I'm trying to open a file, and if the file doesn't exist, I need to create it and open it for writing:
#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?
If you don't need atomicity you can use os module:
import os if not os.path.exists('/tmp/test'): os.mknod('/tmp/test') On macOS for using os.mknod() you need root permissions. On Windows there is no os.mknod() method.
So as an alternative, you may use open() instead of os.mknod()
import os if not os.path.exists('/tmp/test'): with open('/tmp/test', 'w'): pass 
sudo.
exists() and open().
exists() and open() is two separated calls this solution is not atomic. In theory there is a time between this two function calls when another program may also check existance of a file and create it if no file found.
You can achieve the desired behaviour with
file_name = 'my_file.txt' f = open(file_name, 'a+') # open file in append mode f.write('python rules') f.close() These are some valid options for the second parameter mode in open():
""" 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 """ 
r+ does not create any file. As also mentioned here and here(in the description) r+ is for opening a file in reading and writing mode. Correct it as it might confuse people :)
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') 
open(fn, 'a').close(). Is it because the implicit seek in append may be too expensive?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() 
import os must be declared before using this two-liner.Using input() implies Python 3, recent Python 3 versions have made the IOError exception deprecated (it is now an alias for OSError). So assuming you are using Python 3.3 or later:
fn = input('Enter file name: ') try: file = open(fn, 'r') except FileNotFoundError: file = open(fn, 'w') I think this should work:
#open file for reading fn = input("Enter file to open: ") try: fh = open(fn,'r') except: # if file does not exist, create it fh = open(fn,'w') Also, you incorrectly wrote fh = open(fh, "w") when the file you wanted open was fn
except is not a good idea.Be warned, each time the file is opened with this method the old data in the file is destroyed regardless of 'w+' or just 'w'.
with open("file.txt", 'w+') as f: f.write("file is opened for business") First let me mention that you probably don't want to create a file object that eventually can be opened for reading OR writing, depending on a non-reproducible condition. You need to know which methods can be used, reading or writing, which depends on what you want to do with the file object.
That said, you can do it as @That One Random Scrub proposed, using try: ... except:. Actually that is the proposed way, according to the python motto "It's easier to ask for forgiveness than permission".
But you can also easily test for existence:
import os # open file for reading fn = input("Enter file to open: ") if os.path.exists(fn): fh = open(fn, "r") else: fh = open(fn, "w") 
input() and raw_input() only apply to Python 2. Python 3 has replaced raw_input() with input() and the "old" use of input() is gone.