Opening all files in a directory - Python

Question

I found some example code for ClamAV. And it works fine, but it only scans a single file. Here's the code:

import pyclamav
import os

tmpfile = '/home/user/test.txt'
f = open(tmpfile, 'rb')

infected, name = pyclamav.scanfile(tmpfile)
if infected:
    print "File infected with %s Deleting file." %name
    os.unlink(file)
else:
    print "File is clean!"

I'm trying to scan an entire directory, here's my attempt:

import pyclamav
import os
directory = '/home/user/'

for filename in os.listdir(directory):
    f = open(filename, 'rb')
    infected, name = pyclamav.scanfile(filename)
    if infected:
        print "File infected with %s ... Deleting file." %name
        os.unlink(filename)
    else:
        print " %s is clean!" %filename

However, I'm getting the following error:

Traceback (most recent call last):
  File "anti.py", line 7, in <module>
    f = open(filename, 'rb')
IOError: [Errno 21] Is a directory: 'Public'

I'm pretty new to Python, and I've read several similar questions and they do something like what I did, I think.

does it help https://stackoverflow.com/questions/34601758/python-ioerror-errno-21-is-a-directory-home-thomasshera-pictures-star-war ? — Vadim, May 30 '17 at 06:41
1) If you have working code and need to apply that to multiple objects, at least make it into a function and call that for every object. 2) skip directories returned by `listdir()`, as the error clearly indicates there lies the problem — Anthon, May 30 '17 at 06:44

score 3 · Answer 1 · answered May 30 '17 at 07:01

os.listdir("DIRECTORY") returns list of all files/dir in the DIRECTORY . It is just file names not absolute paths. So, if You are executing this program from a different directory it's bound to fail. If you are sure that everything in the directory is a file, no sub directories. You can try following,

def get_abs_names(path):
    for file_name in os.listdir(path):
        yield os.path.join(path, file_name)

Then ,

for file_name in get_abs_names("/home/user/"):
    #Your code goes here.

score 2 · Accepted Answer · edited May 30 '17 at 07:19

2

The following code will go over all your directory file by file. Your error happens because you try to open a directory as if it is a file instead of entering the dir and opening the files inside

for subdir, dirs, files in os.walk(path):   # walks through whole directory
    for file in files:
        filepath = os.path.join(subdir, file)  # path to the file
        #your code here

edited May 30 '17 at 07:19

Ishan Bhatt

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answered May 30 '17 at 06:49

Isdj

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Opening all files in a directory - Python

2 Answers2