经典SQL练习题

向右看齐 2022-07-16 00:46 236阅读 0赞

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。  
  
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');