Python - Get the real path (location) of the source file where I'm coding

Question

From here: Find current directory and file's directory

I tried in spyder (Python IDE):

dir_path = os.path.dirname(os.path.realpath('__file__'))

and:

cwd = os.getcwd()

I only get like just the user directory and not the location of my source file where I'm coding

My source code is here:

C:\Users\username\test\testerrr\test.py

I only get:

C:\Users\username

I use:

os.path.dirname(os.path.realpath('__file__'))

and not:

os.path.dirname(os.path.realpath(__file__))

else I get this:

os.path.dirname(os.path.realpath(__file__))
Traceback (most recent call last):

  File "<ipython-input-18-98004a18344a>", line 1, in <module>
    os.path.dirname(os.path.realpath(__file__))

NameError: name '__file__' is not defined

Well, it seems that's works only when we run its through shell.

So is the output you showed the value of dir_path or cwd? Why do you even retrieve the cwd? — Aran-Fey
– Aran-Fey, Commented Apr 22, 2018 at 12:43

Ishara Dayarathna · Accepted Answer · 2018-04-22 12:47:56Z

1

Try this:

dir_path = os.path.dirname(os.path.realpath(__file__))

not '__file__'. It should be __file__

answered Apr 22, 2018 at 12:47

Ishara Dayarathna

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Dan D. · Accepted Answer · 2018-04-22 12:48:15Z

1

Remove the quotes '__file__' should be __file__.

What you are getting from os.path.realpath('__file__') is $CWD/__file__ which is not what you want. That is why you get $CWD when you call os.path.dirname on that result.

answered Apr 22, 2018 at 12:48

Dan D.

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2 Comments

Dan D. Over a year ago

Where are you running the code? In the REPL? It only works in a module as that is where __file__ is defined by the import process.

pr_charlatan Over a year ago

spyder (Python IDE)

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