Greetings again. This has to be an easy one.
From within my Python-based product's code, how do I get access to the product's directory on the filesystem? os.getcwd() seems to provide the working directory of the shell used to launch Zope, at least when running in debug mode.
I've tried ferreting it out from the source to DTMLFile and etc., but I can't trace it. Thanks, ..Ian -- --- -- --- -- --- -- --- -- --- -- --- -- --- -- --- -- --- -- Dr. Ian Beatty beatty@physics.umass.edu Physics Education Research Group voice: 413.545.9483 Department of Physics fax: 413.545.4884 Univ. of Massachusetts AIM: (available upon request) Amherst, MA 01003-4525 USA http://umperg.physics.umass.edu/ -- --- -- --- -- --- -- --- -- --- -- --- -- --- -- --- -- --- --
On Tue, Mar 09, 2004 at 01:58:17PM -0500, Ian Beatty wrote:
Greetings again.
This has to be an easy one.
From within my Python-based product's code, how do I get access to the product's directory on the filesystem? os.getcwd() seems to provide the working directory of the shell used to launch Zope, at least when running in debug mode. Try os.path.dirname(mymodule.__file__)?
Andreas
On Tuesday 09 March 2004 01:58 pm, Ian Beatty wrote:
This has to be an easy one.
Good, I'll take it. ;-)
From within my Python-based product's code, how do I get access to the product's directory on the filesystem? os.getcwd() seems to provide the working directory of the shell used to launch Zope, at least when running in debug mode.
This isn't actually Zope specific; you can easily get the directory a module lives in using this: import os here = os.path.dirname(os.path.abspath(__file__)) The variable "here" will hold a string with an absolute path to the directory containing the source file for your module (named by __file__). The call to os.path.abspath() isn't strictly necessary, but protects against future calls to os.chdir() (though not past calls!) in case the module was loaded from directory named by a relative directory on sys.path. You can do this just once at module scope; there's no need to recompute this each time you need it. -Fred -- Fred L. Drake, Jr. <fred at zope.com> PythonLabs at Zope Corporation
There is also a convenience function for this in Zope: from Globals import package_home here = package_home(globals()) - C On Tue, 2004-03-09 at 14:16, Fred Drake wrote:
On Tuesday 09 March 2004 01:58 pm, Ian Beatty wrote:
This has to be an easy one.
Good, I'll take it. ;-)
From within my Python-based product's code, how do I get access to the product's directory on the filesystem? os.getcwd() seems to provide the working directory of the shell used to launch Zope, at least when running in debug mode.
This isn't actually Zope specific; you can easily get the directory a module lives in using this:
import os here = os.path.dirname(os.path.abspath(__file__))
The variable "here" will hold a string with an absolute path to the directory containing the source file for your module (named by __file__). The call to os.path.abspath() isn't strictly necessary, but protects against future calls to os.chdir() (though not past calls!) in case the module was loaded from directory named by a relative directory on sys.path.
You can do this just once at module scope; there's no need to recompute this each time you need it.
-Fred
On Tuesday 09 March 2004 02:30 pm, Chris McDonough wrote:
There is also a convenience function for this in Zope:
from Globals import package_home here = package_home(globals())
Maybe I'm just weird, but I generally prefer the general approach when there's not a clear improvement in readability or brevity. Globals.package_home() is definately a more complicated approach to digging out the same information; it might be better for older versions of Python. It's not available in Zope 3. -Fred -- Fred L. Drake, Jr. <fred at zope.com> PythonLabs at Zope Corporation
participants (4)
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Andreas Kostyrka -
Chris McDonough -
Fred Drake -
Ian Beatty