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

Hi, How do I look for only a particular line in a file. For ex: In a particular file error.log, I was look for a particular line "The dir is '/usr' and there are 23 files" and in the line "The dir is '/usr' and there are 23 files" .. I want to store /usr to an array. Thanks

Replies are listed 'Best First'.
Re: look for a word
by TedPride (Priest) on Jun 09, 2005 at 04:07 UTC 
    use strict;
use warnings;

my @matches;
while (<DATA>) {
    push @matches, $1 if m/The dir is '(.*?)'/;
}
print join "\n", @matches;

__DATA__
sdfdfsddf
The dir is '/usr' and there are 23 files
sdfsdfdsdds
The dir is '/bingo' and there are 16 files

    [download]
[reply]
[d/l]
Re: look for a word
by Zaxo (Archbishop) on Jun 09, 2005 at 04:09 UTC

    Do I understand correctly that '/usr' may vary? 

    my @dirs;
while (<>) {
    m/^The dir is '([^']*)' and there are \d* files/
        and push @dirs, $1;
}

    [download]
    If the match is successful, the bit captured by parentheses between quotes is pushed onto @dirs. I've left out file opening and such by using the default diamond op. Read from an open handle for better control.

    After Compline,
    Zaxo

[reply]
[d/l]
      Hello, Thank you for the reply, Yes the '/usr' may vary. It could be any other dir too like '/var'.
[reply]
        Hi again, I am a beginner in perl and am not able to work this out. In the following string "05/27/05 03:03:30 ANS1802E The dir is '/usr' and there are 23 files" I want to be able to push the /usr to an array. I am not sure how to use the search pattern in perl.
[reply]

Back to Seekers of Perl Wisdom