Question

I am having issues with the open() function in Python 3.2.3. The following code works well using 2.7.3, but not with Python 3:

    file = open("text.txt", 'r')

In Python3, it gives me a standard IOError:

    IOError: [Errno 2] No such file or directory: 'text.txt'

Note that the text.txt file that is referenced is in the same directory as the python file.

Any ideas?

Answer 1

The file name is not relative to the directory of the file, but your current working directory (which you can find out with os.getcwd()).

If you want to open a file whose name is relative to your Python file, you can use the magic variable __file__, like this:

import os.path
fn = os.path.join(os.path.dirname(__file__), 'text.txt')
with open(fn, 'r') as file:
   # Do something, like ...
   print(file.read())