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Groups > comp.lang.python > #37234
| From | Terry Reedy <tjreedy@udel.edu> |
|---|---|
| Subject | Re: Windows subprocess.call problem |
| Date | 2013-01-21 18:38 -0500 |
| References | <CAAPnF_XoeAQLqOXSWVF05sazUy6jtqSCSv18r7SL+18AqqbwmQ@mail.gmail.com> <50FD3CCD.4060707@davea.name> <CAAPnF_XpdmDB=kfHg8MS=YWybgCf-ng_NpMhjaTgxnvbxRJC6A@mail.gmail.com> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.769.1358811545.2939.python-list@python.org> (permalink) |
On 1/21/2013 6:22 PM, Tom Borkin wrote: > nobody@nowhere.com <mailto:nobody@nowhere.com> had an excellent > suggestion that worked right off the bat and achieved exactly what I was > after. Thanks all! And what was it? > > On Mon, Jan 21, 2013 at 9:04 AM, Dave Angel <d@davea.name > <mailto:d@davea.name>> wrote: > > On 01/21/2013 06:25 AM, Tom Borkin wrote: > > Hi; > I have this code: > <snip> > > for song in my_songs: > subprocess.call(['notepad.exe'__, '%s.txt' % song]) > print song > > It opens the first song and hangs on subsequent songs. It > doesn't open the > next song or execute the print until I have closed the first > one. I want it > to open all in the list, one after another, so I have all those > songs > available. Please advise. > > > Why not just pass all the filenames as parameters in one invocation > of notepad? Assuming Notepad is written reasonably, that'll give it > all to you in one window, instead of opening many separate ones. -- Terry Jan Reedy
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Re: Windows subprocess.call problem Terry Reedy <tjreedy@udel.edu> - 2013-01-21 18:38 -0500
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