finding the file of a module from inside a class

[Emile van Sebille]

IIRC, the __file__ attribute is what you want, but is not set when the
module is __main__.  One thing you could try is:

#test.py
try:
 print __file__
except:
 print '__name__ is %s ' %  __name__

import test
print test.__file__

HTH,

--

Emile van Sebille
emile at fenx.com

---------
"Thomas Weholt" <thomas at gatsoft.no> wrote in message
news:E_Yr7.373$n5b.170821632 at news.telia.no...
>
> "Duncan Booth" <duncan at NOSPAMrcp.co.uk> wrote in message
> news:Xns91276D78CAF2Dduncanrcpcouk at 127.0.0.1...
> > "Thomas Weholt" <thomas at gatsoft.no> wrote in
> > news:YGXr7.371$n5b.170335232 at news.telia.no:
> >
> > > say I got a module test.py with this content:
> > >
> > > class MyClass:
> > >     def __init__(self):
> > >         pass
> > >     def myfile(self):
> > >         return '' # ????
> >     return os.path.abspath(__file__)
> > >
> > > if I put this into a folder, ex. /home/thomas/dev/test/, how can I get
> > > information about what file the code instance actually is stored in,
> > > from inside my class? I want the myfile-method to return
> > > /home/thomas/dev/test/test.py ( if the class is stored in a module
> > > called test.py in a folder /home/thomas/dev/test/ of course ).
> >
> > __file__ should give you what you need. It may give you a relative path,
> > but you can use os.path.abspath() to make it absolute.
> >
> > --
> > Duncan Booth
duncan at rcp.co.uk
> > int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
> > "\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?
>
> Tried this code :
>
> import os
>
> class MyClass:
>     def __init__(self): pass
>     def myfile(self): return os.path.abspath(__file__)
>
> if __name__ == '__main__':
>     x = MyClass()
>     print x.myfile()
>
> With this result :
>
> Traceback (most recent call last):
>   File "c:\python21\pythonwin\pywin\framework\scriptutils.py", line 298,
in
> RunScript
>     debugger.run(codeObject, __main__.__dict__, start_stepping=0)
>   File "c:\python21\pythonwin\pywin\debugger\__init__.py", line 60, in run
>     _GetCurrentDebugger().run(cmd, globals,locals, start_stepping)
>   File "c:\python21\pythonwin\pywin\debugger\debugger.py", line 582, in
run
>     _doexec(cmd, globals, locals)
>   File "c:\python21\pythonwin\pywin\debugger\debugger.py", line 924, in
> _doexec
>     exec cmd in globals, locals
>   File "C:\Temp\Script10.py", line 9, in ?
>     print x.myfile()
>   File "C:\Temp\Script10.py", line 5, in myfile
>     def myfile(self): return os.path.abspath(__file__)
> NameError: global name '__file__' is not defined
> >>>
>
> Hm ... if it was that simple it would be too good to be true.
>
>
>
>