Open file in Django app

Question

I want to open a file from a Django app using open(). The problem is that open() seems to use whatever directory from which I run the runserver command as the root.

E.g. if I run the server from a directory called foo like this

$pwd /Users/foo $python myapp/manage.py runserver 

open() uses foo as the root directory.

If I do this instead

$cd myapp $pwd /Users/foo/myapp $python manage.py runserver 

myapp will be the root.

Let's say my folder structure looks like this

foo/myapp/anotherapp 

I would like to be able to open a file located at foo/myapp/anotherapp from a script also located at foo/myapp/anotherapp simply by saying

file = open('./baz.txt') 

Now, depending on where I run the server from, I have to say either

file = open('./myapp/anotherapp/baz.txt') 

or

file = open('./anotherapp/baz.txt') 

Answer 1

The solution has been described in the Favorite Django Tips&Tricks question. The solution is as follows:

import os module_dir = os.path.dirname(__file__)  # get current directory file_path = os.path.join(module_dir, 'baz.txt') 

Which does exactly what you mentioned.

Ps. Please do not overwrite file variable, it is one of the builtins.

Answer 2

I think I found the answer through another stack overflow question (yes, I did search before asking...)

I now do this

pwd = os.path.dirname(__file__) file = open(pwd + '/baz.txt')