[Tutor] starting a program from python

johnsonv3 Sat, 22 Jul 2006 09:25:43 -0700

Hi,

Using Windows XP Home & python 234

I am trying to start python programs that run Panda 3D without having to open command prompt and type a path every time I run a program.

This pandastart.bat file works...

cd C:\Panda3D-1.2.3\mystuff

ppython bvd.py

and from python shell this works....

>>> subprocess.Popen("pandastart.bat")
<subprocess.Popen object at 0x00E07DD0>

But it is not flexible to start other file names..

So I would like to write a python script similar to this....

import subprocess
panda = raw_input("Enter name of Panda python program to start (fileName.py)")
subprocess.Popen("C:\panda3D-1.2.3\mystuff\panda")

But this is the result...

>>> subprocess.Popen("C:\panda3D-1.2.3\mystuff\bvd.py")
Traceback (most recent call last):
File "<pyshell#19>", line 1, in ?
    subprocess.Popen("C:\panda3D-1.2.3\mystuff\bvd.py")
File "C:\Python24\lib\subprocess.py", line 542, in __init__
    errread, errwrite)
File "C:\Python24\lib\subprocess.py", line 706, in _execute_child
    startupinfo)
WindowsError: [Errno 123] The filename, directory name, or volume label syntax is incorrect

Also tried the below (change to the path) with result as indicate below....

>>> subprocess.Popen("C:\panda3D-1.2.3\mystuff bvd.py")
Traceback (most recent call last):
File "<pyshell#20>", line 1, in ?
    subprocess.Popen("C:\panda3D-1.2.3\mystuff bvd.py")
File "C:\Python24\lib\subprocess.py", line 542, in __init__
    errread, errwrite)
File "C:\Python24\lib\subprocess.py", line 706, in _execute_child
    startupinfo)
WindowsError: [Errno 5] Access is denied

Have tried using os.system and os.startfile with similar results

[Tutor] starting a program from python

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