Shell Programming and Scripting

Ok, theres a log file containing the below. Lets call the logfile log_fantastic:

2009/03/16 21:42:45 USER: tonnabo - MAC: 0014BF2D385A - STATUS_ID: 30 - STATE: ERROR
2009/03/16 21:42:45 USER: tonnabo - MAC: 001310AC120D - STATUS_ID: 15 - STATE: OK
2009/03/16 21:42:45 USER: tonnabo - MAC: 001DD9277095 - STATUS_ID: 20 - STATE: ERROR


I want to write a perl script that will read the contents of log_fantastic and output only the lines that does not have a status_id of 10?

the below is what i've done. not sure if this is correct:


use Data:: Dumper;

main();


sub main {
my @results;
my $headers = "user,
mac address,
status_id";

my (@log_info, @dircontents);


open (LOG, "</home/jhenson/log_fantastic");
@log_info = <LOG>;
close (LOG);

print Dumper("testing");
foreach my $content (@log_info) {
chomp($content);
my $line = substr ($content, -2);
print Dumper($line);
if ($line eq "OK") {
push @dircontents, $content;
}
}


#push @alarm_info, uc $headers;

print Dumper(@dircontents);

}

you need a perl script to do this? :

Code:

grep -v 'STATUS_ID: 10 ' ${file:-"/home/jhenson/log_fantastic"}

Or is there more to it...?

Try the perl script : -
==================

#!/usr/bin/perl
open(FH,"log_fantastic") or dir ("Couldnt open file");
while(<FH>)
{
$rec=$_;
chomp ($rec);
@arr=split(/:/,$rec);
print $rec,"\n" if ($arr[5]!~/10/);
}
close(FH);

oh theres more to it. i coulda easily done this through shell programming. but this needs to be done in perl. on top of grepping out the wanted status ids, i goto set it up in fields. thanks a million for your suggestion

this works PERFECTLY. thanks a million.

This should be more efficient and more accurate:

Code:

open(FH,"log_fantastic") or dir ("Couldnt open file: $!");
while(my $line = <FH>) {
   next if (/STATUS_ID: 10 /);
   print;
}
close(FH);