Get path from open file in Python

Question

If I have an opened file, is there an os call to get the complete path as a string?

f = open('/Users/Desktop/febROSTER2012.xls')

From f, how would I get "/Users/Desktop/febROSTER2012.xls" ?

Answer 1

The key here is the name attribute of the f object representing the opened file. You get it like that:

>>> f = open('/Users/Desktop/febROSTER2012.xls')
>>> f.name
'/Users/Desktop/febROSTER2012.xls'

Does it help?

Answer 2

I had the exact same issue. If you are using a relative path os.path.dirname(path) will only return the relative path. os.path.realpath does the trick:

>>> import os
>>> f = open('file.txt')
>>> os.path.realpath(f.name)

Answer 3

And if you just want to get the directory name and no need for the filename coming with it, then you can do that in the following conventional way using os Python module.

>>> import os
>>> f = open('/Users/Desktop/febROSTER2012.xls')
>>> os.path.dirname(f.name)
>>> '/Users/Desktop/'

This way you can get hold of the directory structure.

Answer 4

You can get it like this also.

filepath = os.path.abspath(f.name)