Get the filename from a path

[12237ee1]

Iterate over the open file, then you get line by line.
Instead of regex you could use str.rsplit() or str.rpartition().

pdf_files = set()
# https://docs.python.org/3/library/stdtypes.html#set

with open(file) as fd:
    for line in file:
        url, last = line.strip().rsplit("/", 1)
        # https://docs.python.org/3/library/stdtypes.html#str.rsplit
        print(last)
        # then check if last.endswith(".pdf")
        # if true, add it to the set


# Now sort pdf_files with sorted
# https://docs.python.org/3/library/functions.html#sorted
sorted_pdfs = sorted(pdf_files)

PS: If in one line is more than one address, you could split them with the split() method.

I am using linux so where should I put the path of the file in the script
I tried it like this
a= "/home/file.txt"
pdf_files = set(a)
but it didn't work .. also I need the result without brackets only the name of one file in each line
thank you