今天在网上找了几道经典的SQL练习题做了一下，虽然都不难，但是对打基础是很有好处的，在明白的基础上可以进一步做分析，来研究一下各种解法的优劣，甚至进行简单的优化。。
现在将题目和答案分享一下。我使用的是MySQL 5.0，但是绝大部分都是标准SQL。
表结构：
CREATE TABLE STUDENT
(SNO VARCHAR(3) NOT NULL,
SNAME VARCHAR(4) NOT NULL,
SSEX VARCHAR(2) NOT NULL,
SBIRTHDAY DATETIME,
CLASS VARCHAR(5))
go
CREATE TABLE COURSE
(CNO VARCHAR(5) NOT NULL,
CNAME VARCHAR(10) NOT NULL,
TNO VARCHAR(10) NOT NULL)
go
CREATE TABLE SCORE
(SNO VARCHAR(3) NOT NULL,
CNO VARCHAR(5) NOT NULL,
DEGREE NUMERIC(10, 1) NOT NULL)
go
CREATE TABLE TEACHER
(TNO VARCHAR(3) NOT NULL,
TNAME VARCHAR(4) NOT NULL, TSEX VARCHAR(2) NOT NULL,
TBIRTHDAY DATETIME NOT NULL, PROF VARCHAR(6),
DEPART VARCHAR(10) NOT NULL)
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (108 ,'曾华'
,'男' ,1977-09-01,95033);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (105 ,'匡明'
,'男' ,1975-10-02,95031);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (107 ,'王丽'
,'女' ,1976-01-23,95033);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (101 ,'李军'
,'男' ,1976-02-20,95033);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (109 ,'王芳'
,'女' ,1975-02-10,95031);
INSERT INTO STUDENT (SNO,SNAME,SSEX,SBIRTHDAY,CLASS) VALUES (103 ,'陆君'
,'男' ,1974-06-03,95031);
GO
INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('3-105' ,'计算机导论',825)
INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('3-245' ,'操作系统' ,804);
INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('6-166' ,'数据电路' ,856);
INSERT INTO COURSE(CNO,CNAME,TNO)VALUES ('9-888' ,'高等数学' ,100);
GO
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (103,'3-245',86);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (105,'3-245',75);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (109,'3-245',68);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (103,'3-105',92);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (105,'3-105',88);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (109,'3-105',76);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (101,'3-105',64);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (107,'3-105',91);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (108,'3-105',78);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (101,'6-166',85);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (107,'6-106',79);
INSERT INTO SCORE(SNO,CNO,DEGREE)VALUES (108,'6-166',81);
GO
INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
VALUES (804,'李诚','男','1958-12-02','副教授','计算机系');
INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
VALUES (856,'张旭','男','1969-03-12','讲师','电子工程系');
INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
VALUES (825,'王萍','女','1972-05-05','助教','计算机系');
INSERT INTO TEACHER(TNO,TNAME,TSEX,TBIRTHDAY,PROF,DEPART)
VALUES (831,'刘冰','女','1977-08-14','助教','电子工程系');
题目：
1、 查询Student表中的所有记录的Sname、Ssex和Class列。
2、 查询教师所有的单位即不重复的Depart列。
3、 查询Student表的所有记录。
4、 查询Score表中成绩在60到80之间的所有记录。
5、 查询Score表中成绩为85，86或88的记录。
6、 查询Student表中“95031”班或性别为“女”的同学记录。
7、 以Class降序查询Student表的所有记录。
8、 以Cno升序、Degree降序查询Score表的所有记录。
9、 查询“95031”班的学生人数。
10、查询Score表中的最高分的学生学号和课程号。
11、查询‘3-105’号课程的平均分。
12、查询Score表中至少有5名学生选修的并以3开头的课程的平均分数。
13、查询最低分大于70，最高分小于90的Sno列。
14、查询所有学生的Sname、Cno和Degree列。
15、查询所有学生的Sno、Cname和Degree列。
16、查询所有学生的Sname、Cname和Degree列。
17、查询“95033”班所选课程的平均分。
18、假设使用如下命令建立了一个grade表：
create table grade(low number(3,0),upp number(3),rank char(1));
insert into grade values(90,100,’A’);
insert into grade values(80,89,’B’);
insert into grade values(70,79,’C’);
insert into grade values(60,69,’D’);
insert into grade values(0,59,’E’);
commit;
现查询所有同学的Sno、Cno和rank列。
19、查询选修“3-105”课程的成绩高于“109”号同学成绩的所有同学的记录。
20、查询score中选学一门以上课程的同学中分数为非最高分成绩的记录。
21、查询成绩高于学号为“109”、课程号为“3-105”的成绩的所有记录。
22、查询和学号为108的同学同年出生的所有学生的Sno、Sname和Sbirthday列。
23、查询“张旭“教师任课的学生成绩。
24、查询选修某课程的同学人数多于5人的教师姓名。
25、查询95033班和95031班全体学生的记录。
26、查询存在有85分以上成绩的课程Cno.
27、查询出“计算机系“教师所教课程的成绩表。
28、查询“计算机系”与“电子工程系“不同职称的教师的Tname和Prof。
29、查询选修编号为“3-105“课程且成绩至少高于选修编号为“3-245”的同学的Cno、Sno和Degree,并按Degree从高到低次序排序。
30、查询选修编号为“3-105”且成绩高于选修编号为“3-245”课程的同学的Cno、Sno和Degree.
31、查询所有教师和同学的name、sex和birthday.
32、查询所有“女”教师和“女”同学的name、sex和birthday.
33、查询成绩比该课程平均成绩低的同学的成绩表。
34、查询所有任课教师的Tname和Depart.
35 查询所有未讲课的教师的Tname和Depart.
36、查询至少有2名男生的班号。
37、查询Student表中不姓“王”的同学记录。
38、查询Student表中每个学生的姓名和年龄。
39、查询Student表中最大和最小的Sbirthday日期值。
40、以班号和年龄从大到小的顺序查询Student表中的全部记录。
41、查询“男”教师及其所上的课程。
42、查询最高分同学的Sno、Cno和Degree列。
43、查询和“李军”同性别的所有同学的Sname.
44、查询和“李军”同性别并同班的同学Sname.
45、查询所有选修“计算机导论”课程的“男”同学的成绩表
参考答案：
1. SELECT SNAME,SSEX,CLASS FROM STUDENT;
2. SELECT DISTINCT DEPART FROM TEACHER;
3. SELECT * FROM STUDENT;
4. SELECT * FROM SCORE WHERE DEGREE BETWEEN 60 AND 80;
5.SELECT * FROM SCORE WHERE DEGREE IN (85,86,88);
6. SELECT * FROM STUDENT WHERE CLASS='95031' OR SSEX='女';
7.SELECT * FROM STUDENT ORDER BY CLASS DESC;
8.SELECT * FROM SCORE ORDER BY CNO ASC,DEGREE DESC;
9.SELECT COUNT(*) FROM STUDENT WHERE CLASS='95031';
10.SELECT SNO,CNO FROM SCORE WHERE DEGREE=(SELECT MAX(DEGREE) FROM SCORE);
SELECT SNO,CNO FROM SCORE ORDER BY DEGREE DESC LIMIT 1;
11.SELECT AVG(DEGREE) FROM SCORE WHERE CNO='3-105';
12.select avg(degree),cno
from score
where cno like '3%'
group by cno
having count(sno)>= 5;
13.SELECT SNO FROM SCORE GROUP BY SNO HAVING MIN(DEGREE)>70 AND MAX(DEGREE)<90;
14.SELECT A.SNAME,B.CNO,B.DEGREE FROM STUDENT AS A JOIN SCORE AS B ON A.SNO=B.SNO;
15.SELECT A.CNAME, B.SNO,B.DEGREE FROM COURSE AS A JOIN SCORE AS B ON A.CNO=B.CNO ;
16.SELECT A.SNAME,B.CNAME,C.DEGREE FROM STUDENT A JOIN (COURSE B,SCORE C)
ON A.SNO=C.SNO AND B.CNO =C.CNO;
17.SELECT AVG(A.DEGREE) FROM SCORE A JOIN STUDENT B ON A.SNO = B.SNO WHERE B.CLASS='95033';
18.SELECT A.SNO,A.CNO,B.RANK FROM SCORE A,GRADE B WHERE A.DEGREE BETWEEN B.LOW AND B.UPP
ORDER BY RANK;
19.SELECT A.* FROM SCORE A JOIN SCORE B WHERE A.CNO='3-105' AND A.DEGREE>B.DEGREE AND
B.SNO='109' AND B.CNO='3-105';
另一解法：SELECT A.* FROM SCORE A WHERE A.CNO='3-105' AND A.DEGREE>ALL(SELECT DEGREE FROM
SCORE B WHERE B.SNO='109' AND B.CNO='3-105');
20.SELECT * FROM score s WHERE DEGREE<(SELECT MAX(DEGREE) FROM SCORE) GROUP BY SNO HAVING
COUNT(SNO)>1 ORDER BY DEGREE ;
21.见19的第二种解法
22。SELECT SNO,SNAME,SBIRTHDAY FROM STUDENT WHERE YEAR(SBIRTHDAY)=(SELECT YEAR(SBIRTHDAY)
FROM STUDENT WHERE SNO='108');
Oracle:select x.cno,x.Sno,x.degree from score x,score y where x.degree>y.degree and
y.sno='109'and y.cno='3-105';
select cno,sno,degree from score where degree >(select degree from score where sno='109'
and cno='3-105')
23.SELECT A.SNO,A.DEGREE FROM SCORE A JOIN (TEACHER B,COURSE C)
ON A.CNO=C.CNO AND B.TNO=C.TNO
WHERE B.TNAME='张旭';
另一种解法：select cno,sno,degree from score where cno=(select x.cno from course x,teacher y
where x.tno=y.tno and y.tname='张旭');
根据实际EXPLAIN此SELECT语句，第一个的扫描次数要小于第二个
24.SELECT A.TNAME FROM TEACHER A JOIN (COURSE B, SCORE C) ON (A.TNO=B.TNO AND B.CNO=C.CNO)
GROUP BY C.CNO HAVING COUNT(C.CNO)>5;
另一种解法：select tname from teacher where tno in(select x.tno from course x,score y where
x.cno=y.cno group by x.tno having count(x.tno)>5);
实际测试1明显优于2
25。select cno,sno,degree from score where cno=(select x.cno from course x,teacher y where
x.tno=y.tno and y.tname='张旭');
26。SELECT CNO FROM SCORE GROUP BY CNO HAVING MAX(DEGREE)>85;
另一种解法：select distinct cno from score where degree in (select degree from score where
degree>85);
27。SELECT A.* FROM SCORE A JOIN (TEACHER B,COURSE C) ON A.CNO=C.CNO AND B.TNO=C.TNO
WHERE B.DEPART='计算机系';
另一种解法：SELECT * from score where cno in (select a.cno from course a join teacher b on
a.tno=b.tno and b.depart='计算机系');
此时2略好于1，在多连接的境况下性能会迅速下降
28。select tname,prof from teacher where depart='计算机系' and prof not in (select prof from
teacher where depart='电子工程系');
29。SELECT * FROM SCORE WHERE DEGREE>ANY(SELECT DEGREE FROM SCORE WHERE CNO='3-245') ORDER
BY DEGREE DESC;
30。SELECT * FROM SCORE WHERE DEGREE>ALL(SELECT DEGREE FROM SCORE WHERE CNO='3-245') ORDER
BY DEGREE DESC;
31.SELECT SNAME AS NAME, SSEX AS SEX, SBIRTHDAY AS BIRTHDAY FROM STUDENT
UNION
SELECT TNAME AS NAME, TSEX AS SEX, TBIRTHDAY AS BIRTHDAY FROM TEACHER;
32.SELECT SNAME AS NAME, SSEX AS SEX, SBIRTHDAY AS BIRTHDAY FROM STUDENT WHERE SSEX='女'
UNION
SELECT TNAME AS NAME, TSEX AS SEX, TBIRTHDAY AS BIRTHDAY FROM TEACHER WHERE TSEX='女';
33.SELECT A.* FROM SCORE A WHERE DEGREE<(SELECT AVG(DEGREE) FROM SCORE B WHERE A.CNO=B.CNO);
须注意********此题
34。解法一：SELECT A.TNAME,A.DEPART FROM TEACHER A JOIN COURSE B ON A.TNO=B.TNO;
解法二：select tname,depart from teacher a where exists
(select * from course b where a.tno=b.tno);
解法三：SELECT TNAME,DEPART FROM TEACHER WHERE TNO IN (SELECT TNO FROM COURSE);
实际分析，第一种揭发貌似更好，至少扫描次数最少。
35.解法一：SELECT TNAME,DEPART FROM TEACHER A LEFT JOIN COURSE B USING(TNO) WHERE ISNUL
(B.tno);
解法二：select tname,depart from teacher a where not exists
(select * from course b where a.tno=b.tno);
解法三：SELECT TNAME,DEPART FROM TEACHER WHERE TNO NOT IN (SELECT TNO FROM COURSE);
NOT IN的方法效率最差，其余两种差不多
36.SELECT CLASS FROM STUDENT A WHERE SSEX='男' GROUP BY CLASS HAVING COUNT(SSEX)>1;
37.SELECT * FROM STUDENT A WHERE SNAME not like '王%';
38.SELECT SNAME,(YEAR(NOW())-YEAR(SBIRTHDAY)) AS AGE FROM STUDENT;
39.select sname,sbirthday as THEMAX from student where sbirthday =(select min(SBIRTHDAY)
from student)
union
select sname,sbirthday as THEMIN from student where sbirthday =(select max(SBIRTHDAY) from
student);
40.SELECT CLASS,(YEAR(NOW())-YEAR(SBIRTHDAY)) AS AGE FROM STUDENT ORDER BY CLASS DESC,AGE
DESC;
41.SELECT A.TNAME,B.CNAME FROM TEACHER A JOIN COURSE B USING(TNO) WHERE A.TSEX='男';
42.SELECT A.* FROM SCORE A WHERE DEGREE=(SELECT MAX(DEGREE) FROM SCORE B );
43.SELECT SNAME FROM STUDENT A WHERE SSEX=(SELECT SSEX FROM STUDENT B WHERE B.SNAME='李军');
44.SELECT SNAME FROM STUDENT A WHERE SSEX=(SELECT SSEX FROM STUDENT B WHERE B.SNAME='李军' )
AND CLASS=(SELECT CLASS FROM STUDENT C WHERE c.SNAME='李军');
45.解法一：SELECT A.* FROM SCORE A JOIN (STUDENT B,COURSE C) USING(sno,CNO) WHERE B.SSEX='男
' AND C.CNAME='计算机导论';
解法二：select * from score where sno in(select sno from student where
ssex='男') and cno=(select cno from course
where cname='计算机导论');
sql高级查询
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#;
2、查询平均成绩大于60分的同学的学号和平均成绩；
select S#,avg(score)
from sc
group by S# having avg(score) >60;
3、查询所有同学的学号、姓名、选课数、总成绩；
select Student.S#,Student.Sname,count(SC.C#),sum(score)
from Student left Outer join SC on Student.S#=SC.S#
group by Student.S#,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.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');
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');
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；
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;
9、查询所有课程成绩小于60分的同学的学号、姓名；
select S#,Sname
from Student
where S# not in (select Student.S# from Student,SC where S.S#=SC.S# and score>60);
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);
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 S#,Sname from Student,SC where Student.S#=SC.S# and C# in select C# from SC where S#='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.S#,Sname
from Student,SC
where Student.S#=SC.S# and C# in (select C# from SC where S#='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.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and 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；
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');
15、删除学习“叶平”老师课的SC表记录；
Delect SC
from course ,Teacher
where Course.C#=SC.C# and 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；
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)
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 L.C# As 课程ID,L.score AS 最高分,R.score AS 最低分
FROM SC L ,SC AS R
WHERE L.C# = R.C# and
L.score = (SELECT MAX(IL.score)
FROM SC AS IL,Student AS IM
WHERE L.C# = IL.C# and IM.S#=IL.S#
GROUP BY IL.C#)
AND
R.Score = (SELECT MIN(IR.score)
FROM SC AS IR
WHERE R.C# = IR.C#
GROUP BY IR.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顺序
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
21、查询不同老师所教不同课程平均分从高到低显示
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
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
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;
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;
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#;
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;
28、查询男生、女生人数
Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex='男';
Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex='女'；
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(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';
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 ;
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.S# ,avg(score)
from Student,SC
where Student.S#=SC.S# group by SC.S#,Sname having avg(score)>85;
34、查询课程名称为“数据库”，且分数低于60的学生姓名和分数
Select Sname,isnull(score,0)
from Student,SC,Course
where SC.S#=Student.S# and SC.C#=Course.C# and Course.Cname='数据库'and score <60;
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#;
37、查询不及格的课程，并按课程号从大到小排列
select c# from sc where scor e <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 SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='003';
39、求选了课程的学生人数
select count(*) from sc;
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
select Student.Sname,score
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# );
41、查询各个课程及相应的选修人数
select 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
select distinct A.S#,B.score from SC A ,SC B where A.Score=B.Score and A.C# <>B.C# ;
43、查询每门功成绩最好的前两名
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，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#
from sc
group by s#
having count(*) > = 2
46、查询全部学生都选修的课程的课程号和课程名
select C#,Cname
from Course
where C# in (select c# from sc group by c#)
47、查询没学过“叶平”老师讲授的任一门课程的学生姓名
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 S#,avg(isnull(score,0)) from SC where S# in (select S# from SC where score <60 group by S# having count(*)>2)group by S#;
49、检索“004”课程分数小于60，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“002”同学的“001”课程的成绩
delete from Sc where S#='001'and C#='001';
