How to check whether a file is_open and the open_status in python

Question

Is there any python functions such as:

filename = \"a.txt\"
if is_open(filename) and open_status(filename)==\'w\':
   print filename,\" is open for writing\"         


        

Answer 1

Here is an is_open solution for windows using ctypes:

from ctypes import cdll

_sopen = cdll.msvcrt._sopen
_close = cdll.msvcrt._close
_SH_DENYRW = 0x10

def is_open(filename):
    if not os.access(filename, os.F_OK):
        return False # file doesn't exist
    h = _sopen(filename, 0, _SH_DENYRW, 0)
    if h == 3:
        _close(h)
        return False # file is not opened by anyone else
    return True # file is already open

Answer 2

This is not quite what you want, since it just tests whether a given file is write-able. But in case it's helpful:

import os

filename = "a.txt"
if not os.access(filename, os.W_OK):
    print "Write access not permitted on %s" % filename

(I'm not aware of any platform-independent way to do what you ask)

Answer 3

I think the best way to do that is to try.

def is_already_opened_in_write_mode(filename)
     if os.path.exists(filename):
         try:
             f = open(filename, 'a')
             f.close()
         except IOError:
             return True
     return False

Answer 4

I don't think there is an easy way to do what you want, but a start could be to redefine open() and add your own managing code. That said, why do you want to do that?