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Re: Windows subprocess.call problem

References <CAAPnF_XoeAQLqOXSWVF05sazUy6jtqSCSv18r7SL+18AqqbwmQ@mail.gmail.com> <50FD3CCD.4060707@davea.name>
Date 2013-01-21 19:22 -0400
Subject Re: Windows subprocess.call problem
From Tom Borkin <borkintom@gmail.com>
Newsgroups comp.lang.python
Message-ID <mailman.768.1358810538.2939.python-list@python.org> (permalink)

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nobody@nowhere.com had an excellent suggestion that worked right off the
bat and achieved exactly what I was after. Thanks all!
Tom


On Mon, Jan 21, 2013 at 9:04 AM, Dave Angel <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.
>
>
> --
> DaveA
> --
> http://mail.python.org/**mailman/listinfo/python-list<http://mail.python.org/mailman/listinfo/python-list>
>

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Re: Windows subprocess.call problem Tom Borkin <borkintom@gmail.com> - 2013-01-21 19:22 -0400

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