标题打印多次？

def display_books_year_range(): start_year = input("Enter your start year :\n") end_year = input("Enter your end year: \n") book_counter = 0 for book in years: if start_year <= book <= end_year: book_index = years.index(book) print(titles[book_index] + ", by:", authors[book_index]) book_counter = book_counter + 1 if book_counter == 0: print("No books have been found within that range.\n") make_choice()

2条回答

网友

1楼 · 编辑于 2024-05-16 04:10:34

这里的问题是index。记住，它会返回第一个匹配项。如果你有四本2005年的书，那么你会看到第一本书四次

def display_books_year_range():
    start_year = input("Enter your start year :\n")
    end_year = input("Enter your end year: \n")
    book_counter = 0
    for year,title,author in zip(years,titles,authors):
        if start_year <= year <= end_year:
            print(title + ", by:", author)
            book_counter = book_counter + 1
    if book_counter == 0:
        print("No books have been found within that range.\n")
        make_choice()

从长远来看，您不应该将这些信息存储在三个单独的列表中。相反，创建一个“课堂手册”来存储您的三条信息。然后，您可以作为一个单元遍历图书列表

网友

2楼 · 编辑于 2024-05-16 04:10:34

以下是我的解决方案：

def display_books_year_range():
    start_year = int(input("Enter your start year :\n"))
    end_year = int(input("Enter your end year: \n"))
    book_counter = 0
    for book, title, author in zip(years, titles, authors):
        if start_year <= book <= end_year:
                print(f'{title}, by: {author}')
                book_counter = book_counter + 1
    if book_counter == 0:
        print("No books have been found within that range.\n")
        make_choice()

通过使用zip函数，您可以同时循环遍历multiples list的每个元素，而不会遇到list.index（）方法的问题

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