Create a file if it doesn't exist

I'm trying to open a file, and if the file doesn't exist, I need to create it and open it for writing. I have this so far:

#open file for reading
fn = input("Enter file to open: ")
fh = open(fn,'r')
# if file does not exist, create it
if (!fh) 
fh = open ( fh, "w")

The error message says there's an issue on the line if(!fh). Can I use exist like in Perl?

Solution 1:

If you don't need atomicity you can use os module:

import os

if not os.path.exists('/tmp/test'):
    os.mknod('/tmp/test')

UPDATE:

As Cory Klein mentioned, on Mac OS for using os.mknod() you need a root permissions, so if you are Mac OS user, you may use open() instead of os.mknod()

import os

if not os.path.exists('/tmp/test'):
    with open('/tmp/test', 'w'): pass

Solution 2:

'''
w  write mode
r  read mode
a  append mode

w+  create file if it doesn't exist and open it in (over)write mode
    [it overwrites the file if it already exists]
r+  open an existing file in read+write mode
a+  create file if it doesn't exist and open it in append mode
'''

example:

file_name = 'my_file.txt'
f = open(file_name, 'a+')  # open file in append mode
f.write('python rules')
f.close()

I hope this helps. [FYI am using python version 3.6.2]

Solution 3:

Well, first of all, in Python there is no ! operator, that'd be not. But open would not fail silently either - it would throw an exception. And the blocks need to be indented properly - Python uses whitespace to indicate block containment.

Thus we get:

fn = input('Enter file name: ')
try:
    file = open(fn, 'r')
except IOError:
    file = open(fn, 'w')

Solution 4:

Here's a quick two-liner that I use to quickly create a file if it doesn't exists.

if not os.path.exists(filename):
    open(filename, 'w').close()