R
Russell E. Owen
I'm trying to launch an application from Python 2.3 on Windows. The
application is "ds9" (an image viewer), and is installed in C:\Program
Files\ds9\ds9
On unix I just do:
os.popen3("ds9")
and close the returned files and all is good. (I'm not trying to
communicate with the program via popen3 and so had been using
os.spawnlp, but that doesn't exist on Windows.)
On Windows os.popen3("ds9") does nothing, and the stderr returned from
os.popen3 has a message saying the program is unknown.
So I tried being specific:
os.popen3("C:\\Program Files\\ds9\\ds9")
This also fails and the program that is not found is "C:\Program",
suggesting that the space in "Program Files" is causing the problem. The
following failed in exactly the same way:
os.popen3("C:\\Program\ Files\\ds9\\ds9")
Any suggestions?
-- Russell
P.S. I installed the app using its Windows binary installer, which
unpacks a few files in C:\\Program files\. Windows doesn't seem to know
the program exists; it's not in the task bar, for example. But one can
double-click it to run it.
application is "ds9" (an image viewer), and is installed in C:\Program
Files\ds9\ds9
On unix I just do:
os.popen3("ds9")
and close the returned files and all is good. (I'm not trying to
communicate with the program via popen3 and so had been using
os.spawnlp, but that doesn't exist on Windows.)
On Windows os.popen3("ds9") does nothing, and the stderr returned from
os.popen3 has a message saying the program is unknown.
So I tried being specific:
os.popen3("C:\\Program Files\\ds9\\ds9")
This also fails and the program that is not found is "C:\Program",
suggesting that the space in "Program Files" is causing the problem. The
following failed in exactly the same way:
os.popen3("C:\\Program\ Files\\ds9\\ds9")
Any suggestions?
-- Russell
P.S. I installed the app using its Windows binary installer, which
unpacks a few files in C:\\Program files\. Windows doesn't seem to know
the program exists; it's not in the task bar, for example. But one can
double-click it to run it.