Passing variable to zipfile NAME

Question

I want to assign to my zipfile from a variable named file_path, how I can do that? I tried with the code below, and this line is the problem:

with ZipFile('%s.zip' %(file_path,),'w') as zip:

Nothing happens; I get no errors just Python doesn't create the .zip file. Here is all the code.

import os
from zipfile import ZipFile 

file_paths = [] 


basepath = 'my_directory/'
with os.scandir(basepath) as entries:
    for root, directories, files in os.walk(basepath):
        for entry in entries:

            if entry.is_file():
                file_path = os.path.join(entry)
                file_paths.append(file_path)

                with ZipFile('%s.zip' %(file_path,),'w') as zip:
                    print("FILE:", entry.name)
                    for entry in file_paths:
                        zip.write(entry)

Answer 1

I stumbled upon this unanswered query while I was go though something else, a simple work around would be to use .format as below

with ZipFile('{}.zip'.format(d),'w') as zip: