1. 查询学生平均成绩及其名次
    SELECT 1+(SELECT COUNT( distinct 平均成绩) FROM (SELECT `S#`,AVG(score) AS 平均成绩  FROM SC GROUP BY `S#`  ) AS T1  WHERE 平均成绩 > T2.平均成绩) as 名次,  `S#` as 学生学号,平均成绩
   FROM (SELECT `S#`,AVG(score) 平均成绩  FROM SC GROUP BY `S#` ) AS T2  ORDER BY 平均成绩 desc;

修改后：

SELECT  1+(

           SELECT COUNT(DISTINCT 平均成绩) FROM

           (SELECT `S#`,AVG(score) AS 平均成绩  FROM SC GROUP BY `S#`) AS T1  

           WHERE 平均成绩 > T2.平均成绩

         )  AS 名次,

       学生学号,

       平均成绩

FROM ( SELECT `S#` AS 学生学号,AVG(score) 平均成绩  FROM SC GROUP BY `S#` ) 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)
SELECT t1.`S#` as 学生ID,t1.`C#` as 课程ID,Score as 分数  

FROM SC t1

WHERE score IN (SELECT TOP 3 score  FROM SC WHERE t1.`C#`= `C#`  ORDER BY score DESC )  ORDER BY t1.`C#`;

26、查询每门课程被选修的学生数
select `C#`,count(`S#`) from sc group by `C#`;

加课程名：

SELECT s.`C#`,Cname,COUNT(DISTINCT`S#`) 选课人数

FROM course c,sc s WHERE s.`C#`=c.`C#`

GROUP BY s.`C#`;

27、查询出只选修了一门课程的全部学生的学号和姓名 
select SC.`S#`,Student.Sname,count(`C#`) AS 选课数 from SC ,Student
where SC.`S#`=Student.`S#` group by SC.`S#` ,Student.Sname having count(`C#`)=1;

自己写：

SELECT  student.`S#`,Sname,选课数

FROM student,

     (SELECT sc.`S#`,COUNT(sc.`C#`) 选课数 FROM sc GROUP BY sc.`S#`)  AS 选课统计

WHERE student.`S#`=选课统计.`S#` AND 选课数=1

;

28、查询男生、女生人数
Select Ssex, count(Ssex) as 人数

from Student

group by Ssex

29、查询姓“张”的学生名单 
SELECT Sname FROM Student WHERE Sname like '张%';

30、查询同名同性别学生名单，并统计同名人数
select Sname,count(*) from Student group by Sname having count(*)>1;

加性别：

SELECT sname,ssex,COUNT(*) 相同人数 FROM student GROUP BY Sname,Ssex HAVING COUNT(*)>1;

31、1981年出生的学生名单(注：Student表中Sage列的类型是datetime)
select Sname, CONVERT(char (11),DATEPART(year,Sage)) as age
from student where CONVERT(char(11),DATEPART(year,Sage))='1981';

Mysql中没有DATEPART，使用DATE_FORMAT

SELECT `S#`,Sname,DATE_FORMAT(Sage,'%Y') 出生日期,2016-DATE_FORMAT(Sage,'%Y') 年龄 FROM student;

32、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列
Select `C#`,Avg(score) from SC group by `C#` order by Avg(score),`C#` DESC ;

                                             优先排序    二级排序

1. 查询平均成绩大于80的所有学生的学号、姓名和平均成绩
   select Sname,SC.`S#` ,avg(score) from Student,SC
   where Student.`S#`=SC.`S#` group by SC.`S#` having  avg(score)>80;

34、查询课程名称为“数学”，且分数低于60的学生姓名和分数，若没选这节课，用0代替。
 Select Sname,score

 from Student,SC,Course
 where SC.`S#`=Student.`S#` and SC.`C#`=Course.`C#` and Course.Cname='数据库'and score ;

原答案并不能显示成绩为零的记录，根本就不存在

怎么写？

用 student left join 。。。

   
35、查询所有学生的选课情况； 
SELECT SC.`S#`, SC.`C#`, Sname, Cname

FROM SC,Student,Course

where SC.`S#`=Student.`S#` and SC.`C#`=Course.`C#` ;

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；
 SELECT distinct student.`S#`,student.Sname,SC.`C#`,SC.score
 FROM student,Sc
 WHERE SC.score>=70 AND SC.`S#`=student.`S#`;

补充后：

SELECT Sname,Cname,score

FROM Student,Course,score

WHERE SC.`S#`=Student.`S#` AND SC.`C#`=Course.`C#` AND score>70;

37、查询P1004同学不及格的课程，并按课程号从大到小排列
 select `C#`  陆华不及格课程 from sc where S#=’P1004’ score <60 order by `C#` ;

38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名； 
SELECT student.`S#`,student.`Sname`

FROM student,course,sc

WHERE SC.`S#`=Student.`S#` AND SC.`C#`=Course.`C#` AND sc.`C#`='003' AND score>80;

39、求选了’001’课程的学生人数
select count(*) from sc where `C#`=’001’;

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
正确答案：四表联合

SELECT Student.Sname,score 最高分

FROM Student,SC,Course C,Teacher

WHERE Student.`S#`=SC.`S#` AND SC.`C#`=C.`C#` AND C.`T#`=Teacher.`T#`

     AND Teacher.Tname='叶平'

     AND SC.score=(SELECT MAX(score) FROM SC WHERE `C#`=C.`C#` );

说明：sc的别名只能在外侧使用，内测使用而外侧不用，只能查出来一门课程，效果并不是等价的。

 

自己写的有错误：人名只能为各组group的第一行，原因：？

SELECT Sname,Cname 叶平老师课程,最高分       

FROM student,course,

     (SELECT sc3.`S#`,sc3.`C#`,MAX(score) 最高分 FROM sc sc3 GROUP BY sc3.`C#`) rs

WHERE rs.`S#`=Student.`S#` AND course.`C#`=rs.`C#`

      AND rs.`C#` IN

      (SELECT sc2.`C#` FROM course,sc sc2,teacher

       WHERE course.`C#`=sc2.`C#`AND teacher.`T#`=course.`T#`

             AND Tname='叶平' GROUP BY sc2.`C#`

      )

41、查询各个课程及相应的选修人数 
SELECT `C#` ,COUNT(*) FROM sc GROUP BY `C#`;

42、查询不同课程成绩相同的学生的学号、课程号、学生成绩
（没有意义）select distinct A.`S#`,B.score from SC A ,SC B where A.Score=B.Score and A.`C#` <>B.`C#` ;

43、查询每门功课成绩最好的前两名 

（不支持top语句）SELECT t1.`S#` as 学生ID,t1.`C#` as 课程ID,Score as 分数  FROM SC t1
 WHERE score IN (SELECT TOP 2 score   FROM SC  WHERE t1.`C#`= `C#`  ORDER BY score DESC   )   ORDER BY t1.`C#`;

44、统计每门课程的学生选修人数（超过10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
select `C#` as 课程号,count(*) as 人数 from sc group by `C#` order by count(*) desc,`C#`;

45、检索至少选修两门课程的学生学号 
 SELECT `S#`,COUNT(*) 选课数 FROM sc GROUP BY `S#` HAVING COUNT(*)>2;
46、查询每个学生都选修的课程的课程号和课程名
 (错误，查询出来结果是所有课程，因为他们都被选修过，不管多少人)

select `C#`,Cname from Course  where `C#` in (select `C#` from sc group by `C#`)

修改后：

SELECT rs.`C#` FROM

(SELECT `C#`,COUNT(*) 选课数量 FROM sc GROUP BY `C#`) rs

WHERE rs.选课数量=(SELECT COUNT(*) FROM student);

47、查询没学过“叶平”老师讲授的任一门课程的学生姓名

三表联合即可，不必太复杂

select Sname from Student where `S#` not in

(select `S#` from Course,Teacher,SC 

 where Course.`T#`=Teacher.`T#` and SC.`C#`=course.`C#` and Tname='叶平');

48、查询两门以上不及格课程的同学的学号及其平均成绩
SELECT sc.`S#`,AVG(score) 平均成绩

FROM sc

WHERE `S#` IN (SELECT `S#`FROM sc WHERE score<60 GROUP BY `S#` HAVING COUNT(*)>2 )

GROUP BY sc.`S#`;

 

 
49、检索“004”课程分数小于60的学生，按分数降序排列的同学学号
select `S#` from SC

where `C#`='004' and score <60

order by score desc;

50、删除“P1002”同学的“001”课程的成绩
delete from Sc where `S#`='P1001' and `C#`='001';