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Q: The getcwd() doesn't return the directory that the source file is located.

Alex Martelli aleaxit at yahoo.com
Sun Oct 29 06:42:04 EST 2000 
"Dale Strickland-Clark" <dale at out-think.NOSPAMco.uk> wrote in message
news:2hvnvsgs713rj9assjrpefo1ifi7u6gv4q at 4ax.com...
    [snip]
> >When I try the getcwd(), it returns the python2.0 root directory.
> >How to get the current directory that the source file is located ?
    [snip]
> Have you looked in sys.argv?
>
> import sys
> print sys.argv[0]
>
> This prints the path to the source for me.

It could print that path in an implied way, though, depending
on exactly how it was specified on the invocation line.


A more complete way to "get the directory in which the source
file is located" (no "current" about it) could be:

import sys, os.path as p
print p.dirname(p.abspath(sys.argv[0]))


However, this has to do with _the Python script that was
originally loaded_, NOT the one from which instructions
are currently being executed (which could well have been
brought in with an import, quite possibly from a different
directory).  For any module that has been imported, the
__file__ variable of that module should take the place of
sys.argv[0] in the above construct.  The variable will be
undefined (a NameError will result, and can be caught,
just as any other exception) in the main module (the one
named '__main__', and named in sys.argv[0]).

One still needs os.path.abspath to ensure an absolute
path, and os.path.dirname to extract the directory name
from it, of course.


Alex
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