python - 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:junkso>type junksoscriptpathscript1.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:junkso>type python26libsite-packageswhereutils.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:junkso>python26python scriptpathscript1.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'