Trying to solve library management in python core

Question

Know am missing something but can seem to remember
books = {
"Life of Pi": "Adventure Fiction",
"The Three Musketeers": "Historical Adventure",
"Watchmen": "Comics",
"Bird Box": "Horror",
"Harry Potter":"Fantasy Fiction",
"Good Omens": "Comedy"
}

book = input()

#your code goes here
if book not in books:
print("Not found")

Answer 1

it works like this :

books = {
"Life of Pi": "Adventure Fiction",
"The Three Musketeers": "Historical Adventure",
"Watchmen": "Comics",
"Bird Box": "Horror",
"Harry Potter":"Fantasy Fiction",
"Good Omens": "Comedy"
}

book = input()
if book in books:
print(books[book])

#your code goes here
else :

Answer 2

books = {
"Life of Pi": "Adventure Fiction",
"The Three Musketeers": "Historical Adventure",
"Watchmen": "Comics",
"Bird Box": "Horror",
"Harry Potter":"Fantasy Fiction",
"Good Omens": "Comedy"
}

book = input()

#your code goes here
print (books.get(book, "Not found"))

Answer 3

This worked for me(:

books = {
"Life of Pi": "Adventure Fiction",
"The Three Musketeers": "Historical Adventure",
"Watchmen": "Comics",
"Bird Box": "Horror",
"Harry Potter":"Fantasy Fiction",
"Good Omens": "Comedy"
}

book = input()

#change this part to use the .get() method

if book in books:
print(books.get(book))
else:
print("Not found")

Answer 4

i think since its a dictionary, you could compare input to the dictionary keys instead of the key:value pairs.

if found, you'd need to access the dictionary by using the key and print out its genre value.

Answer 5

Thanks

Answer 6

if book in books:
print(books[book])

Answer 7

book = input()
try:
print(books[book])
except:
print("Not found")

Answer 8

#add the print statement in else part
books = {
"Life of Pi": "Adventure Fiction",
"The Three Musketeers": "Historical Adventure",
"Watchmen": "Comics",
"Bird Box": "Horror",
"Harry Potter":"Fantasy Fiction",
"Good Omens": "Comedy"
}

book = input()
if book in books:
print(books[book])
else:
print("Not found")

Answer 9

# Most efficient simplest full answer

books = {
"Life of Pi": "Adventure Fiction",
"The Three Musketeers": "Historical Adventure",
"Watchmen": "Comics",
"Bird Box": "Horror",
"Harry Potter":"Fantasy Fiction",
"Good Omens": "Comedy"
}

book = input()
if book in books:
print(books[book])
else:
print("Not found")

Answer 10

class Book:
def __init__(self, title, author):
self.title = title
self.author = author

class Ebook(Book):
def __init__(self, title, author, file_size):
super().__init__(title, author) # Call the parent class constructor
self.file_size = file_size # Initialize the file_size attribute

# Creating an instance of Ebook
my_ebook = Ebook("1984", "George Orwell", 10)
print(my_ebook.title) # Output: 1984
print(my_ebook.author) # Output: George Orwell
print(my_ebook.file_size, 'MB') # Output: 10 MB

Answer 11

A library management program has a dictionary of books with their corresponding genres.

Take a book name as input and output the genre.

Answer 12

This might help:

print(books.get(input(), "Not found"))

Answer 13

for i in books:
if i==book:
print(books[i])
break
else:
print("Not found")

Answer 14

When the problem is asking to use the .get() method instead of if/else, I used:

print(books.get(book, "Book not found"))

Answer 15

books = {
"Life of Pi": "Adventure Fiction",
"The Three Musketeers": "Historical Adventure",
"Watchmen": "Comics",
"Bird Box": "Horror",
"Harry Potter":"Fantasy Fiction",
"Good Omens": "Comedy"
}

book = input()
print(books.get(book, "Not found"))