• 10
name

A PHP Error was encountered

Severity: Notice

Message: Undefined index: userid

Filename: views/question.php

Line Number: 191

Backtrace:

File: /home/prodcxja/public_html/questions/application/views/question.php
Line: 191
Function: _error_handler

File: /home/prodcxja/public_html/questions/application/controllers/Questions.php
Line: 433
Function: view

File: /home/prodcxja/public_html/questions/index.php
Line: 315
Function: require_once

name Punditsdkoslkdosdkoskdo

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 .

      • 2
    • Perhaps the user you are running Python as does not have permissions to open the file. Run Python as a different user or change the owner/group of the file.
      • 1
    • Could you run ls -l on the file your are trying to access? You will probably see that you do not have read permission on the file (assuming Unix)

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")
  • 28
Reply Report

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.

  • 11
Reply Report
    • i have windows xp , but i don't know what is "another user", and if this is the case , so how can overcome this issue ?
      • 1
    • The file might also already be in use. Some other app might have opened it and not closed it.

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.

  • 4
Reply Report

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
  • 2
Reply Report

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.

  • 1
Reply Report

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.

  • 0
Reply Report

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")   
  • 0
Reply Report

Trending Tags