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Should I close a file with file.close() when there is a FileNotFoundError exception?

I'm trying to open a file in python and print a message when the file doesn't exist. But I'm confused whether to close the file or not when the exception happens.
try:
file = open(sys.argv[1], "r")
file.close() # should I do this?
except OSError:
print(f"{sys.argv[1]} file not found.")

A simpler method of checking if a file exists:
import os
if not os.path.exists(sys.argv[1]):
print(f"{sys.argv[1]} file not found.")
But to answer your question, the ```file.close()`` happens only when the file exists and you successfully open the file. Not when the exception occurs.
Edit:
As pointed out by #ekhumoro, the above has a race condition (when other processes access that file). If no other process accesses that file, then the above code works.
Solution is as #ekhumoro pointed out is to use your original try/except method.

Why a new NamedTemporaryFile object has a path, but a file is not available? [duplicate]

I am attempting to create and write to a temporary file on Windows OS using Python. I have used the Python module tempfile to create a temporary file.
But when I go to write that temporary file I get an error Permission Denied. Am I not allowed to write to temporary files?! Am I doing something wrong? If I want to create and write to a temporary file how should should I do it in Python? I want to create a temporary file in the temp directory for security purposes and not locally (in the dir the .exe is executing).
IOError: [Errno 13] Permission denied: 'c:\\users\\blah~1\\appdata\\local\\temp\\tmpiwz8qw'
temp = tempfile.NamedTemporaryFile().name
f = open(temp, 'w') # error occurs on this line

NamedTemporaryFile actually creates and opens the file for you, there's no need for you to open it again for writing.
In fact, the Python docs state:
Whether the name can be used to open the file a second time, while the named temporary file is still open, varies across platforms (it can be so used on Unix; it cannot on Windows NT or later).
That's why you're getting your permission error. What you're probably after is something like:
f = tempfile.NamedTemporaryFile(mode='w') # open file
temp = f.name # get name (if needed)

Use the delete parameter as below:
tmpf = NamedTemporaryFile(delete=False)
But then you need to manually delete the temporary file once you are done with it.
tmpf.close()
os.unlink(tmpf.name)
Reference for bug: https://github.com/bravoserver/bravo/issues/111
regards,
Vidyesh

Consider using os.path.join(tempfile.gettempdir(), os.urandom(24).hex()) instead. It's reliable, cross-platform, and the only caveat is that it doesn't work on FAT partitions.
NamedTemporaryFile has a number of issues, not the least of which is that it can fail to create files because of a permission error, fail to detect the permission error, and then loop millions of times, hanging your program and your filesystem.

The following custom implementation of named temporary file is expanded on the original answer by Erik Aronesty:
import os
import tempfile
class CustomNamedTemporaryFile:
"""
This custom implementation is needed because of the following limitation of tempfile.NamedTemporaryFile:
> Whether the name can be used to open the file a second time, while the named temporary file is still open,
> varies across platforms (it can be so used on Unix; it cannot on Windows NT or later).
"""
def __init__(self, mode='wb', delete=True):
self._mode = mode
self._delete = delete
def __enter__(self):
# Generate a random temporary file name
file_name = os.path.join(tempfile.gettempdir(), os.urandom(24).hex())
# Ensure the file is created
open(file_name, "x").close()
# Open the file in the given mode
self._tempFile = open(file_name, self._mode)
return self._tempFile
def __exit__(self, exc_type, exc_val, exc_tb):
self._tempFile.close()
if self._delete:
os.remove(self._tempFile.name)

This issue might be more complex than many of you think. Anyway this was my solution:
Make use of atexit module
def delete_files(files):
for file in files:
file.close()
os.unlink(file.name)
Make NamedTemporaryFile delete=False
temp_files = []
result_file = NamedTemporaryFile(dir=tmp_path(), suffix=".xlsx", delete=False)
self.temp_files.append(result_file)
Register delete_files as a clean up function
atexit.register(delete_files, temp_files)

tempfile.NamedTemporaryFile() :
It creates and opens a temporary file for you.
f = open(temp, 'w') :
You are again going to open the file which is already open and that's why you are getting Permission Denied error.
If you really wants to open the file again then you first need to close it which will look something like this-
temp= tempfile.NamedTemporaryFile()
temp.close()
f = open(temp.name, 'w')

Permission was denied because the file is Open during line 2 of your code.
close it with f.close() first then you can start writing on your tempfile

exception stop read files python

I'm trying to control exceptions when reading files, but I have a problem. I'm new to Python, and I am not yet able to control how I can catch an exception and still continue reading text from the files I am accessing. This is my code:
import errno
import sys
class Read:
#FIXME do immutables this 2 const
ROUTE = "d:\Profiles\user\Desktop\\"
EXT = ".txt"
def setFileReaded(self, fileToRead):
content = ""
try:
infile = open(self.ROUTE+fileToRead+self.EXT)
except FileNotFoundError as error:
if error.errno == errno.ENOENT:
print ("File not found, please check the name and try again")
else:
raise
sys.exit()
with infile:
content = infile.read()
infile.close()
return content
And from another class I tell it:
read = Read()
print(read.setFileReaded("verbs"))
print(read.setFileReaded("object"))
print(read.setFileReaded("sites"))
print(read.setFileReaded("texts"))
Buy only print this one:
turn on
connect
plug
File not found, please check the name and try again
And no continue with the next files. How can the program still reading all files?

It's a little difficult to understand exactly what you're asking here, but I'll try and provide some pointers.
sys.exit() will terminate the Python script gracefully. In your code, this is called when the FileNotFoundError exception is caught. Nothing further will be ran after this, because your script will terminate. So none of the other files will be read.
Another thing to point out is that you close the file after reading it, which is not needed when you open it like this:
with open('myfile.txt') as f:
content = f.read()
The file will be closed automatically after the with block.

Detecting for invalid file inputs, Python

I have an assignment to write a Python script which "detects whether the file is readable or not".
I am stuck as to which exceptions I should run. Let's say the input file is intended to be a text file, with extension *.txt
What is the exception I should raise? I suspect there should be multiple. At the moment, I have:
with open('example_file.txt") as textfile:
if not textfile.lower().endswith('.txt'):
raise argparse.ArgumentTypeError(
'Not a text file! Argument filename must be of type *.txt')
return textfile
However, that only checks the file extension. What else could I possibly check? What is the standard for file I/O in Python?

To check whether the file exists:
import os.path
if os.path.exists('example_file.txt'):
print('it exists!')
Beyond this, successfully opening the file will demonstrate readability. The built-in open raises an IOError exception if it fails. Failure can occur for more than one reason, so we must check whether it failed due to readability:
import errno
try:
textfile = open('example_file.txt', 'r')
textfile.close()
print("file is readable")
except IOError as e:
if e.errno == errno.EACCES:
print("file exists, but isn't readable")
elif e.errno == errno.ENOENT:
print("files isn't readable because it isn't there")
The relevant section of the docs on file permissions. Note that the use of os.access to check readability before calling open is discouraged.

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?

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

I assume that you're writing to the file, then closing it (so the user can open it in Excel), and then, before re-opening it for append/write operations, you want to check that the file isn't still open in Excel?
This is how you could do that:
while True: # repeat until the try statement succeeds
try:
myfile = open("myfile.csv", "r+") # or "a+", whatever you need
break # exit the loop
except IOError:
input("Could not open file! Please close Excel. Press Enter to retry.")
# restart the loop
with myfile:
do_stuff()

For windows only
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.')

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

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

Try this method if the above methods corrupt your excel file.
This function attempts to rename the file with its own name. If the file has already been opened, the edit will be reject by the os and an OSError exception will be raised. It does not touch the inner code so it will not corrupt your excel files. LMK if it worked for you.
def check_file_status(self):
try:
os.rename("file1.xlsx", "file1.xlsx")
print("File is closed.")
except OSError:
print("File is opened.")

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

Just use this function. It will close any already opened excel file
import os
def close():
try:
os.system('TASKKILL /F /IM excel.exe')
except Exception:
print("KU")
close()