小白做题 ——数据库系统概念（第六版）第三章习题

——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

update 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 natrual 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。。。。