Re: question about running a system and a perl function background

[hsinclai]

If you have to "go to a system call", you could do a simple way like below for each (by the way there's no error checking in there at all) - but it's highly dependent I would say on what process you'll be running out there and on its particular ways of possibly failing, that will more determine what you'd write.

use strict;

my $joboutputfile = "/tmp/joboutputfile.out";

my $stuff;
#run job
qx!ls -al /var > $joboutputfile!;
( -e $joboutputfile ) && &checkerror;
print $stuff;


sub checkerror
{
  return $stuff = qx!grep "spool" $joboutputfile!;
}

[Anonymous Monk]

In the above code, if the output file exists, then the checkerror function is called. But, is there a way I know when the process had been completed?

This is because I need to wait for all four processes to finish before starting another 4 more simulataneously. And everytime I run 4 processes, they dump their output to the same files everytime.

So,

system 1, ("some process 1 > output1.txt");
system 1, ("some process 2 > output2.txt");
system 1, ("some process 3 > output3.txt");
system 1, ("some process 4 > output4.txt");

Then when I know each is completed, I check for error in those files: 
+If process 1 is completed, I'll check for error in output1.txt etc. W
+hen all are completed, I call next set of 4 processes

system 1, ("some process 5 > output1.txt");
system 1, ("some process 6 > output2.txt");
system 1, ("some process 7 > output3.txt");
system 1, ("some process 8 > output4.txt");

As seen above, the output files remain the same

Thanks.