Asychronous execution *with* return codes?

utabintarbo

I hope I have not overlooked a solution already posted, but I seem to
be unable to suss out a way to achieve both multiple console-less
executions of a given (console) application and gathering the return
code from the application.

What I have found:
<code>
import subprocess

# gives back return code, but does not run asynchronously
retcode = subprocess.call([app, lstArgs])
retcode = subprocess.Popen([app] + lstArgs).wait()
# runs the app async, but only returns the pid (no return code)
pid = subprocess.Popen([app] + lstArgs).pid
</code>

Is there some magic elixir which will get me both?

TIA

Oct 5 '06 #1

MonkeeSage

utabintarbo wrote:

pid = subprocess.Popen([app] + lstArgs).pid

Check out the poll() method and the returncode attribute:
http://docs.python.org/lib/node533.html

Regards,
Jordan

Oct 5 '06 #2

utabintarbo

MonkeeSage wrote:

utabintarbo wrote:

pid = subprocess.Popen([app] + lstArgs).pid

Check out the poll() method and the returncode attribute:
http://docs.python.org/lib/node533.html

Thanks for the reply.

If I understand your meaning, I should do something like this (given I
wish to run an app against several arguments [my use case]):

for lstArgs in pileOflstArgs:
uniqueProcessID = subprocess.Popen([app] + lstArgs)
pid = uniqueProcessID.pid
retcode = uniqueProcessID.poll()
# increment uniqueProcessID
....

If so, how do I handle the poll() on long-running processes? Run a
bunch and then start a check loop? Am I asking too many questions?

Oct 5 '06 #3

Justin

If you're on a POSIX system, you could use the usual fork/exec/wait:

import os
for lstArgs in pileOflstArgs:
pid = os.fork()
if not pid:
os.execv( app, lstArgs )

for i in range(len(pileOflstArgs)):
pid, status = os.wait()

Of couse, os.wait() will block until a child exits. Look at the docs
for the status code it returns, though, as it's not just the return
value of the process.

On Oct 5, 7:43 am, "utabintarbo" <utabinta...@gmail.comwrote:

MonkeeSage wrote:

utabintarbo wrote:

pid = subprocess.Popen([app] + lstArgs).pid

Check out the poll() method and the returncode attribute:
http://docs.python.org/lib/node533.htmlThanks for the reply.

If I understand your meaning, I should do something like this (given I
wish to run an app against several arguments [my use case]):

for lstArgs in pileOflstArgs:
uniqueProcessID = subprocess.Popen([app] + lstArgs)
pid = uniqueProcessID.pid
retcode = uniqueProcessID.poll()
# increment uniqueProcessID
....

If so, how do I handle the poll() on long-running processes? Run a
bunch and then start a check loop? Am I asking too many questions?

Oct 5 '06 #4

utabintarbo

Justin wrote:

If you're on a POSIX system, you could use the usual fork/exec/wait:

Sorry. Win32. We are only allowed spoons - no sharp objects. :-P

Oct 5 '06 #5

Gabriel Genellina

At Thursday 5/10/2006 09:33, utabintarbo wrote:

># gives back return code, but does not run asynchronously
retcode = subprocess.call([app, lstArgs])
retcode = subprocess.Popen([app] + lstArgs).wait()
# runs the app async, but only returns the pid (no return code)
pid = subprocess.Popen([app] + lstArgs).pid
</code>

Is there some magic elixir which will get me both?

Use the async way, and then, os.waitpid()

Gabriel Genellina
Softlab SRL

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Oct 5 '06 #6

Lawrence D'Oliveiro

In message <11**********************@e3g2000cwe.googlegroups. com>,
utabintarbo wrote:

Justin wrote:

>If you're on a POSIX system, you could use the usual fork/exec/wait:

Sorry. Win32. We are only allowed spoons - no sharp objects. :-P

How about installing Cygwin, then, and running under that?

Oct 9 '06 #7

Fredrik Lundh

utabintarbo wrote:

If so, how do I handle the poll() on long-running processes? Run a
bunch and then start a check loop?

or use a thread to keep track of each external process.

</F>

Oct 9 '06 #8

utabintarbo

Fredrik Lundh wrote:

utabintarbo wrote:

If so, how do I handle the poll() on long-running processes? Run a
bunch and then start a check loop?

or use a thread to keep track of each external process.

</F>

This sounds most promising. Might you have a code snippet (or link to
same) illustrating this?

Oct 11 '06 #9