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(10) not null,
depart varchar(10) not null
);
insert into student values
(108,"曾华","男",1977-09-01,95033),
(105,"匡明","男",1975-10-02,95031),
(107,"王丽","女",1976-01-23,95033),
(101,"李军","男",1976-02-20,95033),
(109,"王芳","女",1975-02-10,95031),
(103,"陆君","男",1974-06-03,95031);
insert into course values
("3-105","计算机导论",825),
("3-245","操作系统",825),
("6-166","数据电路",825),
("9-888","高等数学",100);
insert into score values
(103,"3-245",86),
(105,"3-245",75),
(109,"3-245",68),
(103,"3-245",92),
(105,"3-105",88),
(109,"3-105",76),
(101,"3-105",64),
(107,"3-105",91),
(108,"3-105",78),
(101,"6-166",85),
(107,"6-106",79),
(108,"6-166",81);
insert into teacher values
(804,"李诚","男","1958-12-02","副教授","计算机系"),
(856,"张旭","男","1969-03-12","讲师","电子工程系"),
(825,"王萍","女","1972-05-05","助教","计算机系"),
(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。
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 where order by class desc;
8/ select * from student where order by cno asc,degree desc;
9/ select sum(class="95031") from student;
第二中写法
SELECT COUNT(*) FROM STUDENT WHERE CLASS='95031';
10/ select max(degree),min(degree) from score;
select sno,cno from score;
最后结果为： select sno,cno from score where degree=(select max(degree),min(degree) from score);
11、 select avg(degree) from score where cno = "3-105";
12/ select cno,avg(degree) from score where cno like "3%" groupd by cno having count(sno) >= 5;
13/ select sno from score group by sno having min(degree) > 70 and max(degree) < 90;
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
答案：
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');
emp员工表 字段如下：
empno 员工号
ename 员工姓名
job 职位
mgr 上级编号
hiredate入职日期
sal 薪资
comm 提成
deptno 部门编号
--1、选择部门30中的员工；
select *from emp where deptno=30;
--2、查询所有办事员（CLERK）的姓名、编号和部门；
select ename,empno,deptno from emp where job ='CLERK';
--3、找出提成高于工资的员工；
select * from emp where comm>sal;
--4、找出提成高于薪资60%的员工
select * from emp where comm>sal*0.6;
--5、找出部门10中所有经理和部门20中所有办事员的详细资料
select
*
from
emp
where
deptno=10 and job='MANAGER'
or
deptno=20 and job ='CLERK' ;
=============================
--6、找出部门10中所有经理、部门20中所有办事员以及既不是经理又不是办事员但薪资大于或等于2000的所有雇员的详细资料
select
*
from emp
where
deptno=10 and job='MANAGER'
or
deptno=20 and job ='CLERK'
or
sal>=2000 and job not in('MANAGER','CLERK');
第二种方法 （换 where后面第二个or）
or
(job !='MANAGER' and job!='CLERK'and sal>=1200);
--7、找出收取提成的员工的不同职位；
select distinct job
from emp
where comm is not null;
--8、找出不收取提成 或收取提成低于100的员工；
select *
from emp
where comm is null or comm<100;
--9、显示雇员的详细资料，按姓名排序
select * from emp where job!='MANAGER'and job!='CLERK' order by ename asc;
--10、显示员工姓名，根据工作年限将最老的员工排在最前面
select ename,hiredate from emp order by hiredate asc;
--11、显示所有员工的姓名、部门编号和薪资，部门编号降序排列，同部门员工以薪资升序排列
select ename,deptno,sal,job from emp order by deptno desc , sal;
--12、显示姓名字段的任何位置都包含A的所有员工姓名
select ename from emp where ename like '%A%';
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