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"
?
Answers:
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?
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.
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)
You can get it like this also.
filepath = os.path.abspath(f.name)
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"
?
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?
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.
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)
You can get it like this also.
filepath = os.path.abspath(f.name)