50个常用的SQL语句练习.doc

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
7、 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 <score;
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#`;

24、 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

33、 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