2ndSem

3.21

2 min read

Consider the library database of Figure 3.20. Write the following queries in SQL.

a. Find the member number and name of each member who has borrowed at least one book published by “McGraw-Hill”.
b. Find the member number and name of each member who has borrowed every book published by “McGraw-Hill”.
c. For each publisher, find the member number and name of each member who has borrowed more than five books of that publisher.
d. Find the average number of books borrowed per member. Take into account that if a member does not borrow any books, then that member does not appear in the borrowed relation at all, but that member still counts in the average.

a. Find the member number and name of each member who has borrowed at least one book published by “McGraw-Hill”. 

SELECT memb_no, name
FROM member AS m
WHERE EXISTS (
    SELECT * 
    FROM book INNER JOIN borrowed ON book.isbn = borrowed.isbn 
    WHERE book.publisher = 'McGraw-Hill' AND 
            borrowed.memb_no = m.memb_no
)

b. Find the member number and name of each member who has borrowed every book published by “McGraw-Hill”.

SELECT memb_no, name
FROM member AS m 
WHERE NOT EXISTS (
    (
        SELECT isbn
        FROM book
        WHERE publisher = 'McGraw-Hill'
    )
    EXCEPT 
    (
        SELECT isbn
        FROM borrowed
        WHERE memb_no = m.memb_no
    )
)

c. For each publisher, find the member number and name of each member who has borrowed more than five books of that publisher.

WITH member_borrowed_book(memb_no, memb_name,isbn,title,authors,publisher,date) AS (
    SELECT member.memb_no, name, book.isbn, title, authors, publisher, date
    FROM member INNER JOIN borrowed ON member.memb_no = borrowed.memb_no
                INNER JOIN book ON borrowed.isbn = book.isbn
)
SELECT memb_no, memb_name, publisher, COUNT(isbn)
FROM member_borrowed_book
GROUP BY memb_no, memb_name, publisher
HAVING COUNT(isbn) > 5;

d. Find the average number of books borrowed per member. Take into account that if a member does not borrow any books, then that member does not appear in the borrowed relation at all, but that member still counts in the average.

WITH number_of_books_borrowed(memb_no, memb_name, number_of_books) AS (
    SELECT memb_no, name, (
        CASE
            WHEN NOT EXISTS (SELECT * FROM borrowed WHERE borrowed.memb_no = member.memb_no) THEN 0
            ELSE (SELECT COUNT(*) FROM borrowed WHERE borrowed.memb_no = member.memb_no) 
        END
    )
    FROM member
)
SELECT AVG(number_of_books) AS average_number_of_books_borrowed_per_member
FROM number_of_books_borrowed

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