Simple way to get the full path of a running script?

[Benjamin Han]

I know I can do this by get sys.argv[0], tell if it's a full path, and if not,
somehow join the relative path with getcwd(). Just wondering if there's a
simpler way to do this. Thanks!

 

[Benjamin Han]

Duh - the way I described seems to be simple enough:

pathToScript=os.join(os.getcwd(),os.path.split(sys.argv[0])[0])

 

[Benjamin Han]

Should be:

pathToScript=\
os.path.normpath(os.path.join(os.getcwd(),os.path.split(sys.argv[0])[0]))

Duh - the way I described seems to be simple enough:

pathToScript=os.join(os.getcwd(),os.path.split(sys.argv[0])[0])

I know I can do this by get sys.argv[0], tell if it's a full path, and if not,
somehow join the relative path with getcwd(). Just wondering if there's a
simpler way to do this. Thanks!

 

[Mark McEahern]

Duh - the way I described seems to be simple enough:

pathToScript=os.join(os.getcwd(),os.path.split(sys.argv[0])[0])

Or:

#!/usr/bin/env python

import sys
import os

me = sys.argv[0]
print "This is how you invoked me: %s" % (me,)
print "This is the absolute path: %s" % (os.path.abspath(me),)

Usage:

mark@dev /var/tmp/buffer
$ python ../junk.py
This is how you invoked me: ../junk.py
This is the absolute path: /var/tmp/junk.py

Cheers,

// m

 

[Benjamin Han]

Duh - the way I described seems to be simple enough:

pathToScript=os.join(os.getcwd(),os.path.split(sys.argv[0])[0])

me = sys.argv[0]
print "This is the absolute path: %s" % (os.path.abspath(me),)

Thank you - this is even better!

 

[Lee Harr]

#!/usr/bin/env python

import sys
import os

me = sys.argv[0]
print "This is how you invoked me: %s" % (me,)
print "This is the absolute path: %s" % (os.path.abspath(me),)

Usage:

mark@dev /var/tmp/buffer
$ python ../junk.py
This is how you invoked me: ../junk.py
This is the absolute path: /var/tmp/junk.py

Is this going to work well cross platform? How about from IDLE?

I was using sys.path[0] which I thought was working well, but
apparently fails when the script is run from IDLE on windows.

 

[Wolfgang Lipp]

i normally use the following:

import inspect
print inspect.getsourcefile( lambda:None )

the lambda function could be any object; the getsourcefile returns the
path to the file where that object was defined.

_wolf

 

[Hartmut Goebel]

Hi,

import inspect
print inspect.getsourcefile( lambda:None )

Nice idea!

Enhancement: when using inspect.getfile() the snippet will work even for
module without sourcefile, too.

--
Regards
Hartmut Goebel

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