(转) 经典SQL练习题

select Sname,Ssex,Class from student
2、 查询教师所有的单位即不重复的Depart列。

select distinct Depart from teacher
3、 查询Student表的所有记录。

select *from Student
4、 查询Score表中成绩在60到80之间的所有记录。

select*From Score where DEGREE between 60 and 80
5、 查询Score表中成绩为85，86或88的记录。

select *from Score where DEGREE=85 or DEGREE=86 or DEGREE=88

or

select *from Score where DEGREE in (85,86,88)
6、 查询Student表中“95031”班或性别为“女”的同学记录。

select *from student where class=‘95031‘or Ssex=‘女‘
7、 以Class降序查询Student表的所有记录。

select *From Student order by Class  DESC
8、 以Cno升序、Degree降序查询Score表的所有记录。

select *from Score order by  Cno ASC,  Degree DESC
9、 查询“95031”班的学生人数。

select count(*)from student where class=‘95031‘
10、查询Score表中的最高分的学生学号和课程号。

select SNO,CNO from Score where DEGREE=(select max(DEGREE) from Score)
11、查询‘3-105’号课程的平均分。

select avg(DEGREE)from SCORE where CNO=‘3-105‘
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。

select Cno,avg(DEGREE) from Score where Cno in (select Cno from Score where Cno like ‘3%‘ group by Cno having Count(*)>=5) 
13、查询最低分大于70，最高分小于90的Sno列。

select Sno from score group by Sno having max(DEGREE)<90 and min(DEGREE)>70
14、查询所有学生的Sname、Cno和Degree列。

select St.Sname, Cs.Cno, Cs.Degree from Student as St join Score as Sc on  St.Sno=Sc.Sno
16、查询所有学生的Sname、Cname和Degree列。

select A.sname,Bcname,C.degree from student A join (course B,Score C) on A.sno=C.sno and B.cno=C.cno
17、查询“95033”班所选课程的平均分。

SELECT AVG(A.DEGREE) FROM SCORE A JOIN STUDENT B ON A.SNO = B.SNO WHERE B.CLASS=‘95033‘
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。

select A*from Score A  join Score B where A.cno=‘3-105‘ and B.cno=‘3-105‘ and A.DEGREE>B.DEGREE and B.sno=‘109‘
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：

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‘);

(转) 经典SQL练习题

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