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标题: 接上贴,参考答案 [打印本页]

作者: 陈大仙…    时间: 2017-3-17 08:35
标题: 接上贴,参考答案
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');





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