[python] check if a file is open in Python

In my app, I write to an excel file. After writing, the user is able to view the file by opening it. But if the user forgets to close the file before any further writing, a warning message should appear. So I need a way to check this file is open before the writing process. Could you supply me with some python code to do this task?

Using

try:
with open("path", "r") as file:#or just open

may cause some troubles when file is opened by some other processes (i.e. user opened it manually). You can solve your poblem using win32com library. Below code checks if any excel files are opened and if none of them matches the name of your particular one, openes a new one.

import win32com.client as win32
xl = win32.gencache.EnsureDispatch('Excel.Application')

my_workbook = "wb_name.xls"
xlPath="my_wb_path//" + my_workbook


if xl.Workbooks.Count > 0:
    # if none of opened workbooks matches the name, openes my_workbook 
    if not any(i.Name == my_workbook for i in xl.Workbooks): 
        xl.Workbooks.Open(Filename=xlPath)
        xl.Visible = True
#no workbooks found, opening
else:  
    xl.Workbooks.Open(Filename=xlPath)
    xl.Visible = True

'xl.Visible = True is not necessary, used just for convenience'

Hope this will help

If all you care about is the current process, an easy way is to use the file object attribute "closed"

f = open('file.py')
if f.closed:
  print 'file is closed'

This will not detect if the file is open by other processes!

source: http://docs.python.org/2.4/lib/bltin-file-objects.html

You could use with open("path") as file: so that it automatically closes, else if it's open in another process you can maybe try as in Tims example you should use except IOError to not ignore any other problem with your code :)

try:
   with open("path", "r") as file: # or just open
       # Code here
except IOError:
   # raise error or print

if myfile.closed == False:
   print("File is still open ################")

None of the other provided examples would work for me when dealing with this specific issue with excel on windows 10. The only other option I could think of was to try and rename the file or directory containing the file temporarily, then rename it back.

import os

try: 
    os.rename('file.xls', 'tempfile.xls')
    os.rename('tempfile.xls', 'file.xls')
except OSError:
    print('File is still open.')