Os.Path.Dirname(_File_) Returns Empty

os.path.dirname(__file__) returns empty

Because os.path.abspath = os.path.dirname + os.path.basename does not hold. we rather have

os.path.dirname(filename) + os.path.basename(filename) == filename

Both dirname() and basename() only split the passed filename into components without taking into account the current directory. If you want to also consider the current directory, you have to do so explicitly.

To get the dirname of the absolute path, use

os.path.dirname(os.path.abspath(__file__))

os.path.dirname returning empty string in django

The os.path.dirname(os.path.dirname(__file__)) does not make any sense. You need to call os.path.dirname(os.path.abspath(__file__)) as suggested in the duplicate question.

The result is empty because __file__ contains only the file name of the Python file.

os.path.dirname does not resolve the location of the given file but simply strips away the file name from the given string.

os.path.dirname('../foo/bar/baz.txt')
Out[4]: '../foo/bar'

As you can see the path has not been resolved, the bar.txt has simply been removed from the string.

What you're doing is the equivalent of:

__file__ = 'baz.txt'

os.path.dirname(__file__)
Out[6]: ''

Instead you should do:

os.path.dirname(os.path.abspath(__file__))
Out[10]: '/home/noxdafox/foo/bar'

os.path.join(os.path.dirname(__file__)) returns nothing

I guess with nothing you mean an empty string? This could only be the case, if __file__ was an empty string in the first place. Did you accidentally overwrite __file__?

Difference between os.path.dirname(os.path.abspath(__file__)) and os.path.dirname(__file__)

BASE_DIR is pointing to the parent directory of PROJECT_ROOT. You can re-write the two definitions as:

PROJECT_ROOT = os.path.dirname(os.path.abspath(__file__))
BASE_DIR = os.path.dirname(PROJECT_ROOT)

because the os.path.dirname() function simply removes the last segment of a path.

In the above, the __file__ name points to the filename of the current module, see the Python datamodel:

__file__ is the pathname of the file from which the module was loaded, if it was loaded from a file.

However, it can be a relative path, so the os.path.abspath() function is used to turn that into an absolute path before removing just the filename and storing the full path to the directory the module lives in in PROJECT_ROOT.

What does os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir)) mean? python

That is a clever way to refer to paths regardless of the script location. The cryptic line you're referring is:

os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))

There are 3 methods and a 2 constants present:

1. abspath returns absolute path of a path
2. join join to path strings
3. dirname returns the directory of a file
4. __file__ refers to the script's file name
5. pardir returns the representation of a parent directory in the OS (usually ..)

Thus, the expression returns the full path name of the executing script in a multiplatform-safe way. No need to hardwire any directions, that's why it is so useful.

There might be other approaches to get a parent directory of where a file is located, for example, programs have the concept of current working directory, os.getcwd(). So doing os.getcwd()+'/..' might work. But this is very dangerous, because working directories can be changed.

Also, if the file is intended to be imported, the working directory will point to the importing file, not the importee, but __file__ always points to the actual module's file so it is safer.

Hope this helps!

Edit: P.S. - Python 3 greatly simplifies this situation by letting us treat paths in an object-oriented manner, so the above line becomes:

from pathlib import Path
Path(__file__).resolve().parent.parent

How to get the parent dir location

You can apply dirname repeatedly to climb higher: dirname(dirname(file)). This can only go as far as the root package, however. If this is a problem, use os.path.abspath: dirname(dirname(abspath(file))).

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