Python WindowsError: [Error 123] The filename, directory name, or volume label syntax is incorrect:

Question

I am new to programming, this is actually my first work assignment with coding. my code below is throwing an error:

WindowsError: [Error 123] The filename,          


        

Answer 1

I had a related issue working within Spyder, but the problem seems to be the relationship between the escape character ( "\") and the "\" in the path name Here's my illustration and solution (note single \ vs double \\ ):

path =   'C:\Users\myUserName\project\subfolder'
path   # 'C:\\Users\\myUserName\\project\subfolder'
os.listdir(path)              # gives windows error
path =   'C:\\Users\\myUserName\\project\\subfolder'
os.listdir(path)              # gives expected behavior

Answer 2

I had a similar issue while working with Jupyter. I was trying to copy files from one directory to another using copy function of shutil. The problem was that I had forgotten to import the package.(Silly) But instead of python giving import error, it gave this error.

Solved by adding:

from shutil import copy

Answer 3

This is kind of an old question but I wanted to mentioned here the pathlib library in Python3.

If you write:

from pathlib import Path
path: str = 'C:\\Users\\myUserName\\project\\subfolder'
osDir = Path(path)

or

path: str = "C:\\Users\\myUserName\\project\\subfolder"
osDir = Path(path)

osDir will be the same result.

Also if you write it as:

path: str = "subfolder"
osDir = Path(path)
absolutePath: str = str(Path.absolute(osDir))

you will get back the absolute directory as

'C:\\Users\\myUserName\\project\\subfolder'

You can check more for the pathlib library here.

Answer 4

As it solved the problem, I put it as an answer.

Don't use single and double quotes, especially when you define a raw string with r in front of it.

The correct call is then

path = r"C:\Apps\CorVu\DATA\Reports\AlliD\Monthly Commission Reports\Output\pdcom1"

or

path = r'C:\Apps\CorVu\DATA\Reports\AlliD\Monthly Commission Reports\Output\pdcom1'

Answer 5

I was facing same error with Django Rest Framework, It was nothing to do with UI, still was getting this error. I applied below solution, worked for me.

• Restarted Machine.
• Restarted Virtual Environment.

Answer 6

I had this problem with Django and it was because I had forgotten to start the virtual environment on the backend.