Python 中是否有办法以不区分大小写的方式比较变量名和文件名？

Question

I have a function called get_book which opens requested book by its title argument. Function get_book can receive title in any case (eg Foo , FOO , fOo ). While this is not a problem in itself, book folder where the function retrieves a book if the title 's are matching, also has book title s in a mixed case way. The database folder is dynamic and continuous to receive new book file names (like eg Foo.pdf , FOO.pdf , FOo.pdf ).

My question is how can I compare title s and followingly retrieve requested book without changing file names in the database? Is there an efficient way to open files without worrying about case matching?

my code:

def get_book(title):
    """
    Retrieves a book by its title. If no such
    book exists, the function returns None.
    """
    try:
        f = default_storage.open(f"books/{title}.pdf")
        return f.read().decode("utf-8")
    except FileNotFoundError:
        return None

Answer 1

Start with an empty lower_to_name dict. Probe it like this, for each new title:

title = title.lower() + ".pdf"
title = lower_to_name.get(title, title)

Upon getting FileNotFoundError for a title, do a single retry using glob to lookup actual filename:

lower_to_name = {name.lower(): name
                 for name in glob.glob("*.pdf")}
title = lower_to_name.get(title, "no_such_file")

(You may find that os.chdir("books/") is convenient.)

Answer 2

For finding a case-insensitive file name, fnmatch is a good method.

This example is almost literally the example in the documentation:

import fnmatch
import os

title = 'tEst.txt'                    # Try several combinations of upper and lower case
for file in os.listdir('.'):
    if fnmatch.fnmatch(file, title):  # They will be all found here
        print(file)