check for file existence in Python 3 [duplicate]

Answer 1

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).

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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

Answer 2

You can use the os module:

Like this By os.path.isfile():

import os.path
os.path.isfile(file_path)

Or Using os.path.exists():

if os.path.exists(file_name): print("I get the file >hubaa")

if the file doesn't exist :(return False)

if not os.path.exists(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)