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Check the permissions of a file in python

I'm trying to check the readability of a file given the specified path. Here's what I have:

def read_permissions(filepath):
    '''Checks the read permissions of the specified file'''
    try:
        os.access(filepath, os.R_OK) # Find the permissions using os.access
    except IOError:
        return False

    return True

This works and returns True or False as the output when run. However, I want the error messages from errno to accompany it. This is what I think I would have to do (But I know that there is something wrong):

def read_permissions(filepath):
    '''Checks the read permissions of the specified file'''
    try:
        os.access(filepath, os.R_OK) # Find the permissions using os.access
    except IOError as e:
        print(os.strerror(e)) # Print the error message from errno as a string

    print("File exists.")

However, if I were to type in a file that does not exist, it tells me that the file exists. Can someone help me as to what I have done wrong (and what I can do to stay away from this issue in the future)? I haven't seen anyone try this using os.access. I'm also open to other options to test the permissions of a file. Can someone help me in how to raise the appropriate error message when something goes wrong?

Also, this would likely apply to my other functions (They still use os.access when checking other things, such as the existence of a file using os.F_OK and the write permissions of a file using os.W_OK). Here is an example of the kind of thing that I am trying to simulate:

>>> read_permissions("located-in-restricted-directory.txt") # Because of a permission error (perhaps due to the directory)
[errno 13] Permission Denied
>>> read_permissions("does-not-exist.txt") # File does not exist
[errno 2] No such file or directory

This is the kind of thing that I am trying to simulate, by returning the appropriate error message to the issue. I hope that this will help avoid any confusion about my question.

I should probably point out that while I have read the os.access documentation, I am not trying to open the file later. I am simply trying to create a module in which some of the components are to check the permissions of a particular file. I have a baseline (The first piece of code that I had mentioned) which serves as a decision maker for the rest of my code. Here, I am simply trying to write it again, but in a user-friendly way (not just True or just False, but rather with complete messages). Since the IOError can be brought up a couple different ways (such as permission denied, or non-existent directory), I am trying to get my module to identify and publish the issue. I hope that this helps you to help me determine any possible solutions.

like image 580
Zizouz212 Avatar asked Dec 11 '14 23:12

Zizouz212


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1 Answers

os.access returns False when the file does not exist, regardless of the mode parameter passed.

This isn't stated explicitly in the documentation for os.access but it's not terribly shocking behavior; after all, if a file doesn't exist, you can't possibly access it. Checking the access(2) man page as suggested by the docs gives another clue, in that access returns -1 in a wide variety of conditions. In any case, you can simply do as I did and check the return value in IDLE:

>>> import os
>>> os.access('does_not_exist.txt', os.R_OK)
False

In Python it's generally discouraged to go around checking types and such before trying to actually do useful things. This philosophy is often expressed with the initialism EAFP, which stands for Easier to Ask Forgiveness than Permission. If you refer to the docs again, you'll see this is particularly relevant in the present case:

Note: Using access() to check if a user is authorized to e.g. open a file before actually doing so using open() creates a security hole, because the user might exploit the short time interval between checking and opening the file to manipulate it. It’s preferable to use EAFP techniques. For example:

if os.access("myfile", os.R_OK):
    with open("myfile") as fp:
        return fp.read()
return "some default data"

is better written as:

try:
    fp = open("myfile")
except IOError as e:
    if e.errno == errno.EACCES:
        return "some default data"
    # Not a permission error.
    raise
else:
    with fp:
        return fp.read()

If you have other reasons for checking permissions than second-guessing the user before calling open(), you could look to How do I check whether a file exists using Python? for some suggestions. Remember that if you really need an exception to be raised, you can always raise it yourself; no need to go hunting for one in the wild.


Since the IOError can be brought up a couple different ways (such as permission denied, or non-existent directory), I am trying to get my module to identify and publish the issue.

That's what the second approach above does. See:

>>> try:
...     open('file_no_existy.gif')
... except IOError as e:
...     pass
...
>>> e.args
(2, 'No such file or directory')
>>> try:
...     open('unreadable.txt')
... except IOError as e:
...     pass
...
>>> e.args
(13, 'Permission denied')
>>> e.args == (e.errno, e.strerror)
True

But you need to pick one approach or the other. If you're asking forgiveness, do the thing (open the file) in a try-except block and deal with the consequences appropriately. If you succeed, then you know you succeeded because there's no exception.

On the other hand, if you ask permission (aka LBYL, Look Before You Leap) in this that and the other way, you still don't know if you can successfully open the file until you actually do it. What if the file gets moved after you check its permissions? What if there's a problem you didn't think to check for?

If you still want to ask permission, don't use try-except; you're not doing the thing so you're not going to throw errors. Instead, use conditional statements with calls to os.access as the condition.

like image 96
Air Avatar answered Oct 06 '22 00:10

Air