On Thu, Apr 16, 2009 at 8:32 AM, Michael Alipio <daem0n...@yahoo.com> wrote:
>
> Hi,
>
> I have this code:
>
> die "Could not fork command1!" unless defined (my $command1_pid = fork);
> if ($command1_pid == 0){
> open EXTERNAL_PROG1, "external_prog1 |" or die "Can't run external_prog1";
>
> while (<EXTERNAL_PROG1>){
> if (/pattern/){
> die "Could not fork command2" unless defined (my $command2_pid = fork);
> die "Could not fork command3" unless defined (my $command3_pid = fork);
>
> if ($command2_pid == 0){
> `external_prog2`
> }
> }
> waitpid($command2_pid,0);
>
> if ($command3_pid == 0){
> `external_prog3`;
> }
> waitpid($command3_pid, 0);
>
> }
> }
>
>
>
> That is, run external_prog1 and watch the output.. as soon as it sees a
> particular pattern, run two other external commands at the same time.
> However, with the above code, I keep getting duplicated external_prog2
> process everytime i run the script. I need to track both PIDs of
> external_prog2 and 3 as I need to kill them later in the code. However, i'm
> having trouble running them at the same time.
> if I will put a "kill 15, $command3_pid" right after the first waitpid, i
> will surely kill one of those two command 2 process. I tried several places
> to put "die" lines and "waitpids" but I couldn't get it right. It's either
> they don't run at the same time or they will run but command 2 has a twin
> brother.
>
> Any idea how to solve this chicken and egg problem?
>
You have problems with scope.
At the top of your program, add
use warnings;
use strict;
The messages that are printed should solve your problem.
HTH,
-- j
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