How to Check If Directory Exists in Python

How do I check if directory exists in Python?

Use os.path.isdir for directories only:

>>> import os
>>> os.path.isdir('new_folder')
True

Use os.path.exists for both files and directories:

>>> import os
>>> os.path.exists(os.path.join(os.getcwd(), 'new_folder', 'file.txt'))
False

Alternatively, you can use pathlib:

 >>> from pathlib import Path
 >>> Path('new_folder').is_dir()
 True
 >>> (Path.cwd() / 'new_folder' / 'file.txt').exists()
 False

How can I safely create a nested directory?

On Python ≥ 3.5, use pathlib.Path.mkdir:

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:

Try os.path.exists, and consider os.makedirs for the creation.

import os
if not os.path.exists(directory):
    os.makedirs(directory)

As noted in comments and elsewhere, there's a race condition – if the directory is created between the os.path.exists and the os.makedirs calls, the os.makedirs will fail with an OSError. Unfortunately, blanket-catching OSError and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.

One option would be to trap the OSError and examine the embedded error code (see Is there a cross-platform way of getting information from Python’s OSError):

import os, errno

try:
    os.makedirs(directory)
except OSError as e:
    if e.errno != errno.EEXIST:
        raise

Alternatively, there could be a second os.path.exists, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.

Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.

Modern versions of Python improve this code quite a bit, both by exposing FileExistsError (in 3.3+)...

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

...and by allowing a keyword argument to os.makedirs called exist_ok (in 3.2+).

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

Python: Check if a directory exists using os module

Have you read the docs for the os module?

Check out the following two links:

os.path.exists()

Return True if path refers to an existing path.

os.path.isdir()

Return True if path is an existing directory.

How do I check if a directory exists and then, if so, ask for a new directory name in Python?

Problem

When you are calling the function recursively, you are not returning from the previously called function.

What happens is, when you call the casefile first time and enter an existing name i.e Case1, an entry is created in the stack for it. Lets call it casefile{1}. At the line casefile() in the else statement, the function will be called recursively.

Now, due to this recursive call, another entry for the function is created in the stack. Lets call this casefile{2}. Now the controlled would be passed to this function, but note that casefile{1} has not yet run its course - you have not returned from it.

So now the control is at casefile{2}. This time you enter a non-existing name i.e Case2. The casefile{2} does not go to the else branch and thus no more recursive calls would be made. So not it prints the name just like you asked it to (which in the scope of casefile{2} happens to be Case2, and returns the name.

But now the control goes back to casefile{1}, which will now exit the else statement and run the next line: print case_name, which for the scope of casefile{1} happens to be Case1. So it prints it and returns.

Solution

You should return after calling the function recursively.
Please change the code to this:

def casefile():
    print '\n\nPlease enter a case name/reference: '
    case_name = raw_input()
    if not os.path.exists('C:\\Users\\laura17\\Documents\\ProgrammeOutput\\{}'.format(case_name)):
        os.mkdir('C:\\Users\\laura17\\Documents\\ProgrammeOutput\\{}'.format(case_name))
        print '\n{} case file created. All logs will be saved to C:\Users\laura17\Documents\ProgrammeOutput\{}' \
            .format(case_name, case_name)
    else:
        print '\n**Case already exists. Please choose a different case name.**'
        return casefile()
    print case_name
    return case_name

Please note that i have changed the line casefile() in the else statement to return casefile().

This would only print/return the name being passed to it by its recursively called functions.

Hope this helps. Cheers!

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