oracle sql试题
转载
数据准备
create table student(
sno varchar2(10) primary key,
sname varchar2(20),
sage number(3),
ssex varchar2(5)
);
create table teacher(
tno varchar2(10) primary key,
tname varchar2(20)
);
create table course(
cno varchar2(10),
cname varchar2(20),
tno varchar2(20),
constraint pk_course primary key (cno,tno)
);
create table sc(
sno varchar2(10),
cno varchar2(10),
score number(4,2),
constraint pk_sc primary key (sno,cno)
);
/*******初始化学生表的数据******/
insert into student values ('s001','张三',23,'男');
insert into student values ('s002','李四',23,'男');
insert into student values ('s003','吴鹏',25,'男');
insert into student values ('s004','琴沁',20,'女');
insert into student values ('s005','王丽',20,'女');
insert into student values ('s006','李波',21,'男');
insert into student values ('s007','刘玉',21,'男');
insert into student values ('s008','萧蓉',21,'女');
insert into student values ('s009','陈萧晓',23,'女');
insert into student values ('s010','陈美',22,'女');
commit;
/******************初始化教师表***********************/
insert into teacher values ('t001', '刘阳');
insert into teacher values ('t002', '谌燕');
insert into teacher values ('t003', '胡明星');
commit;
/***************初始化课程表****************************/
insert into course values ('c001','J2SE','t002');
insert into course values ('c002','Java Web','t002');
insert into course values ('c003','SSH','t001');
insert into course values ('c004','Oracle','t001');
insert into course values ('c005','SQL SERVER 2005','t003');
insert into course values ('c006','C#','t003');
insert into course values ('c007','JavaScript','t002');
insert into course values ('c008','DIV+CSS','t001');
insert into course values ('c009','PHP','t003');
insert into course values ('c010','EJB3.0','t002');
commit;
/***************初始化成绩表***********************/
insert into sc values ('s001','c001',78.9);
insert into sc values ('s002','c001',80.9);
insert into sc values ('s003','c001',81.9);
insert into sc values ('s004','c001',60.9);
insert into sc values ('s001','c002',82.9);
insert into sc values ('s002','c002',72.9);
insert into sc values ('s003','c002',81.9);
insert into sc values ('s001','c003','59');
commit;
student table
sno |
sname |
sage |
ssex |
s001 |
张三 |
23 |
男 |
s002 |
李四 |
23 |
男 |
s003 |
吴鹏 |
25 |
男 |
s004 |
琴沁 |
20 |
女 |
s005 |
王丽 |
20 |
女 |
s006 |
李波 |
21 |
男 |
s007 |
刘玉 |
21 |
男 |
s008 |
萧蓉 |
21 |
女 |
s009 |
陈萧晓 |
23 |
女 |
s010 |
陈美 |
22 |
女 |
teacher table
tno |
tname |
t001 |
刘阳 |
t002 |
胡燕 |
t003 |
胡明星 |
course table
cno |
cname |
tno |
c001 |
J2SE |
t002 |
c002 |
Java Web |
t002 |
c003 |
SSH |
t001 |
c004 |
Oracle |
t001 |
c005 |
SQL SERVER 2005 |
t003 |
c006 |
C# |
t003 |
c007 |
JavaScript |
t002 |
c008 |
DIV+CSS |
t001 |
c009 |
PHP |
t003 |
c010 |
EJB3.0 |
t002 |
score table
sno |
cno |
score |
s001 |
c001 |
78.9 |
s002 |
c001 |
80.9 |
s003 |
c001 |
81.9 |
s004 |
c001 |
60.9 |
s001 |
c002 |
82.9 |
s002 |
c002 |
72.9 |
s003 |
c002 |
81.9 |
s001 |
c003 |
59 |
试题
1、查询“c001”课程比“c002”课程成绩高的所有学生的学号;
2、查询平均成绩大于60分的同学的学号和平均成绩;
3、查询所有同学的学号、姓名、选课数、总成绩;
4、查询姓“刘”的老师的个数;
5、查询没学过“谌燕”老师课的同学的学号、姓名;
6、查询学过“c001”并且也学过编号“c002”课程的同学的学号、姓名;
7、查询学过“谌燕”老师所教的所有课的同学的学号、姓名;
8、查询课程编号“c002”的成绩比课程编号“c001”课程低的所有同学的学号、姓名;
9、查询所有课程成绩小于60 分的同学的学号、姓名;
10、查询没有学全所有课的同学的学号、姓名;
11、查询至少有一门课与学号为“s001”的同学所学相同的同学的学号和姓名;
12、查询至少学过学号为“s001”同学所有一门课的其他同学学号和姓名;
13、把“SC”表中“谌燕”老师教的课的成绩都更改为此课程的平均成绩;
14、查询和“s001”号的同学学习的课程完全相同的其他同学学号和姓名;
15、删除学习“谌燕”老师课的SC 表记录;
16、向SC 表中插入一些记录,这些记录要求符合以下条件:没有上过编号“c002”课程的同学学号、“c002”号课的平均成绩;
17、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分
18、按各科平均成绩从低到高和及格率的百分数从高到低顺序
19、查询不同老师所教不同课程平均分从高到低显示
20、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
22、查询每门课程被选修的学生数
23、查询出只选修了一门课程的全部学生的学号和姓名
24、查询男生、女生人数
25、查询姓“张”的学生名单
26、查询同名同性学生名单,并统计同名人数
27、1981 年出生的学生名单(注:Student 表中Sage 列的类型是number)
28、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列
29、查询平均成绩大于85 的所有学生的学号、姓名和平均成绩
30、查询课程名称为“数据库”,且分数低于60 的学生姓名和分数
31、查询所有学生的选课情况;
32、查询任何一门课程成绩在70 分以上的姓名、课程名称和分数;
33、查询不及格的课程,并按课程号从大到小排列
34、查询课程编号为c001 且课程成绩在80 分以上的学生的学号和姓名;
35、求选了课程的学生人数
36、查询选修“谌燕”老师所授课程的学生中,成绩最高的学生姓名及其成绩
37、查询各个课程及相应的选修人数
38、查询不同课程成绩相同的学生的学号、课程号、学生成绩
39、查询每门功课成绩最好的前两名
40、统计每门课程的学生选修人数(超过10 人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,若人数相同,按课程号升序排列
41、检索至少选修两门课程的学生学号
42、查询全部学生都选修的课程的课程号和课程名
43、查询没学过“谌燕”老师讲授的任一门课程的学生姓名
44、查询两门以上不及格课程的同学的学号及其平均成绩
45、检索“c004”课程分数小于60,按分数降序排列的同学学号
46、删除“s002”同学的“c001”课程的成绩
答案
- 1.
- *********************************
- select a.* from
- (select * from sc a where a.cno='c001') a,
- (select * from sc b where b.cno='c002') b
- where a.sno=b.sno and a.score > b.score;
- *********************************
- select * from sc a
- where a.cno='c001'
- and exists(select * from sc b where b.cno='c002' and a.score>b.score
- and a.sno = b.sno)
- *********************************
- 2.
- *********************************
- select sno,avg(score) from sc group by sno having avg(score)>60;
- *********************************
- 3.
- *********************************
- select a.*,s.sname from (select sno,sum(score),count(cno) from sc group by sno) a ,student s where a.sno=s.sno
- *********************************
- 4.
- *********************************
- select count(*) from teacher where tname like '刘%';
- *********************************
- 5.
- *********************************
- select a.sno,a.sname from student a
- where a.sno
- not in
- (select distinct s.sno
- from sc s,
- (select c.*
- from course c ,
- (select tno
- from teacher t
- where tname='谌燕')t
- where c.tno=t.tno) b
- where s.cno = b.cno )
- *********************************
- select * from student st where st.sno not in
- (select distinct sno from sc s join course c on s.cno=c.cno
- join teacher t on c.tno=t.tno where tname='谌燕')
- *********************************
- 6.
- *********************************
- select st.* from sc a
- join sc b on a.sno=b.sno
- join student st
- on st.sno=a.sno
- where a.cno='c001' and b.cno='c002' and st.sno=a.sno;
- *********************************
- 7.
- *********************************
- select st.* from student st join sc s on st.sno=s.sno
- join course c on s.cno=c.cno
- join teacher t on c.tno=t.tno
- where t.tname='谌燕'
- *********************************
- 8.
- *********************************
- select * from student st
- join sc a on st.sno=a.sno
- join sc b on st.sno=b.sno
- where a.cno='c002' and b.cno='c001' and a.score < b.score
- *********************************
- 9.
- *********************************
- select st.*,s.score from student st
- join sc s on st.sno=s.sno
- join course c on s.cno=c.cno
- where s.score <60
- *********************************
- 10.
- *********************************
- select stu.sno,stu.sname,count(sc.cno) from student stu
- left join sc on stu.sno=sc.sno
- group by stu.sno,stu.sname
- having count(sc.cno)<(select count(distinct cno)from course)
- ===================================
- select * from student where sno in
- (select sno from
- (select stu.sno,c.cno from student stu
- cross join course c
- minus
- select sno,cno from sc)
- )
- ===================================
- *********************************
- 11.
- *********************************
- select st.* from student st,
- (select distinct a.sno from
- (select * from sc) a,
- (select * from sc where sc.sno='s001') b
- where a.cno=b.cno) h
- where st.sno=h.sno and st.sno<>'s001'
- *********************************
- 12.
- *********************************
- select * from sc
- left join student st
- on st.sno=sc.sno
- where sc.sno<>'s001'
- and sc.cno in
- (select cno from sc
- where sno='s001')
- *********************************
- 13.
- *********************************
- update sc c set score=(select avg(c.score) from course a,teacher b
- where a.tno=b.tno
- and b.tname='谌燕'
- and a.cno=c.cno
- group by c.cno)
- where cno in(
- select cno from course a,teacher b
- where a.tno=b.tno
- and b.tname='谌燕')
- *********************************
- 14.
- *********************************
- select* from sc where sno<>'s001'
- minus
- (
- select* from sc
- minus
- select * from sc where sno='s001'
- )
- *********************************
- 15.
- *********************************
- delete from sc
- where sc.cno in
- (
- select cno from course c
- left join teacher t on c.tno=t.tno
- where t.tname='谌燕'
- )
- *********************************
- 16.
- *********************************
- insert into sc (sno,cno,score)
- select distinct st.sno,sc.cno,(select avg(score)from sc where cno='c002')
- from student st,sc
- where not exists
- (select * from sc where cno='c002' and sc.sno=st.sno) and sc.cno='c002';
- *********************************
- 17.
- *********************************
- select cno ,max(score),min(score) from sc group by cno;
- *********************************
- 18.
- *********************************
- select cno,avg(score),sum(case when score>=60 then 1 else 0 end)/count(*)
- as 及格率
- from sc group by cno
- order by avg(score) , 及格率desc
- *********************************
- 19.
- *********************************
- select max(t.tno),max(t.tname),max(c.cno),max(c.cname),c.cno,avg(score) from sc , course c,teacher t
- where sc.cno=c.cno and c.tno=t.tno
- group by c.cno
- order by avg(score) desc
- *********************************
- 20.
- *********************************
- select sc.cno,c.cname,
- sum(case when score between 85 and 100 then 1 else 0 end) AS "[100-85]",
- sum(case when score between 70 and 85 then 1 else 0 end) AS "[85-70]",
- sum(case when score between 60 and 70 then 1 else 0 end) AS "[70-60]",
- sum(case when score <60 then 1 else 0 end) AS "[<60]"
- from sc, course c
- where sc.cno=c.cno
- group by sc.cno ,c.cname;
- *********************************
- 21.
- *********************************
- select * from
- (select sno,cno,score,row_number()over(partition by cno order by score desc) rn from sc)
- where rn<4
- *********************************
- 22.
- *********************************
- select cno,count(sno)from sc group by cno;
- *********************************
- 23.
- *********************************
- select sc.sno,st.sname,count(cno) from student st
- left join sc
- on sc.sno=st.sno
- group by st.sname,sc.sno having count(cno)=1;
- *********************************
- 24.
- *********************************
- select ssex,count(*)from student group by ssex;
- *********************************
- 25.
- *********************************
- select * from student where sname like '张%';
- *********************************
- 26.
- *********************************
- select sname,count(*)from student group by sname having count(*)>1;
- *********************************
- 27.
- *********************************
- select sno,sname,sage,ssex from student t where to_char(sysdate,'yyyy')-sage =1988
- *********************************
- 28.
- *********************************
- select cno,avg(score) from sc group by cno order by avg(score)asc,cno desc;
- *********************************
- 29.
- *********************************
- select st.sno,st.sname,avg(score) from student st
- left join sc
- on sc.sno=st.sno
- group by st.sno,st.sname having avg(score)>85;
- *********************************
- 30.
- *********************************
- select sname,score from student st,sc,course c
- where st.sno=sc.sno and sc.cno=c.cno and c.cname='Oracle' and sc.score<60
- *********************************
- 31.
- *********************************
- select st.sno,st.sname,c.cname from student st,sc,course c
- where sc.sno=st.sno and sc.cno=c.cno;
- *********************************
- 32.
- *********************************
- select st.sname,c.cname,sc.score from student st,sc,course c
- where sc.sno=st.sno and sc.cno=c.cno and sc.score>70
- *********************************
- 33.
- *********************************
- select sc.sno,c.cname,sc.score from sc,course c
- where sc.cno=c.cno and sc.score<60 order by sc.cno desc;
- *********************************
- 34.
- *********************************
- select st.sno,st.sname,sc.score from sc,student st
- where sc.sno=st.sno and cno='c001' and score>80;
- *********************************
- 35.
- *********************************
- select count(distinct sno) from sc;
- *********************************
- 36.
- *********************************
- select st.sname,score from student st,sc ,course c,teacher t
- where
- st.sno=sc.sno and sc.cno=c.cno and c.tno=t.tno
- and t.tname='谌燕' and sc.score=
- (select max(score)from sc where sc.cno=c.cno)
- *********************************
- 37.
- *********************************
- select cno,count(sno) from sc group by cno;
- *********************************
- 38.
- *********************************
- select a.* from sc a ,sc b where a.score=b.score and a.cno<>b.cno
- *********************************
- 39.
- *********************************
- select * from (
- select sno,cno,score,row_number()over(partition by cno order by score desc) my_rn from sc t
- )
- where my_rn<=2
- *********************************
- 40.
- *********************************
- select cno,count(sno) from sc group by cno
- having count(sno)>10
- order by count(sno) desc,cno asc;
- *********************************
- 41.
- *********************************
- select sno from sc group by sno having count(cno)>1;
- ||
- select sno from sc group by sno having count(sno)>1;
- *********************************
- 42.
- *********************************
- select distinct(c.cno),c.cname from course c ,sc
- where sc.cno=c.cno
- ||
- select cno,cname from course c
- where c.cno in
- (select cno from sc group by cno)
- *********************************
- 43.
- *********************************
- select st.sname from student st
- where st.sno not in
- (select distinct sc.sno from sc,course c,teacher t
- where sc.cno=c.cno and c.tno=t.tno and t.tname='谌燕')
- *********************************
- 44.
- *********************************
- select sno,avg(score)from sc
- where sno in
- (select sno from sc where sc.score<60
- group by sno having count(sno)>1
- ) group by sno
- *********************************
- 45.
- *********************************
- select sno from sc where cno='c004' and score<90 order by score desc;
- *********************************
- 46.
- *********************************
- delete from sc where sno='s002' and cno='c001';
- *********************************
oracle sql试题的更多相关文章
- 部分常见ORACLE面试题以及SQL注意事项
部分常见ORACLE面试题以及SQL注意事项 一.表的创建: 一个通过单列外键联系起父表和子表的简单例子如下: CREATE TABLE parent(id INT NOT NULL, PRIMARY ...
- oracle 笔试题
ORACLE笔试题一.单选题1.在Oracle中,以下不属于集合操作符的是( ). A. UNION B. SUM C. MINUS D. INTERSECT2.在Oracle中,执行下面的语句:SE ...
- Oracle SQL Developer 连接 MySQL
1. 在ORACLE官网下载Oracle SQL Developer第三方数据库驱动 下载页面:http://www.oracle.com/technetwork/developer-tools/sq ...
- Oracle sql连接
inner-join left-outer-join right-outer-join full- ...
- 解决Oracle SQL Developer无法连接远程服务器的问题
在使用Oracle SQL Developer连接远程服务器的时候,出现如下的错误 在服务器本地是可以正常连接的.这个让人想起来,跟SQL Server的一些设计有些类似,服务器估计默认只在本地监听, ...
- Oracle sql语句执行顺序
sql语法的分析是从右到左 一.sql语句的执行步骤: 1)词法分析,词法分析阶段是编译过程的第一个阶段.这个阶段的任务是从左到右一个字符一个字符地读入源程序,即对构成源程序的字符流进行扫描然后根据构 ...
- Oracle SQL explain/execution Plan
From http://blog.csdn.net/wujiandao/article/details/6621073 1. Four ways to get execution plan(anyti ...
- 处理 Oracle SQL in 超过1000 的解决方案
处理oracle sql 语句in子句中(where id in (1, 2, ..., 1000, 1001)),如果子句中超过1000项就会报错.这主要是oracle考虑性能问题做的限制.如果要解 ...
- Oracle sql develpoer
Oracle SQL Developer是针对Oracle数据库的交互式开发环境(IDE) Oracle SQL Developer简化了Oracle数据库的开发和管理. SQL Develo ...
随机推荐
- [中山市选2011][bzoj2440] 完全平方数 [二分+莫比乌斯容斥]
题面 传送门 思路 新姿势get 莫比乌斯容斥 $\sum_{i=1}{n}\mu(i)f(i)$ 这个东西可以把所有没有平方质因子的东西表示出来,还能容斥掉重复的项 证明是根据莫比乌斯函数的定义,显 ...
- cdoj 1259 线段树+bitset 区间更新/查询
Description 昊昊喜欢运动 他N天内会参加M种运动(每种运动用一个[1,m]的整数表示) 现在有Q个操作,操作描述如下 昊昊把第l天到第r天的运动全部换成了x(x∈[1,m]) 问昊昊第l天 ...
- [ CodeVS冲杯之路 ] P1295
不充钱,你怎么AC? 题目:http://codevs.cn/problem/1295/ 数据很小,直接DFS,加上剪枝 剪枝其实就是判重,首先深度是行下标,这里自带不重复,枚举的列下标,用 f 记录 ...
- 错误:'nasm' 不是内部或外部命令,也不是可运行的程序
原文转自 http://blog.csdn.net/alexcrazy/article/details/7183312 >正在执行自定义生成步骤 >'nasm' 不是内部或外部命令,也不是 ...
- matlab默认字体设置
Monospaced Plain 10 SansSerif Plain 10 这是默认设置.希望能帮到你!
- windows安装scrapy
1.安装Twisted 直接pip install Twisted 然后报错 error: Microsoft Visual C++ 14.0 is required. Get it with &qu ...
- LVM更换硬盘
#检测坏道 smartctl -a /dev/sdd #硬盘检测 e2fsck -f /dev/mapper/vg_root-lv_data #重新定义空间大小,将原来的大小上减去要移走的硬盘 res ...
- MySQL -MMM 学习整理
一. 规划 1.主机规划 服务器 IP 作用 monitor 10.0.0.10 监控服务器 master-01 10.0.0.5 读写主机01 master-02 10.0.0.6 读写主机02 s ...
- 模拟【p2239】 螺旋矩阵
顾z 你没有发现两个字里的blog都不一样嘛 qwq 题目描述--->p2239 螺旋矩阵 看到题,很明显,如果直接模拟的话,复杂度为\(O(n^2)\)过不去.(这个复杂度应该不正确,我不会分 ...
- 加强版dd工具dc3dd
加强版dd工具dc3dd dd是Linux最常用的磁盘备份工具,但缺少渗透测试常用的数据校验.hash等重要功能.Kali Linux提供的一款专用工具dc3dd.该工具是dd的加强版.它在dd的 ...