Regular Expression Question

[Anonymous Monk]

I have a file that looks like this

box1
Filesystem           1k-blocks      Used Available Use% Mounted on
/dev/hda1               124427     54401     63602  46% /
/dev/hda6              5289348   3682364   1338300  73% /usr

box2
Filesystem           1k-blocks      Used Available Use% Mounted on
/dev/hda1               124427     54401     63602  46% /
/dev/hda6              5289348   3682364   1338300  73% /usr

box3
Filesystem           1k-blocks      Used Available Use% Mounted on
/dev/hda1               124427     54401     63602  46% /
/dev/hda6              5289348   3682364   1338300  73% /usr

This is the output from df -l. The info I need is every box that is over 88% full.Thank You in advance

[mirod]

The answer is that none of them is over 88% full.

You are welcome.

[boo_radley]

While I am annoyed at the presentation of your question, it gave me a chance to abuse $/ in an entertaining fashion. I think I'm either thinking more perly, or going insane. :)

use strict;
open LOG, "C:/fs.log";
$/="\n\n";
while(<LOG>){
    m |^(\w+)\n|; print "Machine $1\n";
    while (m|\n(/.*?\s).*?(\d+)%|gi) {
        print "\t",$2 > 88? "WARNING --> ":"","$1 at $2 %\n"
    }
}

does the undef need to be there?
Update No, it can be replaced by the "" you see in the ?: .

[Masem]

OOTOMH...

while (<FILE>) {
    my $boxname;
    if ( ! ( /^Filesystem/ ) ) {
        $boxname = $_;
    } else {
        while (<FILE>) {
            my ( $title, $blocks, $used, $avail, $percent, $mounted ) 
+= split /\s+/;
            last if ( !defined( $title ) );
            my $number = ( $percent =~ /(\d*)\%/ );
            if ( $number > 88 ) {
                print "$boxname's $title is at $percent!\n";
            }
        }
    }
}

Update Should be split \/s+\ rather than \/s*\

---

Dr. Michael K. Neylon - [e-mail] || "You've left the lens cap of your mind on again, Pinky" - The Brain

[MrNobo1024]

Did you mean split(/\s+/)? split(/\s*/) will split it into individual characters because /\s*/ can match a 0 character string.

[japhy]

Do you mean box, or filesystem? I'll do both. (I'm good like that.)

$/ = "";
while (<>) {
  my ($box) = /(.*)/;
  my ($t_used, $t_avail);

  for (split /\n/) {
    my ($fd,$used,$avail,$per) =
      /^(\S+)\s+(\d+)\s+(\d+)\s+(\d+)\%/ or next;
    $t_used += $used;
    $t_avail += $avail;
    warn "$box:$fd at $per%\n" if $per > 88;
  }

  my $t_per = int(100 * $t_used / $t_avail);
  warn "$box at $t_per%\n" if $t_per > 88;
}



japhy -- Perl and Regex Hacker

[Chady]

read on at this

---

He who asks will be a fool for five minutes, but he who doesn't ask will remain a fool for life.

[SilverB1rd]

Please tell us exactly what you want to do? If your trying to pull the information out of that output try using a regex.

Something like this,


m/(\/.+) +(\d) +(\d) +(\d+) +(\d)% +(.+)/


You would then be able to test $5 to see if it is over 80.

NOTE My regex is not in anyway the best and I'm not even sure if that would work. But I hope you get the idea. If you want to know more read up on regex. UPDATE My post was outdated before I even post it :)

------
The Price of Freedom is Eternal Vigilance

[larryk]

you are right to suspect that this would not work - the single \d as opposed to \d+ before the % would mean that the regex would fail to match unless the % was between 0 and 9.

what I expect you are looking for is for(<>) { /(\d+)%/; print if $1 >= 88 }

[SilverB1rd]

Thanks larry. Typo on my part. It was supposed to be (\d+)%

------
The Price of Freedom is Eternal Vigilance