Error while using listdir in Python

Question

I\'m trying to get the list of files in a particular directory and count the number of files in the directory. I always get the following error:

WindowsError         


        

Answer 1

I decided to change the code into:

def numOfFiles(path):
    return len(next(os.walk(path))[2])

and use the following the call the code:

print numOfFiles("client_side")

Many thanks to everyone who told me how to pass the windows directory correctly in Python and to nrao91 in here for providing the function code.

EDIT: Thank you eryksun for correcting my code!

Answer 2

Checking for existence is subject to a race. Better to handle the error (beg forgiveness instead of ask permission). Plus, in Python 3 you can suppress errors. Use suppress from contextlib:

 with suppress(FileNotFoundError):
     for name in os.listdir('foo'):
         print(name)

Answer 3

Two things:

1. os.listdir() does not do a glob pattern matching, use the glob module for that
2. probably you do not have a directory called '/client_side/*.*', but maybe one without the . in the name

The syntax you used works fine, if the directory you look for exists, but there is no directory called '/client_side/.'.

In addition, be careful if using Python 2.x and os.listdir, as the results on windows are different when you use u'/client_side/' and just '/client_side'.

Answer 4

As I can see a WindowsError, Just wondering if this has something to do with the '/' in windows ! Ideally, on windows, you should have something like os.path.join('C:','client_side')

Answer 5

You can do just

os.listdir('client_side')

without slashes.

Answer 6

You want:

print len([name for name in os.listdir('./client_side/') if os.path.isfile(name)])

with a "." before "/client_side/".

The dot means the current path where you are working (i.e. from where you are calling your code), so "./client_side/" represents the path you want, which is specified relatively to your current directory.

If you write only "/client_side/", in unix, the program would look for a folder in the root of the system, instead of the folder that you want.