How to open multiple files in loop, in python

Question

I'm new to python. I'm want to open multiple files in Python. I'm can open each of them with open() function. I'm not sure about formatting.

with open("/test/filename.css", "r") as f:
     s = f.readlines()
     print(s)

I'm can open one file, but I'm not sure how to open multiple files. This is the code that I have. In live_filename() function have many files.

inputfiles = live_filename()
    for live in inputfiles:
        with open("/test/............. .css, "r") as f:

I don't know how to put code formatting in space. and I think the live variable is a tuple can't concatenate str. what should i do?

Answer 1

Open each just as you did for one, and then append them to a list:

import os

folderpath = r"D:\my_data" # make sure to put the 'r' in front
filepaths  = [os.path.join(folderpath, name) for name in os.listdir(folderpath)]
all_files = []

for path in filepaths:
    with open(path, 'r') as f:
        file = f.readlines()
        all_files.append(file)

Now, all_files[0] holds the first file loaded, all_files[1] the second , and so on.

---

UPDATE : for all files in the same folder: first, get the folder path (on Windows, like this ). Suppose it's "D:\my_data" . Then, you can get all the filepaths of the files as in the script above.

Answer 2

You can do like this:

folder = "..." # Absolute path to folder in which the files reside
files_to_read = [("filename.css","FileName"),( "filename2.css","Filename2")]
for (file, _) in files_to_read:
    filepath = os.path.join(folder, file)
    with open(filepath, "r") as f:
        s = f.readlines()
        print(s)