如何从 txt 文档中获取文本并创建新目录？

Question

The challenge is to take a name from a text file and create a new folder in a different directory. My issue is that when I attempt to do so it mentions that the

join() argument must be str, bytes, or os.PathLike object, not 'list'

Is there a way of converting a list to do this or is there another way of doing this I'm not seeing?

import os

clientnames = "/home/michael/tafe/Customer Service Team/Customer Service Team new client names" #filepath/filename of new client names.

#Error handeling, if the txt file is missing then it will say the file is unavaliable.
if (os.path.isfile(clientnames)) == True:
    print("The file is avaliable, folder creation will now start.")
else:
    print("The file is unavaliable, please provide Customer Service Team new client names .txt file")

folderdir = "/home/michael/tafe/FS1/Administration/New_Customers" #filepath of where the new folders are to be created in.

#take name from txt file.
with open(clientnames, "r") as newfolder:
    for line in newfolder:
        newfolder = line.strip().split()
        created_folder = os.path.join(folderdir, newfolder)

os.mkdir(created_folder)
print("Directory '% s' created" % newfolder)

I am pretty new to learning python but feel like I'm close to figuring this out once I find how to get the directory created. (there are some other parts to the challenge as well but are not related to this...).

Using Python 3.8.7 64-bit on Visual Studio Code.

Answer 1

The problem lies in the for loop of:

with open(clientnames, "r") as newfolder:
    for line in newfolder:
        newfolder = line.strip().split()
        created_folder = os.path.join(folderdir, newfolder)

star.split() returns an object of type list . Knowing that newFolder will be a list ,

created_folder = os.path.join(folderdir, newfolder)

Is like

created_folder = os.path.join("/home/michael/tafe/FS1/Administration/New_Customers", ["Tom", "Lam"])

How would python join the string with the list?

Remove the .split() and see what happens.

Answer 2

os.path.join accepts only strings, and you are passing a list as a second argument. Eg lets assume your newfolder looks like this: ['folder1', 'folder2', 'etc'] , your call to this function looks like:

>>> os.path.join(folderdir, ['folder1', 'folder2', 'etc'])
# error

Instead you have to pass only strings, like:

os.path.join(folderdir, 'folder1', 'folder2', 'etc')
# 'folderdir/folder1/folder2/etc'

To do this, you can use splat operator, to unpack argument lists:

os.path.join(folderdir, *newfolder)
# 'folderdir/folder1/folder2/etc'

Update the other problem as pointed by @Ann Zen is the for loop. Each iteration changes created_folder variable, which is used only after loop. You need to read all text from file (as list), then after loop join it into path using *newfolder .

with open(clientnames, "r") as newfolder:
    newfolder = newfolder.read()
    newfolder = [t for t in newfolder.split(' ') if len(t)>0]
created_folder = os.path.join(folderdir, *newfolder)

Update 2 Another problem is when you are trying to create multiple nested folders in a path, eg create folder test1 in folder test2, but folder test2 doesn't exist yet, so need to be created first. Assuming each folder is named in another line in file clientnames :

with open(clientnames, "r") as newfolder:
    folderpath = folderdir
    for line in newfolder.readlines():
        foldername = line.strip()
        folderpath = os.path.join(folderpath, foldername)
        os.mkdir(folderpath)

Then if the file clientnames looks like this:

test1
test2
test3

it will first create folderdir/test1 , then folderdir/test1/test2 , and at the end folderdir/test1/test2/test3 , so not to throw FileNotFoundError

Update 3 If you want to create folder for each line of text from clientnames , all rooted in folderdir simply don't stack the path, and create new one for each line.

with open(clientnames, "r") as newfolder:
    for line in newfolder.readlines():
        foldername = line.strip()
        folderpath = os.path.join(folderdir, foldername)
        os.mkdir(folderpath)