小白做题
——数据库系统概念（第六版）第三章习题
数据库版本：oracle 11g； 工具：PL/ SQL
有问题下面评论处提
3.11 使用大学模式，用SQL写出如下查询
a.超出所有选修了至少一门Comp. SCI.课程的学生姓名，保证结果中没有重复的姓名。
SELECT DISTINCT NAME FROM student NATURAL JOIN takes NATURAL JOIN course WHERE dept_name = 'Comp. Sci.'
b.找出所有没有选修在2009年春季之前开设的任何课程的学生的ID和姓名
SELECT ID,NAME FROM student WHERE ID NOT IN (
SELECT ID FROM takes WHERE YEAR < 2009 )
c.找出每个系教师的最高工资。可以假设每个系至少有一位教师。
SELECT dept_name,MAX(salary) FROM instructor GROUP BY dept_name
d.从前述查询所计算出的每个系最高工资中选出最低值
WITH max_salary ( dept_name,max_salary) AS (
SELECT dept_name,MAX(salary) FROM instructor GROUP BY dept_name)
SELECT MIN(max_salary) FROM max_salary
3.12使用大学模式用SQL写出如下查询
a.创建一门课程“CS-001”，其名称为“Weekly Seminar”，学分为0
INSERT INTO course VALUES ('CS-001','Weekly Seminar','Comp. Sci.',0);
与之前类似的，网站的DDL对course表的credits项有检查不能为0，故需用去掉该项检查。
b.创建该课程在2009年秋季的一个课程段，sec_id为1.
INSERT INTO SECTION(course_id,sec_id,Semester,Year) VALUES('CS-001','1','Fall','2009');
c.让Comp. Sci.系的每个学生都选修上述课程段
INSERT INTO takes (ID,course_id,sec_id,semester,year) SELECT ID,'CS-001','1','Fall','2009'FROM student WHERE dept_name='Comp. Sci.' ;
d.删除名为Chavez的学生选修上述课程段的信息
DELETE FROM takes WHERE ID=(
SELECT ID FROM student WHERE NAME='Chavez'
)
AND course_id ='CS-001'
e.删除课程CS-001。如果在运行此语句之前没有先删除这门课程授课信息（课程段），会发生什么事情。
DELETE FROM course WHERE course_id = 'CS-001'
// 由于DDL 中设定的course_id外健为级联删除,故section表中对应的CS-001元组会一同被删除
f.删除课程名中包含“database”的任意课程的任意课程段所对应的所有takes元组，在课程名的匹配中忽略大小写
DELETE FROM takes WHERE course_id =(
SELECT course_id FROM course WHERE LOWER(title) LIKE '%database%'
)
3.13 写出对应于图3-18中模式的SQL DDL。在数据类型上做合理的假设，确保声明主码和外码。
CREATE TABLE person
(driver_id NUMBER(13,0),
NAME VARCHAR2(20) NOT NULL,
address VARCHAR2(100),
PRIMARY KEY (driver_id)
);
CREATE TABLE car
(license VARCHAR2(8),
MODEL VARCHAR2(10),
YEAR NUMBER(4,0),
PRIMARY KEY (license)
);
CREATE TABLE accident
(report_number NUMBER(6,0),
datetime DATE,
LOCATION VARCHAR2(100),
PRIMARY KEY (report_number)
);
CREATE TABLE owns
(driver_id NUMBER(13,0),
license VARCHAR2(8),
PRIMARY KEY (driver_id),
FOREIGN KEY (driver_id) REFERENCES person
on delete CASCADE,
FOREIGN KEY (license) REFERENCES car
on delete set NULL
);
CREATE TABLE participated
(report_number NUMBER(6,0),
license VARCHAR2(8),
driver_id NUMBER(13,0),
damage_amount NUMBER(15,0),
PRIMARY KEY (report_number ,license),
FOREIGN KEY (report_number) REFERENCES accident
ON DELETE CASCADE,
FOREIGN KEY (license) REFERENCES car
ON DELETE SET NULL
)
3.14考虑图3-18中的保险公司数据库，其中加下划线的是主码。对这个关系数据库构造如下的SQL查询。
a.找出和“John Smith”的车有关的交通事故数量
select count(report_number) from participated natural join person
where name = 'John Smith'
b.对事故报告编号为“AR2197”中的车牌是“AABB2000”的车辆损坏保险费用更新到3000美元
updata participated set damage_amount = '3000'
where report_number='AR2197' and license='AABB2000'
3.15考虑图3-19中的银行数据库，其中加下划线的是主码。为这个关系数据库构造出如下SQL查询：
a.找出在“Brooklyn”的所有支行都有账户的所有客户。
WITH s_br(num1) AS (
SELECT COUNT(1) FROM branch WHERE branch_city='Brooklyn'
)
(SELECT customer_name FROM customer a
WHERE (
SELECT COUNT(DISTINCT branch_name) FROM
(SELECT * from customer NATURAL JOIN depositor NATURAL JOIN ACCOUNT NATURAL JOIN branch
) b
WHERE a.customer_name=b.customer_name AND branch_city='Brooklyn'
)=(
SELECT num1 FROM s_br))
b.找出银行的所有贷款额的总和
select sum(amount) from loan
c.找出总资产至少比位于Brooklyn的某一家支行要多的所有支行名字
select customer_name from depositor a
where Brooklyn_count =(
select count(distinct branch_name) from (SELECT * FROM customer natural join depositor narual join account natural join branch) b
where a.name=b.name)
3.16考虑图3-20中的雇员数据库，其中加下划线的是主码。给出下面每个查询对应的SQL语句
c.找错所有为“First Bank Corporation”工作的雇员的名字
select employee_name from works where company_name='First Bank Corporation'
b.找出数据库中所有居住城市和公司所在城市相同的雇员
with em_c(name,city) as
(select employee_name,city from works natural join company)
select employee_name from employee,em_c
where employee.employee_name=em_c.name and employee.city=em_c.city
c.找出数据库中所有居住的街道和城市与其经理相同的雇员
select a.employee_name from employee a,employee b,managers
where a.employee_name=managers.employee_name
and managers.manager_name = b.managers
and a.city=b.city
and a.street =b.street
d.找出工资高于其所在公司雇员平均工资工资的所有雇员
select employee_name from works w1
where salary >
(select avg(salary) from works group by company_name
having works.company_name =w1.company_name
)
e.找出工资总和最小的公司
with salary_sum (name,numb) as
(select company_name,sum(salary) from works group by company_name )
select name from salary_sum where numb =
( select min(numb) from salary_sum )
3.17 考虑图3-20中的关系数据库。给出下面每个查询对应的SQL表达式。
a.为“First Bank Corporation”的所有雇员增长10%的工资
update works set salary=salary*1.1
where company_name = 'First Bank Corporation'
b.为“First Bank Corporation”的所有经理增长10%的工资
update works set salary=salary*1.1
where company_name = 'First Bank Corporation'
and employee_name in (select manager_name from managers )
c.删除“Small Bank Corporation”的雇员在works关系中的所有元组
delete from works where company_name = 'Small Bank Corporation'
3.18 列出两个原因，说明为什么空值可以被引入数据库
客观世界中，有时描述事物的信息并不全面，引入空值就可以表示记录的这个属性缺乏，便于数据库的维护和管理，那么引入空值就有两个原因：属性未知，属性不存在
3.19 证明在SQL中，<>all 等价于not in
在SQL中，<>all 即表示与全部元组均不相符，not in表示在全部元组中没有相匹配的，故<>all与not in 等价
3.20 给出图3-20中雇员数据库的SQL模式定义。为每个属性选择合适的域，并为每个关系模式选择合适的主码。
CREATE TABLE employee
(employee_name VARCHAR2(20),
street VARCHAR2(20),
city VARCHAR2(20),
PRIMARY KEY (employee_name)
);
CREATE TABLE company
(company_name VARCHAR2(20),
city VARCHAR2(20),
PRIMARY KEY (company_name)
);
CREATE TABLE works
(employee_name VARCHAR2(20),
company_name VARCHAR2(20),
salary NUMBER(8,2),
PRIMARY KEY (employee_name),
FOREIGN KEY (employee_name) REFERENCES employee ON DELETE CASCADE,
FOREIGN KEY (company_name) REFERENCES company ON DELETE CASCADE
);
CREATE TABLE managers
(employee_name VARCHAR2(20),
manager_name VARCHAR2(20),
PRIMARY KEY (employee_name),
FOREIGN KEY (employee_name) REFERENCES employee ON DELETE CASCADE,
FOREIGN KEY (manager_name) REFERENCES employee ON DELETE SET NULL
)
3.21考虑图3-21中图书馆数据库。用SQL写出如下查询
a.打印借阅了任意由“McGraw-Hill”出版的书的会员名字
select distinct name from member natural join borrowed natural join book
where publisher ='McGraw-Hill'
b.打印借阅了所有由“McGraw-Hill”出版的书的会员名字
with Mc_num(m_no,is) as
(select memb_no,isbn from borrowed natural join
where publisher ='McGraw-Hill')
select name from member where memb_no in
(select m_no from Mc_num
group by m_no having count(is) =
(select count(isbn) from book where publisher= 'McGraw-Hill')
)
c.对于每个出版商，打印借阅了多于5本由该出版商出版的书的会员名字
select name,publisher from member natural join book natural join borrowed
group by name,publisher having count(isbn)>5
d.打印每位会员借阅书籍数量的平均值。考虑这样的情况，如果某会员没有借阅任何书籍，那么该会员根本不会出现在borrowed关系中
select(select count(*) from borrowed)/ (select count(*)from member) FROM dual
3.22不使用unique结构，重写下面的where子句
where unique (select title from course)
where 1>=(select count(title) from course)
3.23考虑查询：
select course_id,semeter,year,sec_id,avg(tot_cred)
from takes natural join student
where year=2000
group by course_id,semester,year,sec_id
having count(ID) >=2
解释为什么在from子句中还加上与section的连接不会改变查询结果
takes 和 section 是通过一些共同的外键相连系，每一个takes的某一个元组不会因为增加额外的元祖。
3.24 考虑查询：
with dept_total (dept_name,value) as
(select dept_name,sum(salary)
from instructor
group by dept_name),
dept_total_avg (value) as
(select avg(value)
from dept_total)
select dept_name
from dept_total,dept_total_avg
where dept_total.value>=dept_total_avg.value
不使用with结构，重写此查询
SELECT dept_name
FROM (
SELECT dept_name,SUM(salary) sa_sum
FROM instructor
GROUP BY dept_name
)
WHERE sa_sum>=(
SELECT AVG(SUM(salary))
FROM instructor
GROUP BY dept_name
)
终于码完了。。。。