check for file existence in Python 3

Question

I would like to know how I could check to see if a file exist based on user input and if it does not, then i would like to create one.

so I would have something like:

file_name = input('what is the file name? ')

then I would like to check to see if file name exist.

If file name does exist, open file to write or read, if file name does not exist, create the file based on user input.

I know the very basic about files but not how to use user input to check or create a file.

What do you mean by "open it to write or read" - do you mean you want to do both, i.e. `r+` mode? — Tim Pietzcker, May 04 '14 at 17:14
I just meant that i needed to do it in a way that will allow me to do either. — krona, May 04 '14 at 17:15
Well, you can always open a file for writing - if it exists, it will be cleared, if it doesn't exist, it will be created. — Tim Pietzcker, May 04 '14 at 17:18

score 52 · Answer 1 · edited Dec 05 '21 at 15:00

52

if the file is not a file :(return False)

import os.path
if not os.path.isFile(file_name):
      print("The File s% it's not created "%file_name)
      os.touch(file_name)
      print("The file s% has been Created ..."%file_name)

And you can write a simple code based on (try,Except):

try:
    my_file = open(file_name)
except IOError:
    os.touch(file_name)

edited Dec 05 '21 at 15:00

rofrol

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answered May 04 '14 at 17:46

saudi_Dev

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What is os.touch? I can't find it at https://docs.python.org/3/library/os.html or https://docs.python.org/2/library/os.html . Also "s%" should be "%s". – Kivi Shapiro Feb 01 '17 at 20:38
isFile() -> isfile() – Mark Andersen Jan 21 '22 at 20:27

Amber · Answer 2 · 2014-05-04T17:21:55.887

5

To do exactly what you asked:

You're going to want to look at the os.path.isfile() function, and then the open() function (passing your variable as the argument).

However, as noted by @LukasGraf, this is generally considered less than ideal because it introduces a race condition if something else were you create the file in the time between when you check to see if it exists and when you go to open it.

Instead, the usual preferred method is to just try and open it, and catch the exception that's generated if you can't:

try:
    my_file = open(file_name)
except IOError:
    # file couldn't be opened, perhaps you need to create it

edited May 04 '14 at 17:21

answered May 04 '14 at 17:14

Amber

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Hmm, what if the user enters the name of a directory? Wouldn't `.exists()` return `True`, but opening would then fail? – Tim Pietzcker May 04 '14 at 17:15
@TimPietzcker Good call; updated to `isfile()`. – Amber May 04 '14 at 17:16
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That's exactly what you should **not** do, because that introduces a [race condition](http://stackoverflow.com/questions/14574518/how-does-using-the-try-statement-avoid-a-race-condition). – Lukas Graf May 04 '14 at 17:17
@LukasGraf *sigh* we're going down this again. Fine, I'll write up an entire explanation of exceptions as well for a user that is still figuring out how to use user input. Stand by... – Amber May 04 '14 at 17:18
@Amber no, just mark it as a duplicate, since this has been answered over and over again. – Lukas Graf May 04 '14 at 17:19

check for file existence in Python 3

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