CODEWITHSUNDEEP
python
mzn.rft
mzn.rft

Reputation: 839

How to read a file in other directory in python

I have a file named 5_1.txt in a directory named direct, how can I read that file using read?

I verified the path using :

import os
os.getcwd()
os.path.exists(direct)

the result was
True 

x_file=open(direct,'r')

and I got this error : 

Traceback (most recent call last):
File "<pyshell#17>", line 1, in <module>
x_file=open(direct,'r')
IOError: [Errno 13] Permission denied

I don't know why I can't read the file ? Any suggestions?

thanks .

Upvotes: 33

Views: 212646

Answers (7)

Deepak Yadav
Deepak Yadav

Reputation: 640

For windows you can either use the full path with '\\' ('/' for Linux and Mac) as separator of you can use os.getcwd to get the current working directory and give path in reference to the current working directory

data_dir = os.getcwd()+'\\child_directory'
file = open(data_dir+'\\filename.txt', 'r')

When I tried to give the path of child_diectory entirely it resulted in error. For e.g. in this case:

file = open('child_directory\\filename.txt', 'r')

Resulted in error. But I think it must work or I am doing it somewhat wrong way but it doesn't work for me. The about way always works.

Upvotes: 1

jarvis
jarvis

Reputation: 1183

For folks like me looking at the accepted answer, and not understanding why it's not working, you need to add quotes around your sub directory, in the green checked example,

x_file = open(os.path.join(direct, "5_1.txt"), "r")

should actually be

x_file = open(os.path.join('direct', "5_1.txt"), "r")

Upvotes: 1

mzn.rft
mzn.rft

Reputation: 839

i found this way useful also.

import tkinter.filedialog
from_filename = tkinter.filedialog.askopenfilename()

here a window will appear so you can browse till you find the file , you click on it then you can continue using open , and read . 

from_file = open(from_filename, 'r')
contents = from_file.read()
contents

Upvotes: 2

alexis
alexis

Reputation: 50190

Looks like you are trying to open a directory for reading as if it's a regular file. Many OSs won't let you do that. You don't need to anyway, because what you want (judging from your description) is 

x_file = open(os.path.join(direct, "5_1.txt"), "r")

or simply

x_file = open(direct+"/5_1.txt", "r")

Upvotes: 31

Thomas Orozco
Thomas Orozco

Reputation: 55197

You can't "open" a directory using the open function. This function is meant to be used to open files.

Here, what you want to do is open the file that's in the directory. The first thing you must do is compute this file's path. The os.path.join function will let you do that by joining parts of the path (the directory and the file name):

fpath = os.path.join(direct, "5_1.txt")

You can then open the file:

f = open(fpath)

And read its content:

content = f.read()

Additionally, I believe that on Windows, using open on a directory does return a PermissionDenied exception, although that's not really the case.

Upvotes: 5

braunmagrin
braunmagrin

Reputation: 886

In case you're not in the specified directory (i.e. direct), you should use (in linux):

x_file = open('path/to/direct/filename.txt')

Note the quotes and the relative path to the directory.

This may be your problem, but you also don't have permission to access that file. Maybe you're trying to open it as another user.

Upvotes: 11

shalakhin
shalakhin

Reputation: 4866

As error message said your application has no permissions to read from the directory. It can be the case when you created the directory as one user and run script as another user.

Upvotes: 0

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