CODEWITHSUNDEEP
python
user3579482
user3579482

Reputation: 7

Python absolute path with __file__ for a module called different ways

In my app I have a setup python script (/root/ha/setup.py) which looks through a directory called modules and runs setup.py in each of the subdirectories.

The relevant code is this:

""" Exec the module's own setup file """
if os.path.isfile(os.path.join(root, module, "setup.py")):
    execfile(os.path.join(root, module, "setup.py"))

The thing is I want /root/ha/modules/modulename/setup.py to work no matter where it's called from.

If I am in modules/modulename and run python setup.py it's fine but if I run it from the directory above modules/ i get this error

idFile = open(os.path.dirname(os.path.abspath(__file__)) + "/id.txt", "r").read()
IOError: [Errno 2] No such file or directory: '/root/ha/id.txt'

as you can see it is getting the path of the script that is calling it instead of the script that is running. It should be trying to read /root/ha/modules/modulename/id.txt

I've tried using different methods to get the path but all end up with this error...

Upvotes: 0

Views: 930

Answers (2)

Jan Vlcinsky
Jan Vlcinsky

Reputation: 44112

If what you need is to access a file from some of your packages, then consider using pkg_resources as documented here: http://pythonhosted.org/setuptools/pkg_resources.html#basic-resource-access

Example of getting content of a file stored as part of package named package is in this SO answer

Upvotes: 0

Antoine
Antoine

Reputation: 4029

execfile does not modify the globals (as __file__) so the exectued script will indeed take an incorrect path.

You can pass global variables to execfile so you can modify its __file__ variable:

script = os.path.join(root, module, "setup.py")
if os.path.isfile(script):
    g = globals().copy()
    g['__file__'] = script
    execfile(script, g)

Upvotes: 1

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