很好的练习基础SQL操作的习题

时间：2015-10-20 12:16:44 阅读：226 评论：0 收藏：0 [点我收藏+]

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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))

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)

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);

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

46、列出部门中工资高于本部门平均工资的的员工数和对应的部门编号，并按部门号进行降序排序。

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‘);

46.第一种子查询:select count(*), dept_1.departno from department dept_1 where dept_1.salary>(select avg(dept_2.salary) from department dept_2 where dept_1.departno=dept_2.departno) group by dept_1.departno

order by dept_1.departno desc;

第二种关联查询(临时表): select count(*) from department dept_1,(select departno,avg(salary) from department

group by departno) dept_2 where dept_1.departno=dept_2.departno and dept_1.salary>dept_2.salary group by

dept_1.departnorder by dept_1.departno desc;

//下面是本人的练习记录:仅供参考

create table student (sno varchar(3) not null, sname varchar(4) not null, ssex varchar(2) not null, sbirthday datetime, class varchar(5))
create table course (cno varchar(5) not null, cname varchar(10) not null, tno varchar(10) not null)
create table score (sno varchar(3) not null, cno varchar(5) not null, degree numeric(10, 1) not null)
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 (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);
create table grade(low decimal(3,0),upp decimal(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‘);

18,20,23,30,33,35,42

select sname,ssex,class from student;
select * from score where degree between 60 and 80;
select * from score where degree in(85,86,88);
select * from student where class=‘95031‘ or ssex=‘女‘;
select * from student order by class desc;
select * from score order by cno asc,degree desc;
select count(sno) num from student where class=‘95031‘;
select cno,sno from score where degree=(select max(degree) from score);
select avg(degree) from score group by cno having cno=‘3-105‘;
select avg(degree) from score where cno like ‘3%‘ group by cno having count(sno)>=5;
select sno from score group by cno having min(degree)>70 and max(degree)<90;
select student.sname,score.cno,score.degree from student,score where student.sno=score.sno;
select student.sno,course.cname,score.degree from student,score,course where student.sno=score.sno and score.cno=course.cno;
select student.sname,course.cname,score.degree from student,score,course where student.sno=score.sno and score.cno=course.cno;
select score.cno,avg(score.degree) from score,student where score.sno=student.sno and student.class=‘95033‘ group by student.class,score.cno
//select student.sno,score.cno,degree from student inner join score on student.sno=score.sno;
//select student.sno,score.cno,case from student,score where student.sno=score.sno;
select sno,cno,degree,rank from score,grade where degree between low and upp order by rank desc
select student.* from student,score where student.sno=score.sno and score.cno=‘3-105‘ and score.degree>all(select score2.degree from score score2 where score2.cno=‘3-105‘ and score2.sno=109);
select score.*,count(score.*) from score group by sno having count(cno)>1 and max(degree)<(select max(score2.)
select score.* from score where degree<(select max(degree) from score) group by sno having count(cno)>1 order by degree desc; ******
select s2.* from score s2 where s2.sno in(select s1.*,count(s1.degree) from score s1 group by s1.sno having count(s1.cno)>1 and s1.degree<(select max(s.degree) from score s where s.cno=s1.cno) order by s1.degree desc;
select score.* from score where score.degree>(select score2.degree from score score2 where score2.sno= 109 and score2.cno=‘3-105‘);
select sno,sname,sbirthday from student where SUBSTRING(sbirthday,1,4)=(select SUBSTRING(stu.sbirthday,1,4) from student stu where stu.sno=108)
select score.* from score inner join course on score.cno=course.cno inner join teacher on course.tno=teacher.tno where teacher.tname=‘张旭‘;
select teacher.tname from score inner join course on score.cno=course.cno inner join teacher on course.tno=teacher.tno group by score.cno having count(sno)>5;
select student.* from student where student.class in(‘95033‘,‘95031‘);
select score.cno from score group by score.cno having max(score.degree)>85
select score.* from score,teacher,course where score.cno=course.cno and course.tno=teacher.tno and teacher.depart=‘计算机系‘
select tname,prof from teacher where depart in(‘计算机系‘,‘电子工程系‘);
select tname,prof from teacher where depart in(‘计算机系‘) and prof not in(select t.prof from teacher t where t.depart=‘电子工程系‘);
select score.* from score where cno=‘3-105‘ and degree>(select max(score2.degree) from score score2 where score2.cno=‘3-245‘) order by degree desc;
select score.* from score group by cno having degree<avg(degree); ********** union
select score.* from score where score.degree<(select avg(score2.degree) from score score2 where score.cno=score2.cno);

select t.tname,t.tsex,t.tbirthday,stu.sname,stu.ssex,stu.sbirthday from teacher t right join course c on t.tno=c.tno right join score s on s.cno=c.cno right join student stu on stu.sno=s.sno
select t.tname,t.tsex,t.tbirthday,stu.sname,stu.ssex,stu.sbirthday from teacher t right join course c on t.tno=c.tno right join score s on s.cno=c.cno right join student stu on stu.sno=s.sno where t.tsex=‘女‘ and stu.ssex=‘女‘; **************
select score.* ,avg(degree) from score group by cno having avg(degree)>=degree ************
select t.tname,t.depart from teacher t where exist (select t2.* from class t2 where t2.tno=t.tno);************
select tname,depart from teacher,course where teacher.tno exists(select distinct course.tno from course); *******
select class from student where ssex=‘男‘ group by class having count(sno)>=2;
select * from student where sname not like ‘王%‘;
select sname,(2014-substring(sbirthday,1,4)) sage from student;
select max(sbirthday),min(sbirthday) from student; *************
select sname,sbirthday from student where sbirthday=(select max(sbirthday) from student) or sbirthday=(select min(sbirthday) from student);
select * from student order by class desc,(2014-substring(sbirthday,1,4)) desc;
select t.*,c.cno,c.cname from teacher t,course c where t.tsex=‘男‘ and t.tno=c.tno; *******
select score.* from score group by cno having max(degree)=degree;
select sname from student where ssex=(select ssex from student stu where stu.sname=‘李军‘);
select sname from student where ssex=(select ssex from student stu where stu.sname=‘李军‘) and class=(select class from student stu where stu.sname=‘李军‘);
select s.* from score s,course c,student t where t.sno=s.sno and s.cno=c.cno and t.ssex=‘男‘ and c.cname=‘计算机导论‘;

很好的练习基础SQL操作的习题

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原文地址：http://www.cnblogs.com/aipiaoborensheng/p/4894306.html

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