Find path to currently running file

Question

How can I find the full path to the currently running Python script? That is to say, what do I have to do to achieve this:

$ pwd /tmp $ python baz.py running from /tmp  file is baz.py 

Answer 1

__file__ is NOT what you are looking for. Don't use accidental side-effects

sys.argv[0] is always the path to the script (if in fact a script has been invoked) -- see http://docs.python.org/library/sys.html#sys.argv

__file__ is the path of the currently executing file (script or module). This is accidentally the same as the script if it is accessed from the script! If you want to put useful things like locating resource files relative to the script location into a library, then you must use sys.argv[0].

Example:

C:\junk\so>type \junk\so\scriptpath\script1.py import sys, os print "script: sys.argv[0] is", repr(sys.argv[0]) print "script: __file__ is", repr(__file__) print "script: cwd is", repr(os.getcwd()) import whereutils whereutils.show_where()  C:\junk\so>type \python26\lib\site-packages\whereutils.py import sys, os def show_where():     print "show_where: sys.argv[0] is", repr(sys.argv[0])     print "show_where: __file__ is", repr(__file__)     print "show_where: cwd is", repr(os.getcwd())  C:\junk\so>\python26\python scriptpath\script1.py script: sys.argv[0] is 'scriptpath\\script1.py' script: __file__ is 'scriptpath\\script1.py' script: cwd is 'C:\\junk\\so' show_where: sys.argv[0] is 'scriptpath\\script1.py' show_where: __file__ is 'C:\\python26\\lib\\site-packages\\whereutils.pyc' show_where: cwd is 'C:\\junk\\so'