Problem with subprocess.call and windows schtasks
ramit.prasad at jpmorgan.com
Fri Nov 23 23:32:48 CET 2012
Dave Angel wrote:
> On 11/20/2012 06:41 PM, Tom Borkin wrote:
> (Please don't top-post. Now we lose all the context)
> > Using shlex, I now have this:
> > #!\Python27\python
> > import os, subprocess
> > path = os.path.join("C:\\", "Program Files", "Apache Group", "Apache2",
> > "htdocs", "ccc", "run_alert.py")
> > #subprocess.call(['SchTasks', '/Create', '/SC', 'ONCE', '/TN', '"test"',
> > '/TR', path, '/ST', '23:50'])
> > subprocess.call(['SchTasks', '/Create', '/SC', 'ONCE', '/TN', '"test"',
> > '/TR', 'run_alert.py', '/ST', '23:50'])
> > Both of the above commands throw the same error:
> > ERROR: The filename, directory name or volume label syntax is incorrect.
> I don't use Windows, but doesn't a Windows program usually have an .exe
> extension? So why would you expect it to find SchTasks ? Adding
> extensions is a shell feature, and you're not using the shell.
> Also, you should take a look at the value "path". On Linux, it shows up as:
> C:\\/Program Files/Apache Group/Apache2/htdocs/ccc/run_alert.py
> It'll be different under Windows, but probably still wrong.
Windows 7 + Python 2.6
>>> os.path.join("C:\\", "Program Files", "Apache Group", "Apache2",
... "htdocs", "ccc", "run_alert.py")
'C:\\Program Files\\Apache Group\\Apache2\\htdocs\\ccc\\run_alert.py'
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