python NameError: global name '__file__' is not defined

When I run this code in python 2.7, I get this error:

Traceback (most recent call last):
File "C:\Python26\Lib\site-packages\pyutilib.subprocess-3.5.4\setup.py", line 30, in <module>
    long_description = read('README.txt'),
  File "C:\Python26\Lib\site-packages\pyutilib.subprocess-3.5.4\setup.py", line 19, in read
    return open(os.path.join(os.path.dirname(__file__), *rnames)).read()
NameError: global name '__file__' is not defined

code is:

import os
from setuptools import setup


def read(*rnames):
    return open(os.path.join(os.path.dirname(__file__), *rnames)).read()


setup(name="pyutilib.subprocess",
    version='3.5.4',
    maintainer='William E. Hart',
    maintainer_email='[email protected]',
    url = 'https://software.sandia.gov/svn/public/pyutilib/pyutilib.subprocess',
    license = 'BSD',
    platforms = ["any"],
    description = 'PyUtilib utilites for managing subprocesses.',
    long_description = read('README.txt'),
    classifiers = [
        'Development Status :: 4 - Beta',
        'Intended Audience :: End Users/Desktop',
        'License :: OSI Approved :: BSD License',
        'Natural Language :: English',
        'Operating System :: Microsoft :: Windows',
        'Operating System :: Unix',
        'Programming Language :: Python',
        'Programming Language :: Unix Shell',
        'Topic :: Scientific/Engineering :: Mathematics',
        'Topic :: Software Development :: Libraries :: Python Modules'],
      packages=['pyutilib', 'pyutilib.subprocess', 'pyutilib.subprocess.tests'],
      keywords=['utility'],
      namespace_packages=['pyutilib'],
      install_requires=['pyutilib.common', 'pyutilib.services']
      )

This error comes when you append this line os.path.join(os.path.dirname(__file__)) in python interactive shell.

Python Shell doesn't detect current file path in __file__ and it's related to your filepath in which you added this line

So you should write this line os.path.join(os.path.dirname(__file__)) in file.py. and then run python file.py, It works because it takes your filepath.

I had the same problem with PyInstaller and Py2exe so I came across the resolution on the FAQ from cx-freeze.

When using your script from the console or as an application, the functions hereunder will deliver you the "execution path", not the "actual file path":

print(os.getcwd())
print(sys.argv[0])
print(os.path.dirname(os.path.realpath('__file__')))

Source:
http://cx-freeze.readthedocs.org/en/latest/faq.html

Your old line (initial question):

def read(*rnames):
return open(os.path.join(os.path.dirname(__file__), *rnames)).read()

Substitute your line of code with the following snippet.

def find_data_file(filename):
    if getattr(sys, 'frozen', False):
        # The application is frozen
        datadir = os.path.dirname(sys.executable)
    else:
        # The application is not frozen
        # Change this bit to match where you store your data files:
        datadir = os.path.dirname(__file__)

    return os.path.join(datadir, filename)

With the above code you could add your application to the path of your os, you could execute it anywhere without the problem that your app is unable to find it's data/configuration files.

Tested with python:

• 3.3.4
• 2.7.13

I've run into cases where __file__ doesn't work as expected. But the following hasn't failed me so far:

import inspect
src_file_path = inspect.getfile(lambda: None)

This is the closest thing to a Python analog to C's __FILE__.

The behavior of Python's __file__ is much different than C's __FILE__. The C version will give you the original path of the source file. This is useful in logging errors and knowing which source file has the bug.

Python's __file__ only gives you the name of the currently executing file, which may not be very useful in log output.