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Regular Expression Question

by Anonymous Monk
on Mar 27, 2001 at 21:17 UTC ( #67540=perlquestion: print w/replies, xml ) Need Help??

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

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

Replies are listed 'Best First'.
Re: Regular Expression Question
by mirod (Canon) on Mar 27, 2001 at 21:25 UTC

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

    You are welcome.

(boo) Regexp'ing through a logfile
by boo_radley (Parson) on Mar 27, 2001 at 22:27 UTC
    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 ?: .
Re: Regular Expression Question
by japhy (Canon) on Mar 27, 2001 at 23:11 UTC
    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
Re: Regular Expression Question
by Masem (Monsignor) on Mar 27, 2001 at 21:30 UTC
    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 - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain
      Did you mean split(/\s+/)? split(/\s*/) will split it into individual characters because /\s*/ can match a 0 character string.
Re: Regular Expression Question
by Chady (Priest) on Mar 27, 2001 at 21:23 UTC

    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.
Re: Regular Expression Question
by SilverB1rd (Scribe) on Mar 27, 2001 at 21:39 UTC
    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
      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 }

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

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

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