第3章 SQL 习题 - 3.21

3.21 考虑图3-21中的图书馆数据库。用SQL写出如下查询：

member(memb_no, name, age)

book(isbn, title, authors, publishers)

borrowed(memb_no, isbn, date)

图3-21 习题3.21的图书馆数据库

为了解题方便，在数据库中创建这些关系：

create table member (
    memb_no varchar,
    name varchar,
    age int,
    primary key (memb_no)
);

create table book(
    isbn varchar,
    title varchar,
    authors varchar,
    publishers varchar,
    primary key (isbn)
);

create table borrowed (
    memb_no varchar,
    isbn varchar,
    date varchar,
    primary key (memb_no, isbn),
    foreign key (memb_no) references member on delete no action,
    foreign key (isbn) references book on delete no action
);

insert into member values ('会员号_1', '会员_1', 18);
insert into member values ('会员号_2', '会员_2', 19);
insert into member values ('会员号_3', '会员_3', 20);
insert into member values ('会员号_4', '会员_4', 21);

insert into book values ('ISBN_1', '书本_1', '作者_1', '出版商_1');
insert into book values ('ISBN_2', '书本_2', '作者_1', '出版商_1');
insert into book values ('ISBN_3', '书本_3', '作者_1', '出版商_1');
insert into book values ('ISBN_4', '书本_4', '作者_1', '出版商_1');
insert into book values ('ISBN_5', '书本_5', '作者_1', '出版商_1');
insert into book values ('ISBN_6', '书本_6', '作者_1', '出版商_1');
insert into book values ('ISBN_7', '书本_7', '作者_1', '出版商_2');
insert into book values ('ISBN_8', '书本_8', '作者_1', '出版商_2');
insert into book values ('ISBN_9', '书本_9', '作者_1', '出版商_2');
insert into book values ('ISBN_10', '书本_10', '作者_1', '出版商_3');

insert into borrowed values ('会员号_1', 'ISBN_1', '');
insert into borrowed values ('会员号_1', 'ISBN_2', '');
insert into borrowed values ('会员号_2', 'ISBN_3', '');
insert into borrowed values ('会员号_3', 'ISBN_4', '');
insert into borrowed values ('会员号_3', 'ISBN_5', '');
insert into borrowed values ('会员号_3', 'ISBN_6', '');
insert into borrowed values ('会员号_2', 'ISBN_10', '');

a.打印借阅了任意由“出版商_1”出版的书的会员名字。

select distinct name from member natural join book
natural join borrowed where publishers = '出版商_1';

  name  
--------
 会员_1
 会员_2
 会员_3
(3 rows)

b.打印借阅了所有由“出版商_2”出版的书的会员名字。

我们再添加一些数据，让会员_5借阅所有由“出版商_2”出版的书。

insert into member values ('会员号_5', '会员_5', 22);
insert into borrowed values ('会员号_5', 'ISBN_7', '');
insert into borrowed values ('会员号_5', 'ISBN_8', '');
insert into borrowed values ('会员号_5', 'ISBN_9', '');

现在开始实现本题要求的SQL：

with publisher_book(isbn) as (
	select isbn from book where publishers = '出版商_2'
) 
select distinct name from member
natural join borrowed as B
where not exists (
	(select isbn from borrowed where memb_no = B.memb_no)
	except
	(select isbn from publisher_book)
);

  name  
--------
 会员_5
(1 row)

c.对于每个出版商，打印借阅了多于五本由该出版商出版的书的会员名字。

我们的数据库中的数据没有多于5本借阅的，咱们选择多玩2本的吧。

with borrowed_book(memb_no, name, isbn, publishers) as (
	select member.memb_no, name, book.isbn, publishers from
	member natural join borrowed natural join book
)
select distinct name from borrowed_book as BB
where exists(
	select count(isbn) from borrowed_book
	where memb_no = BB.memb_no
	group by (memb_no, publishers)
	having count(isbn) > 2
)

  name  
--------
 会员_3
 会员_5
(2 rows)

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 cast(cast(count(isbn) as numeric(18,1))/count(distinct memb_no) as numeric(18,1)) from
borrowed;

 numeric 
---------
     2.5
(1 row)