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。
CREATE TABLE STUDENT
(SNO VARCHAR(3) NOT NULL,
SNAME VARCHAR(4) NOT NULL,
SSEX VARCHAR(2) NOT NULL,
SBIRTHDAY DATETIME,
CLASS VARCHAR(5))ENGINE=MyISAM AUTO_INCREMENT=348 DEFAULT CHARSET=utf8 ;;
CREATE TABLE COURSE
(CNO VARCHAR(5) NOT NULL,
CNAME VARCHAR(10) NOT NULL,
TNO VARCHAR(10) NOT NULL)ENGINE=MyISAM AUTO_INCREMENT=348 DEFAULT CHARSET=utf8 ;;
CREATE TABLE SCORE
(SNO VARCHAR(3) NOT NULL,
CNO VARCHAR(5) NOT NULL,
DEGREE NUMERIC(10, 1) NOT NULL)ENGINE=MyISAM AUTO_INCREMENT=348 DEFAULT CHARSET=utf8 ;;
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)ENGINE=MyISAM AUTO_INCREMENT=348 DEFAULT CHARSET=utf8 ;;
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);
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);
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);
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‘);