Is there any way that I can find the path of the main .py file of my
application?
For example, I have an application with some resources which are in a
subdirectory:
myPythonApp.py
/resources
image1
image2
etc.
If i just do a call to os.getcwd() I get back the directory I was in when I
typed 'python myPythonApp.py' which could be any directory. What I want is
the directory that contains the file myPythonApp.py. Then I can use this
directory to construct the path to the resources directory.
(Actually, the app I'm writing has several subdirectories with stuff that it
needs such as plugins, configuration files, workspaces, etc.
Thanks for the help.
Ron
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On Dec 11, 10:08 am, "ron.longo" <long...@npt.nu wc.navy.milwrot e:
Is there any way that I can find the path of the main .py file of my
application?
For example, I have an application with some resources which are in a
subdirectory:
myPythonApp.py
/resources
image1
image2
etc.
I just put the reference in my module. Don't hard code an absolute
path, use the environment tools.
app_path = os.getenv('HOME ') + "/your_sub_dir"
resources_path = os.getenv('HOME ') + "/your_sub_dir/resources"
If there's another way, someone else will jump in.
rd
Some usage of __file__ will always get what you want in various situations:
print __file__
print modulename.__fi le__
print os.getcwd() + "/" + __file__
Rick Dooling wrote:
On Dec 11, 10:08 am, "ron.longo" <long...@npt.nu wc.navy.milwrot e:
>Is there any way that I can find the path of the main .py file of my
application?
For example, I have an application with some resources which are in a
subdirectory :
myPythonApp.py
/resources
image1
image2
etc.
I just put the reference in my module. Don't hard code an absolute
path, use the environment tools.
app_path = os.getenv('HOME ') + "/your_sub_dir"
resources_path = os.getenv('HOME ') + "/your_sub_dir/resources"
If there's another way, someone else will jump in.
rd
--
Shane Geiger
IT Director
National Council on Economic Education sg*****@ncee.ne t | 402-438-8958 | http://www.ncee.net
Leading the Campaign for Economic and Financial Literacy
Nope, maybe I'm not explaining myself well.
When I do os.getenv('HOME ') I get back None.
According to the docs, 'HOME' is the user's home directory on some
platforms. Which is not what I want.
What I want is the directory in which an application's main .py file
resides. That is, when I type: python MyApp.py, I want to know in which
directory does MyApp.py reside?
Thanks,
Ron
Rick Dooling-2 wrote:
>
On Dec 11, 10:08 am, "ron.longo" <long...@npt.nu wc.navy.milwrot e:
>Is there any way that I can find the path of the main .py file of my
application?
For example, I have an application with some resources which are in a
subdirectory :
myPythonApp.py
/resources
image1
image2
etc.
I just put the reference in my module. Don't hard code an absolute
path, use the environment tools.
app_path = os.getenv('HOME ') + "/your_sub_dir"
resources_path = os.getenv('HOME ') + "/your_sub_dir/resources"
If there's another way, someone else will jump in.
rd
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Sent from the Python - python-list mailing list archive at Nabble.com.
ron.longo wrote:
Nope, maybe I'm not explaining myself well.
When I do os.getenv('HOME ') I get back None.
According to the docs, 'HOME' is the user's home directory on some
platforms. Which is not what I want.
What I want is the directory in which an application's main .py file
resides. That is, when I type: python MyApp.py, I want to know in which
directory does MyApp.py reside?
Shane is right.
>>print __file__
>>print modulename.__fi le__
Just call os.path.dirname () on __file__ to get the directory.
--
Shane Geiger <sg*****@ncee.n etwrites:
Some usage of __file__ will always get what you want in various situations:
print __file__
print modulename.__fi le__
print os.getcwd() + "/" + __file__
Rick Dooling wrote:
>On Dec 11, 10:08 am, "ron.longo" <long...@npt.nu wc.navy.milwrot e:
>>Is there any way that I can find the path of the main .py file of my
application ?
For example, I have an application with some resources which are in a
subdirector y:
myPythonApp.py
/resources
image1
image2
etc.
The following two versions are working for me:
import os, sys
print os.path.abspath (os.path.dirnam e(sys.argv[0]))
print os.path.abspath (os.path.dirnam e(__file__))
Stefan.
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